Probability

 Mind Maps

Class X - maths: Probability
Q) P(E1⋃ E2) = P(E1) +______ P(E1 ⋂ E2)?

Q) An operation which can produce some well defined outcomes is called______________?

Q) All the possible outcomes of the experiments are known as __________________?

Q) If there are more than one element of the sample space in the set representing an event, then this event is called ________________?

Q) An event which is sure to occur at every performance of an experiment is called a _____________?

Q) The outcomes which make necessary the happening of an event in a trial are called ____________?

Q) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a king of red colour ?

Q) A die is thrown twice. What is the probability that 5 will come up at least once?

Q) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting, a face card?

Q) If X and Y are two mutually exclusive events then X ⋂ Y =?

Q) The probability that a non-leap year has 53 sundays,is

Q) If P(E) = 0.05,then P(not E) =?

Q) An operation which can produce some well defined outcomes is called an?

Q) The outcomes which make necessary the happening of an event in a trial are called ?

Q) If P(E) = 0.06, what is the probability of 'not' E?

Q) A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy?

Q) Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Q) A die is thrown once. Find the probability of getting (i) a prime number (ii) a number lying between 2 and 6 (iii) an odd number.

Q) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (ii) not red?

Q) From five cards, if a queen is drawn and put aside, what is the probability that the second card picked up is an ace?

Q) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

Q) All the possible outcomes of the experiments are known as

Q) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) Red? (ii) White? (iii) Not green?

Q) A piggy bank contains hundred 60p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 2 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 60 p coin? (ii) Will not be a Rs.2 coin?

Q) A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at (i) 8 ? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

Q) Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Q) A box contains 100 discs which are numbered from 1 to 100. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 3.

Q) Suppose you drop a die at random on the rectangular region shown below. What is the probability that it will land inside the circle with diameter 1m?

Q) A lot consists of 122 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it? (ii) She will not buy it?

Q) A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws: What is the probability that total score is (i) even? (ii) 4? (iii) at least 4?

Q) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be white is ___?

Q) The probability of guessing the correct answer to a certain test question is x / 12.If the probability of not guessing the correct answer to this question is 2 / 3,then x=?

Q) A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out an orange flavoured candy?

Q) A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a two-digit number?

#### content

• Introduction
• Some Terms Related to Probability
• problems on probability of Mutually Exclusive Events
• Events in Probability
• Equivalent Events or Identical Events
• Complementary Event
• example on Probability
• Experimental Probability
• Odds and Probability
• Problems on Odds and Probability
• Probability of Tossing Three Coins
• problems on probability involving tossing or throwing or flipping three coins
• Playing Cards Probability
• problems on Playing cards probability
• Probability for Rolling Two Dice
• problems involving probability for rolling two dice
• Probability for Rolling Three Dice
• problems involving probability for rolling three dice
• Coin toss Probability
• Word problems on coin toss probability
• Theoretical Probability
• Random Experiments
• Empirical Probability
• Problems on empirical probability
Formulae

Experiment:
An operation which can produce some well defined outcomes is called an experiment.
Random Experiment:
In an experiment where all possible outcomes are known and in advance if the exact outcome cannot be predicted, is called a random experiment.
Sample space:
A sample space of an experiment is the set of all possible results of that random experiment.
Event:
Out of the total results obtained from a certain experiment, the set of those results which are in favour of a definite result is called the event and it is denoted as E
Equally Likely Events:
When there is no reason to expect the happening of one event in preference to the other, then the events are known equally likely events.
Exhaustive Events:
All the possible outcomes of the experiments are known as exhaustive events.
Favourable Events:
The outcomes which make necessary the happening of an event in a trial are called favourable events.
If E1 and E2 be any two events (not necessarily mutually exclusive events), then

P(E1⋃E2) = P(E1) + P(E2)- P(E1 ⋂ E2)

Mutually Exclusive Events
Definition of Mutually Exclusive Events:
If two events are such that they cannot occur simultaneously for any random experiment are said to be mutually exclusive events.
If X and Y are two mutually exclusive events then X⋂ Y = O/
problems on probability of Mutually Exclusive Events
Simple or Elementary Event:
If there be only one element of the sample space in the set representing an event, then this event is called a simple or elementary event.
Compound Event:
If there are more than one element of the sample space in the set representing an event, then this event is called a compound event.
Certain Events:
An event which is sure to occur at every performance of an experiment is called a certain event connected with the experiment.
Impossible Event:
An event which cannot occur at any performance of the experiment is called an possible event.
Equally Likely Events:
When there is no reason to expect the happening of one event in preference to the other, then the events are known equally likely events.
Exhaustive Events:
All the possible outcomes of the experiments are known as exhaustive events.
Favourable Events:
The outcomes which make necessary the happening of an event in a trial are called Favourable events.
Mutually Exclusive Events:
If there be no element common between two or more events, i.e., between two or more subsets of the sample space, then these events are called mutually exclusive events.
If E1 and E2 are two mutually exclusive events, then E1 ⋃ E2 = O/
Complementary Event:
An event which consists in the negation of another event is called complementary event of their event.
Event points, Even Space:
Let an experiment be donated by E. The simple events connected with E will be called even points: and the set S of all possible even points is called event space of E.
Compound Event:
If there are more than one element of the sample space in the set representing an event, then this event is called a compound event.
odds against an event A= number of unFavourable events/number of Favourable events.
Probability of Occurrence of an Event:
The probability of occurrence of an event is defined as:
                             Number of trials in which event occurred
P(occurrence of an event)= _________________________________________
Total number of trails

Experimental Probability
Definition of experiment: A process which can produce some well defined results (outcomes) is called an experiment.
Definition of Odds:
Odds in probability of a particular event, means the ratio between the number of Favourable outcomes to the number of unFavourable outcomes.
Odds in favour:
Odds in favour of a particular event are given by Number of Favourable outcomes to Number of unFavourable outcomes.
          Number of Favourable outcomes
P(A) =  ______________________________
Number of unFavourable outcomes


Odds against:
Odds against is given by Number of unFavourable outcomes to number of Favourable outcomes.
           Number of unFavourable outcomes
P(A) =  ______________________________
Number of Favourable outcomes

                         Number of Favourable outcomes
P(Event) =  ___________________________________________________
no.of Favourable outcomes+ no.of unFavourable outcomes


What is the difference between odds and probability?

success
Odds= ________________
(success+failure)
Definition of random experiment
An experiment for which we know the set of all different results but it is not possible to predict which one of the set will occur at any particular execution of the experiment is called a random experiment.
Definition of Empirical Probability
The experimental probability of occurring of an event is the ratio of the number of trials in which the event occurred to the total number of trials.
The empirical probability of the occurrence of an event E is defined as:
P(E) = Number of trials in which event occurred / Total number of trials

Probability

Probability in everyday life, we come across statements such as:

• Most probably it will rain today.
• Chances are high that the prices of petrol will go up.
• I doubt that he will win the race.
• The words 'most probably','chances','doubt' etc., show the probability of occurrence of an event.
Some Terms Related to Probability

Experiment:
• An operation which can produce some well defined outcomes is called an experiment. Each outcome is called an event.

Random Experiment:

• In an experiment where all possible outcomes are known and in advance if the exact outcome cannot be predicted, is called a random experiment.
• Thus, when we throw a coin we know that all possible outcomes are Head and Tail.
• But, if we throw a coin at random, we cannot predict in advance whether its upper face will show a head or a tail.
• So, tossing a coin is a random experiment.
Similarly, throwing a dice is a random experiment.

#### Trial:

• By a trial, we mean performing a random experiment.
For example: throwing a die or tossing a coin etc.

#### Sample space:

A sample space of an experiment is the set of all possible results of that random experiment. For example; in throwing a die possible results are {1, 2, 3, 4, 5, 6}.

#### Event:

Out of the total results obtained from a certain experiment, the set of those results which are in favour of a definite result is called the event and it is denoted as E.

#### Equally Likely Events:

When there is no reason to expect the happening of one event in preference to the other, then the events are known equally likely events.
For example: when an unbiased coin is tossed the chances of getting a head or a tail are the same.

#### Exhaustive Events:

All the possible outcomes of the experiments are known as exhaustive events.
For example; in throwing a die there are 6 exhaustive events in a trial.

#### Favourable Events:

The outcomes which make necessary the happening of an event in a trial are called Favourable events.
For example: if two dice are thrown, the number of Favourable events of getting a sum 5 is four, i.e., (1, 4), (2, 3), (3, 2) and (4, 1).

#### If E1 and E2 be any two events (not necessarily mutually exclusive events), then

P(E1⋃E2) = P(E1) + P(E2)- P(E1 ⋂ E2)

#### Mutually Exclusive Events

Definition of Mutually Exclusive Events:

• If two events are such that they cannot occur simultaneously for any random experiment are said to be mutually exclusive events.
• If X and Y are two mutually exclusive events, then X⋂ Y = O/
• For example, events in rolling of a die are "even face" and "odd face" which are known as mutually exclusive events.
• But"odd face" and "multiple of 3" are not mutually exclusive, because when "face-3" occurs both the events "odd face" and "multiply of 3" are said to be occurred simultaneously.
• We see that two simple events are always mutually exclusive while two compound events may or may not mutually exclusive.

#### Addition Theorem Based on Mutually Exclusive Events:

If X and Y are two mutually exclusive events, then the probability of 'X union Y' is the sum of the probability of X and the probability of Y and represented as,
P(X ⋃ Y) = P(X) + P(Y)
Proof:
Let E be a random experiment and N(X) be the number of frequency of the event X in E. Since X and Y are two mutually exclusive events then;

N(X ⋃ Y) = N(X) + N(Y)
or, N(X ⋃ Y)/N = N(X)/N + N(Y)/N;
Dividing both the sides by N.
P(X ⋃Y) = P(X) + P(Y)

Worked out problems on probability of Mutually Exclusive Events:

1. One card is drawn from a well shuffled deck of 52 cards. What is the probability of getting a king or an ace?
Solution:

• Let X be the event of 'getting a king' and,
• Y be the event of 'getting an ace'
• We know that, in a well shuffled deck of 52 cards there are 4 kings and 4 aces.
• Therefore, probability of getting a king from well shuffled deck of 52 cards
= P(X) = 4/52 = 1/13
• Similarly, probability of getting an ace from well shuffled deck of 52 cards
= P(Y) = 4/52 = 1/13
• According to the definition of mutually exclusive we know that, drawing of a well shuffled deck of 52 cards 'getting a king' and 'getting an ace' are known as mutually exclusive events.
• We have to find out P(King or ace).
• So according to the addition theorem for mutually exclusive events, we get;
• P(X ⋂ Y) = P(X) + P(Y)
• Therefore, P(X ⋂ Y)
• = 1/13 + 1/13
= (1 + 1)/13
= 2/13

Hence, probability of getting a king or an ace from a well shuffled deck of 52 cards = 2/13

#### 2. A bag contains 8 black pens and 2 red pens and if a pen is drawn at random. What is the probability that it is black pen or red pen?

Solution:

• Let X be the event of 'getting a black pen' and,
• Y be the event of 'getting a red pen'.
• We know that, there are 8 black pens and 2 red pens.
• Therefore, probability of getting a black pen = P(X) = 8/10 = 4/5
• Similarly, probability of getting a red pen = P(Y) = 2/10 = 1/5
• According to the definition of mutually exclusive we know that, the event of 'getting a black pen' and 'getting a red pen' from a bag are known as mutually exclusive event.
• We have to find out P(getting a black pen or getting a red pen).
• So according to the addition theorem for mutually exclusive events, we get;
• P(X ⋂ Y) = P(X) + P(Y)
• Therefore, P(X ⋂ Y)
• = 4/5 + 1/5
= 5/5
= 1
Hence, probability of getting 'a black pen' or 'a red pen' = 1

Events in Probability

The outcomes of a random experiment are called events connected with the experiment.
For example: 'head' and 'tail' are the outcomes of the random experiment of throwing a coin and hence are events connected with it.

### Now we can distinguish between two types of events.

• simple event
• compound event

#### Simple or Elementary Event:

If there be only one element of the sample space in the set representing an event, then this event is called a simple or elementary event.
For example: if we throw a die,then the sample space, S = {1, 2, 3, 4, 5, 6}. Now the event of 2 appearing on the die is simple and is given by E = {2}.

#### Compound Event:

If there are more than one element of the sample space in the set representing an event, then this event is called a compound event.
For example: if we throw a die, having S = {1, 2, 3, 4, 5, 6}, the event of a odd number being shown is given by E = {1, 3, 5}.

• Odd in favour of an event A is defined as; number of Favourable events/number of unFavourable events.
• Similarly, odds against an event A = number of unFavourable events/number of Favourable events.

#### Certain Events:

An event which is sure to occur at every performance of an experiment is called a certain event connected with the experiment.
For example: "Head or Tail" is a certain event connected with tossing a coin.

Face-1 or face-2, face-3,....., face-6 is a certain event connected with throwing a die.

#### Impossible Event:

An event which cannot occur at any performance of the experiment is called an possible event. Following are such examples

• 'Seven' in case of throwing a die.
• 'Sum-13' in case of throwing a pair of dice.

Equivalent Events or Identical Events:

Two events are said to be equivalent or identical if one of them implies and implied by other. That is, the occurrence of one event implies the occurrence of the other and vice versa.
For example: "even face" and "face-2" or "face-4" or "face-6" are two identical events.

#### Equally Likely Events:

When there is no reason to expect the happening of one event in preference to the other, then the events are known equally likely events.
For example; when an unbiased coin is tossed the chances of getting a head or a tail are the same.

#### Exhaustive Events:

All the possible outcomes of the experiments are known as exhaustive events.
For example: in throwing a die there are 6 exhaustive events in a trial.

#### Favourable Events:

The outcomes which make necessary the happening of an event in a trial are called Favourable events.
For example; if two dice are thrown, the number of Favourable events of getting a sum 5 is four,
i.e., (1, 4), (2, 3), (3, 2) and (4, 1).

#### Mutually Exclusive Events:

If there be no element common between two or more events, i.e., between two or more subsets of the sample space, then these events are called mutually exclusive events.

• If E1 and E2 are two mutually exclusive events, then E1 ⋃ E2 = O/
For example: in connection with throw a die "even face" and "odd face" are mutually exclusive.
• But"odd-face" and "multiple of 3" are not mutually exclusive, because when "face-3" occurs both the events "odd face" and "multiply of 3" are said to be occurred simultaneously.
• We see that two simple-events are always mutually exclusive while two compound events may or may not mutually exclusive.

Complementary Event:

An event which consists in the negation of another event is called complementary event of their event. In case of throwing a die, 'even face' and 'odd face' are complementary to each other. "Multiple of 3" ant "Not multiple of 3" are complementary events of each other.
Not happening of the event E is called the complementary event of the event E. It is denoted by E' or bar(E).

Note : complementary event of certain event is an impossible event and vice versa

### Event points, Even Space:

Let an experiment be donated by E. The simple events connected with E will be called even points: and the set S of all possible even points is called event space of E.

• Any subset A of S is obviously an event. If A contains single point then it is a simple event, if A contains more than one point of S then A is compound event.
• Then entire space S is certain event and empty set O/ is impossible event.
Compound Event:

If there are more than one element of the sample space in the set representing an event, then this event is called a compound event.
For example; if we throw a die, having S = {1, 2, 3, 4, 5, 6}, the event of a odd number being shown is given by E = {1, 3, 5}.

• Odd in favour of an event A is defined as the ratio of number of Favourable events and number of unFavourable events.
• odds against an event A= number of unFavourable events/number of Favourable events.

• An event which is sure to occur at every performance of an experiment is called a certain event connected with the experiment.
• For example, "Head or Tail" is a certain event connected with tossing a coin.
• Face-1 or face-2, face-3, ....., face-6is a certain event connected with throwing a die.

We see that two simple events are always mutually exclusive while two compound events may or may not mutually exclusive.

### Probability of Occurrence of an Event:

The probability of occurrence of an event is defined as:


Number of trials in which event occurred
P(occurrence of an event)=  _________________________________________
Total number of trails


Solved example on Probability:

1. A dice is thrown 65 times and 4 appeared 21 times. Now, in a random throw of a dice, what is the probability of getting a 4?
Solution:
Total number of tria1s = 65.
Number of times 4 appeared = 21.
Probability of getting a 4 = Number of times 4 appeared/Total number of trials
= 21/65

Experimental Probability

Definition of experiment: A process which can produce some well defined results (outcomes) is called an experiment.
Some Experiments and their outcomes:

Tossing a coin:

Suppose we toss a coin and let it fall flat on the ground. Its upper face will show either Head (H) or Tail (T).

1. Whatever comes up, is called an outcome.
2. All possible outcomes are Head (H) and Tail (T).

Throwing a dice:

A dice is a solid cube having 6 faces, marked as 1, 2, 3, 4, 5, 6 respectively.
Suppose we throw a dice and let it fall flat on the ground. Its upper face will show one of the numbers 1, 2, 3, 4, 5, 6.

1. Whatever comes up, is called an outcome.
2. All possible outcomes are 1, 2, 3, 4, 5, 6.

The act of tossing a coin or throwing a dice is called an experiment. Whatever comes up, is called an outcome.
In an experiment, all possible outcomes are known.
The plural of die is dice.

Drawing a card from a well shuffled deck of 52 cards:

A deck of playing cards has in all 52 cards.

1. It has 13 cards each of four suits, namely spades, clubs, hearts and diamonds.
• Cards of spades and clubs are black cards.
• Cards of hearts and diamonds are red cards.
2. Kings, queens and jacks (or knaves) are known as face cards. Thus, there are 12 face cards in all.

Definition of Experimental Probability:
The experimental probability of happening of an event is the ratio of the number of trials in which the event happened to the total number of trials.
The experimental probability of the occurrence of an event E is defined as:

  Number of trials in which event happened P(E)= __________________________________________ Total number of trials 

Odds and Probability

#### Definition of Odds:

• Odds in probability of a particular event, means the ratio between the number of Favourable outcomes to the number of unFavourable outcomes.
Odds in favour and odds in against probability:

#### Odds in favour:

Odds in favour of a particular event are given by Number of Favourable outcomes to Number of unFavourable outcomes.

  Number of Favourable outcomes P(A) = ______________________________ Number of unFavourable outcomes 

#### Example; 1.Find the odds in favour of throwing a die to get "3 dots". Solution:

Total number of outcomes in throwing a die = 6
Number of Favourable outcomes = 1
Number of unFavourable outcomes = (6 - 1) = 5
Therefore, odds in favour of throwing a die to get "3 dots"is 1 : 5 or 1/5

#### Odds against:

Odds against is given by Number of unFavourable outcomes to number of Favourable outcomes.

  Number of unFavourable outcomes P(A) = ______________________________ Number of Favourable outcomes 

#### Example: Find the odds in against of throwing a die to get "3 dots". Solution:

Total number of outcomes in throwing a die = 6
Number of Favourable outcomes = 1
Number of unFavourable outcomes = (6 - 1) = 5
Therefore, odds in against of throwing a die to get "3 dots" is 5 : 1 or 5/1
Then,

  Number of Favourable outcomes P(Event) = ___________________________________________________ no.of Favourable outcomes+ no.of unFavourable outcomes 

Worked out Problems on Odds and Probability

#### 1. If odds in favour of X solving a problem are 4 to 3 and odds against Y solving the same problem are 2 to 6. Find probability for: (i) X solving the problem (ii) Y solving the problem Solution:

                          Number of Favourable outcomes
P(Event) =  ___________________________________________________
no.of Favourable outcomes+ no.of unFavourable outcomes


Given odds in favour of X solving a problem are 4 to 3.
Number of Favourable outcomes = 4
Number of unFavourable outcomes = 3
(i) X solving the problem
P(X) = P(solving the problem) = 4/(4 + 3)
= 4/7
Given odds against Y solving the problem are 2 to 6
Number of Favourable outcomes = 6
Number of unFavourable outcomes = 2
(ii) Y solving the problem
P(Y) = P(solving the problem) = 6/(2 + 6)
= 6/8
= 3/4

#### 2. What is the difference between odds and probability? Solution:

The difference between odds and probability are:
Odds of an event are the ratio of the success to the failure.
Probability of an event is the ratio of the success to the sum of success and failure.

  success Odds= ________________ (success+failure) 

Probability of Tossing Two Coins

Let us take the experiment of tossing two coins simultaneously

• When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.
• Therefore, total numbers of outcome are 22 = 4
• The above explanation will help us to solve the problems on finding the probability of tossing two coins.
Worked out problems on probability involving tossing or flipping two coins

#### 1. Two different coins are tossed randomly. Find the probability of (i) getting two heads (ii) getting two tails (iii) getting one tail (iv) getting no head (v) getting no tail (vi) getting at least 1 head (vii) getting at least 1 tail (viii) getting atmost 1 tail (ix) getting 1 head and 1 tail Solution:

When two different coins are tossed randomly, the sample space is given by
S = {HH, HT, TH, TT}
Therefore, n(S) = 4.
Let E1 = event of getting 2 heads. Then,
E1 = {HH} and, therefore, n(E1) = 1.
Therefore, P(getting 2 heads) = P(E1)
= n(E1)/n(S)
= 1/4.

(ii) getting two tails:
Let E2 = event of getting 2 tails. Then,
E2 = {TT} and, therefore, n(E2) = 1.
Therefore, P(getting 2 tails) = P(E2)
= n(E2)/n(S)
= 1/4.

(iii) getting one tail:
Let E3 = event of getting 1 tail. Then,
E3 = {TH, HT} and, therefore, n(E3) = 2.
Therefore, P(getting 1 tail) = P(E3)
= n(E3)/n(S)
= 2/4 = 1/2

Let E4 = event of getting no head. Then,
E4 = {TT} and, therefore, n(E4) = 1.
Therefore, P(getting no head) = P(E4)
= n(E4)/n(S)
=1/4.

(v) getting no tail:
Let E5= event of getting no tail. Then,
E5= {HH} and, therefore, n(E5) = 1.
Therefore, P(getting no tail) = P(E5)
= n(E5)/n(S) = 1/4.

(vi) getting at least 1 head:
Let E6= event of getting at least 1 head. Then,
E6 = {HT, TH, HH} and, therefore, n(E6) = 3.
Therefore, P(getting at least 1 head) = P(E6)
= n(E6)/n(S)
= 3/4.

(vii) getting at least 1 tail:
Let E7 = event of getting at least 1 tail. Then,
E7 = {TH, HT, TT} and, therefore, n(E7) = 3.
Therefore, P(getting at least 1 tail) = P(E7)
= n(E7)/n(S)
= 3/4.

(viii) getting atmost 1 tail:
Let E8= event of getting atmost 1 tail. Then,
E8 = {TH, HT, HH} and, therefore, n(E8) = 3.
Therefore, P(getting atmost 1 tail) = P(E8)
= n(E8)/n(S)
= 3/4.

(IX) Getting 1 Head And 1 Tail:
Let E9= event of getting 1 head and 1 tail. Then,
E9 = {HT, TH } and, therefore, n(E9) = 2.
Therefore, P(getting 1 head and 1 tail) = P(E9)
= n(E9)/n(S)
= 2/4 = 1/2.

Probability of Tossing Three Coins

Let us take the experiment of tossing three coins simultaneously:
When we toss three coins simultaneously then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail.
Therefore, total numbers of outcome are 23 = 8
The above explanation will help us to solve the problems on finding the probability of tossing three coins.

Worked out problems on probability involving tossing or throwing or flipping three coins

#### 1. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times. If three coins are tossed simultaneously at random, find the probability of: (i) getting three heads, (ii) getting two heads, (iii) getting one head, (iv) getting no head Solution:

Total number of trials = 250
Number of times three heads appeared = 70
Number of times two heads appeared = 55
Number of times one head appeared = 75
Number of times no head appeared = 50
In a random toss of 3 coins, let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and 0 head respectively. Then,
=Number of times three heads appeared/ Total number of trials
= 70/250
= 0.28

P(getting two heads) = P(E2) =Number of times three heads appeared/ Total number of trials
= 55/250
= 0.22

=Number of times three heads appeared/ Total number of trials
= 75/250
= 0.30

=Number of times three heads appeared/ Total number of trials
= 50/250
= 0.20

Note:In tossing 3 coins simultaneously, the only possible outcomes are E1, E2, E3 E4 and
P(E1) + P(E2) + P(E3) + P(E4)

= (0.28 + 0.22 + 0.30 + 0.20)
= 1

#### 2. When 3 Unbiased Coins Are Tossed Once. What Is The Probability Of: (I) Getting All Heads (Ii) Getting Two Heads (Iii) Getting One Head (Iv) Getting At Least 1 Head (V) Getting At Least 2 Heads (Vi) Getting Atmost 2 Heads Solution:

In tossing three coins, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
And, therefore, n(S) = 8.
Let E1= event of getting all heads. Then,
E1= {HHH}
and, therefore, n(E1) = 1.
Therefore, P(getting all heads) = P(E1)
= n(E1)/n(S) = 1/8.

Let E2 = event of getting 2 heads. Then,
E2 = {HHT, HTH, THH}
and, therefore, n(E2) = 3.
Therefore, P(getting 2 heads) = P(E2)
= n(E2)/n(S) = 3/8.

Let E3 = event of getting 1 head. Then,
E3 = {HTT, THT, TTH} and, therefore,
n(E3) = 3.
Therefore, P(getting 1 head) = P(E3)
= n(E3)/n(S) = 3/8.

(iv) getting at least 1 head
Let E4 = event of getting at least 1 head. Then,
E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH}
and, therefore, n(E4) = 7.
Therefore, P(getting at least 1 head) = P(E4)
= n(E4)/n(S) = 7/8.

(v) getting at least 2 heads
Let E5 = event of getting at least 2 heads. Then,
E5 = {HHT, HTH, THH, HHH}
and, therefore, n(E5) = 4.
Therefore, P(getting at least 2 heads) = P(E5)
= n(E5)/n(S) = 4/8 = 1/2.

Let E6 = event of getting atmost 2 heads. Then,
E6 = {HHT, HTH, HTT, THH, THT, TTH, TTT}
and, therefore, n(E6) = 7.
Therefore, P(getting atmost 2 heads) = P(E6)
= n(E6)/n(S) = 7/8

#### Example 1.Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail. Solution :

In the experiment of tossing a coin once, the number of possible outcomes is two - Head (H) and Tail (T). Let E be the event 'getting a head'. The number of outcomes favourable to E, (i.e., of getting a head) is 1.
P(E) = P (head)=[Number of outcomes favourable to E]/[Number of all possible outcomes]
= 1/ 2
if F is the event 'getting a tail',
P(F) = P(tail) = 1 /2

#### Example 2 : A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the (i) yellow ball? (ii) red ball? (iii) blue ball? Solution :

Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.
Let Y be the event 'the ball taken out is yellow', B be the event 'the ball taken out is blue', and R be the event 'the ball taken out is red'.
Now, the number of possible outcomes = 3.
(i) The number of outcomes favourable to the event Y = 1
P(Y) = 1/ 3
(ii) P(R) = 1 /3 and
(iii) P(B) = 1/ 3

Remarks : 1. An event having only one outcome of the experiment is called an elementary event. In Example 1, both the events E and F are elementary events. Similarly, in Example 2, all the three events, Y, B and R are elementary events. 2.
In Example 1, we note that : P(E) + P(F) = 1
In Example 2, we note that : P(Y) + P(R) + P(B) =1
The sum of the probabilities of all the elementary events of an experiment is 1.

#### Example 3 : Suppose we throw a die once. (i) What is the probability of getting a number greater than 4 ? (ii) What is the probability of getting a number less than or equal to 4 ? Solution :

(i) Here, let E be the event 'getting a number greater than 4'. The number of possible outcomes is six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6.
Therefore, the number of outcomes favourable to E is 2.
P(E) = P(number greater than 4) = 2/ 6 = 1/ 3
(ii) Let F be the event 'getting a number less than or equal to 4'. Number of possible outcomes = 6 Outcomes favourable to the event F are 1, 2, 3, 4.
So, the number of outcomes favourable to F is 4.
P(F) = 4/ 6 = 2/ 3
Are the events E and F in the example above elementary events they are not because the event E has 2 outcomes and the event F has 4 outcomes.
Remarks: From Example 1, we note that
P(E) + P(F) = (1/2)+(1 /2)=1
where E is the event 'getting a head' and F is the event 'getting a tail'. From (i) and (ii) of Example 3, we also get
P(E) + P(F) = (1/3)+2/3=1

E is the event 'getting a number >4' and F is the event 'getting a number <= 4'. Note that getting a number not greater than 4 is same as getting a number less than or equal to 4, and vice versa. In (1) and (2) above, is F not the same as not E? Yes, it is. We denote the event 'not E' by E .
So, P(E) + P(not E) = 1
P(E) + P( E ) = 1.
P( E ) = 1 - P(E).
The event E , representing 'not E', is called the complement of the event E. We also say that E and E are complementary events.

#### (i) What is the probability of getting a number 8 in a single throw of a die?

• We know that there are only six possible outcomes in a single throw of a die. These outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8,
• so there is no outcome favourable to 8, i.e., the number of such outcomes is zero. In other words, getting 8 in a single throw of a die, is impossible. So, P(getting 8) = 0/ 6 = 0.
• The probability of an event which is impossible to occur is 0. Such an event is called an impossible event.

#### (ii) What is the probability of getting a number less than 7 in a single throw of a die?

• every face of a die is marked with a number less than 7, it is sure that we will always get a number less than 7 when it is thrown once. So, the number of favourable outcomes is the same as the number of all possible outcomes, which is 6. Therefore, P(E) = P(getting a number less than 7) = 6/ 6 = 1
• So, the probability of an event which is sure (or certain) to occur is 1.
• Such an event is called a sure event or a certain event.
• Note : From the definition of the probability P(E), we see that the numerator (number of outcomes favourable to the event E) is always less than or equal to the denominator (the number of all possible outcomes). Therefore, 0 <= P(E) <= 1
• let us take an example related to playing cards. Have you seen a deck of playing cards.
• It consists of 52 cards which are divided into 4 suits of 13 cards each - spades (♠), hearts (♥), diamonds (♦) and clubs (♣).
• Clubs and spades are of black colour, while hearts and diamonds are of red colour. The cards in each suit are ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face cards.

#### Example 4 : One card is drawn from a well shuffled deck of 52 cards. Calculate the probability that the card will (i) be an ace, (ii) not be an ace. Solution :

Well shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck. Let E be the event 'the card is an ace'.
The number of outcomes favourable to E = 4
The number of possible outcomes = 52 (Why)
Therefore, P(E) =4 /52
=1/13
(ii) Let F be the event 'card drawn is not an ace'.
The number of outcomes favourable to the event F = 52 - 4 = 48
The number of possible outcomes = 52
Therefore, P(F) =48/52
= 12/13
Remark : Note that F is nothing but E .
Therefore, we can also calculate P(F) as follows:
P(F) = P( E ) = 1 - P(E)
=1-(1/13) =12/13.

#### Example 5 : Two players, Sangeeta and Reshma, play a tenn is match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match? Solution :

Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
The probability of Sangeeta's winning = P(S) = 0.62 (given)
The probability of Reshma's winning = P(R) = 1 -P(S) [As the events R and S are complementary]
= 1 -0.62 = 0.38.

#### Example 6 : Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year). Solution :

Out of the two friends, one girl, say, Savita's birthday can be any day of the year. Now, Hamida's birthday can also be any day of 365 days in the year.
We assume that these 365 outcomes are equally likely.
(i) If Hamida's birthday is different from Savita's,
the number of favourable outcomes for her birthday is 365 - 1 = 364
So, P (Hamida's birthday is different from Savita's birthday) =364/365
(ii) P(Savita and Hamida have the same birthday)
= 1 - P (both have different birthdays)
= 1-364/365 [Using P( E ) = 1 - P(E)]
= 1/365

#### Example 7 : There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy? Solution :

There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25 (Why?)
Therefore, P (card with name of a girl) = P(Girl) =25/40
=5/8
(ii) The number of outcomes favourable for a card with the name of a boy = 15 (Why?)
Therefore, P(card with name of a boy) = P(Boy) =15/40 =3/8
Note : We can also determine P(Boy), by taking
P(Boy) = 1 - P(not Boy)
= 1 - P(Girl)
=1- 5/8=3/8

#### Example 8 : A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be (i) white? (ii) blue? (iii) red? Solution :

Saying that a marble is drawn at random is a short way of saying that all the marbles are equally likely to be drawn.
Therefore, the number of possible outcomes = 3 +2 + 4 = 9 (Why?)
Let W denote the event 'the marble is white', B denote the event 'the marble is blue' and R denote the event 'marble is red'.
(i) The number of outcomes favourable to the event W = 2
P(W) =2/9
Similarly, (ii) P(B) =3/9=1/3
(iii) P(R) =4/9
Note that P(W) + P(B) + P(R) =1

#### Example 9 : Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that she gets at least one head? Solution :

• We write H for 'head' and T for 'tail'. When two coins are tossed simultaneously,
• The possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely.
Here (H, H) means head up on the first coin (say on Re 1) and head up on the second coin (Rs 2).
• Similarly (H, T) means head up on the first coin and tail up on the second coin and so on. The outcomes favourable to the event E, 'at least one head' are (H, H), (H, T) and (T, H).

So, the number of outcomes favourable to E is 3.
Therefore, P(E) =3/4
i.e., the probability that Harpreet gets at least one head is3/4
Note : You can also find P(E) as follows:
P (E) = 1 -P(E) [Since P(E) = P(no head) = 1/4]
= 1-1/4=3/4

#### Example 10 : In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing.What is the probability that the music will stop within the first half-minute after starting? Solution :

Here the possible outcomes are all the numbers between 0 and 2. This is the portion of the number line from 0 to 2.9(see fig 15.1)

                       _______________________
0     1/2      1      2


• Let E be the event that 'the music is stopped within the first half minute'.
• The outcomes favourable to E are points on the number line from 0 to 1/2.
• The distance from 0 to 2 is 2, while the distance from 0 to 1/2 is1/2 .
• Since all the outcomes are equally likely, we can argue that, of the total distance of 2, the distance favourable to the event E is1/2 .

P(E) = Distance favourable to the event E
Total distance in which outcomes can lie
=(1/2)/ 2
=1/ 4

#### Example 11 : A missing helicopter is reported to have crashed somewhere in the rectangular region . What is the probability that it crashed inside the lake shown in the figure? Solution :

The helicopter is equally likely to crash anywhere in the region.
Area of the entire region where the helicopter can crash= (4.5 xx 9) km^2
= 40.5 km
Area of the lake = (2.5 xx 3) km^2= 7.5 km^2
Therefore, P (helicopter crashed in the lake) =(7.5)/(40.5)
=(75)/(405)
=5/(27)

#### Example 12 : A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that (i) it is acceptable to Jimmy? (ii) it is acceptable to Sujatha? Solution :

One shirt is drawn at random from the carton of 100 shirts. Therefore,there are 100 equally likely outcomes.
(i) The number of outcomes favourable (i.e., acceptable) to Jimmy = 88 Therefore,
P (shirt is acceptable to Jimmy) =88/100=0.88
(ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96
So, P (shirt is acceptable to Sujatha) =96/100=0.96

#### Example 13 : Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8? (ii) 13? (iii) less than or equal to 12? Solution :

When the blue die shows '1', the grey die could show any one of the
numbers 1, 2, 3, 4, 5, 6. The same is true when the blue die shows '2', '3', '4', '5' or '6'.
The possible outcomes of the experiment are listed .The first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die.
Note that the pair (1, 4) is different from (4, 1).
So, the number of possible outcomes = 6 * 6 = 36.
(i) The outcomes favourable to the event 'the sum of the two numbers is 8' denoted by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
i.e., the number of outcomes favourable to E = 5.
Hence, P(E) =5/36
So, P(G) =36/36=1


12      3      456
______________________________________________
1    (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2    (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3    (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4    (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5    (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6    (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

fig 15.3


(ii) As you can see from Fig. 15.3, there is no outcome favourable to the event F,
'the sum of two numbers is 13'.
So, P(F) =0/36 =0 (iii) As you can see from Fig. 15.3, all the outcomes are favourable to the event G,
'sum of two numbers <= 12'.

Playing Cards Probability

#### Basic concept on drawing a card

In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each
i.e. spades ♠, hearts ♥ , diamonds ♦, clubs ♣ .
Cards of Spades and clubs are black cards.
Cards of hearts and diamonds are red cards.
The card in each suit, are ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3 and 2.
King, Queen and Jack (or Knaves) are face cards. So, there are 12 face cards in the deck of 52 playing cards.

Worked out problems on Playing cards probability

#### 1. A card is drawn from a well shuffled pack of 52 cards. Find the probability of: (i) 2 of spades (ii) a jack (iii) a king of red colour (iv) a card of diamond (v) a king or a queen (vi) a non face card (vii) a black face card (viii) a black card (ix) a non ace (x) non face card of black colour (xi) neither a spade nor a jack (xii) neither a heart nor a red king

Solution:
In a playing card there are 52 cards.
Therefore the total number of possible outcomes = 52
Number of favourable outcomes i.e. 2 of spades is 1 out of 52 cards.
Therefore, probability of getting 2 of spade
P(A) = Number of Favourable outcomes / Total number of possible outcome
= 1/52
(ii) a jack
Number of favourable outcomes i.e. 'a jack' is 4 out of 52 cards.
Therefore, probability of getting 'a jack'
P(B) = Number of Favourable outcomes / Total number of possible outcome
= 4/52
= 1/13
(iii) a king of red colour
Number of favourable outcomes i.e. 'a king of red colour' is 2 out of 52 cards.
Therefore, probability of getting 'a king of red colour'
P(C) = Number of Favourable outcomes / Total number of possible outcome
= 2/52
= 1/26
(iv) a card of diamond
Number of favourable outcomes i.e. 'a card of diamond' is 13 out of 52 cards.
Therefore, probability of getting 'a card of diamond'
P(D) = Number of Favourable outcomes / Total number of possible outcome
= 13/52
= 1/4
(v) a king or a queen
Total number of king is 4 out of 52 cards.
Total number of queen is 4 out of 52 cards
Number of favourable outcomes i.e. 'a king or a queen' is 4 + 4 = 8 out of 52 cards.
Therefore, probability of getting 'a king or a queen'
P(E) = Number of Favourable outcomes / Total number of possible outcome
= 8/52 = 2/13
(vi) a non face card
Total number of face card out of 52 cards = 3 times 4 = 12
Total number of non face card out of 52 cards = 52 - 12 = 40
Therefore, probability of getting 'a non-face card'
P(F) =Number of Favourable outcomes / Total number of possible outcome
= 40/52 = 10/13
(vii) a black face card:
Cards of Spades and Clubs are black cards.
Number of face card in spades (king, queen and jack or knaves) = 3
Number of face card in clubs (king, queen and jack or knaves) = 3
Therefore, total number of black face card out of 52 cards = 3 + 3 = 6
Therefore, probability of getting 'a black face card'
P(G) = Number of Favourable outcomes / Total number of possible outcome
= 6/52 = 3/26
(viii) a black card:
Cards of spades and clubs are black cards.
Number of clubs = 13
Therefore, total number of black card out of 52 cards = 13 + 13 = 26
Therefore, probability of getting 'a black card'
P(H) = Number of Favourable outcomes / Total number of possible outcome
= 26/52 = 1/2
(ix) a non ace:
Number of ace cards in each of four suits namely spades, hearts, diamonds and clubs = 1
Therefore, total number of ace cards out of 52 cards = 4
Thus, total number of non ace cards out of 52 cards = 52 - 4
= 48
Therefore, probability of getting 'a non-ace'
P(I) = Number of Favourable outcomes / Total number of possible outcome
= 48/52 = 12/13
(x) non face card of black colour:
Cards of spades and clubs are black cards.
Number of clubs = 13
Therefore, total number of black card out of 52 cards = 13 + 13 = 26
Number of face cards in each suits namely spades and clubs = 3 + 3 = 6
Therefore, total number of non face card of black colour out of 52 cards = 26 - 6 = 20
Therefore, probability of getting 'non face card of black colour'
P(J) = Number of Favourable outcomes / Total number of possible outcome
= 20/52
= 5/13
(xi) neither a spade nor a jack
Total number of non spades out of 52 cards = 52 - 13 = 39
Number of jack out of 52 cards = 4
Number of jack in each of three suits namely hearts, diamonds and clubs = 3
[Since, 1 jack is already included in the 13 spades so, here we will take number of jacks is 3]
Neither a spade nor a jack = 39 - 3 = 36
Therefore, probability of getting 'neither a spade nor a jack'
P(K) = Number of Favourable outcomes / Total number of possible outcome
= 36/52
= 9/13
(xii) neither a heart nor a red king
Number of hearts = 13
Total number of no hearts out of 52 cards = 52 - 13 = 39
Therefore, spades, clubs and diamonds are the 39 cards.
Cards of hearts and diamonds are red cards.
Number of red kings in red cards = 2
Therefore, neither a heart nor a red king = 39 - 1 = 38
[Since, 1 red king is already included in the 13 hearts so, here we will take number of red kings is 1] Therefore, probability of getting 'neither a heart nor a red king'
P(L) = Number of Favourable outcomes / Total number of possible outcome
= 38/52
= 19/26

#### 2. A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is: (i) a red face card (ii) neither a club nor a spade (iii) neither an ace nor a king of red color (iv) neither a red card nor a queen (v) neither a red card nor a black king.

Solution: Total number of card in a pack of well shuffled cards = 52
(i) a red face card
Cards of hearts and diamonds are red cards.
Number of face card in hearts = 3
Number of face card in diamonds = 3
Total number of red face card out of 52 cards = 3 + 3 = 6
Therefore, the probability of getting 'a red face card'
P(A) = Number of Favourable outcomes / Total number of possible outcome
= 6/52
= 3/26

(ii) neither a club nor a spade
Number of clubs = 13
Number of club and spade = 13 + 13 = 26
Number of card which is neither a club nor a spade = 52 - 26 = 26
Therefore, the probability of getting 'neither a club nor a spade'
P(B) = Number of Favourable outcomes / Total number of possible outcome
= 26/52
= 1/2

(iii) neither an ace nor a king of red color
Number of ace in a deck 52 cards = 4
Number of king of red color in a deck 52 cards
= (1 diamond king + 1 heart king) = 2
Number of ace and king of red color = 4 + 2 = 6
Number of card which is neither an ace nor a king of red color = 52 - 6 = 46
Therefore, the probability of getting 'neither an ace nor a king of red color'
P(C) = Number of Favourable outcomes / Total number of possible outcome
= 46/52
= 23/26

(iv) neither a red card nor a queen
Number of hearts in a deck 52 cards = 13
Number of diamonds in a deck 52 cards = 13
Number of queen in a deck 52 cards = 4
Total number of red card and queen = 13 + 13 + 2 = 28,
[since queen of heart and queen of diamond are removed]
Number of card which is neither a red card nor a queen = 52 - 28 = 24
Therefore, the probability of getting 'neither a red card nor a queen'
P(D) =Number of Favourable outcomes / Total number of possible outcome
= 24/52
= 6/13

(v) neither a red card nor a black king.
Number of hearts in a deck 52 cards = 13
Number of diamonds in a deck 52 cards = 13
Number of black king in a deck 52 cards
= (1 king of spade + 1 king of club) = 2
Total number of red card and black king = 13 + 13 + 2 = 28
Number of card which is neither a red card nor a black king = 52 - 28 = 24
Therefore, the probability of getting 'neither a red card nor a black king'
P(E) = Number of Favourable outcomes / Total number of possible outcome
= 24/52
= 6/13

Probability for Rolling Two Dice

• Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die.
• When two dice are thrown simultaneously, thus number of event can be 62 = 36 because each die has 1 to 6 number on its faces. Then the possible outcomes are shown in the below table.

#### Probability Sample space for two dice (outcomes): Note:

(i) The outcomes (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) are called doublets.
(ii) The pair (1, 2) and (2, 1) are different outcomes.

Worked out problems involving probability for rolling two dice

#### 1. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of 4 respectively. Then, show that (i) A is a simple event (ii) B and C are compound events (iii) A and B are mutually exclusive

Solution:
Clearly, we have
A = {(1, 1)}, B = {(1, 2), (2, 1)} and C = {(1, 3), (3, 1), (2, 2)}.
(i) Since A consists of a single sample point, it is a simple event.
(ii) Since both B and C contain more than one sample point, each one of them is a compound event.
(iii) Since A ⋂ B = O/ , A and B are mutually exclusive.

#### 2. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5, and B is the event that at least one of the dice shows up a 3. Are the two events (i) mutually exclusive, (ii) exhaustive?

Solution:
When two dice are rolled, we have n(S) = (6 * 6) = 36.
Now, A = {(1, 4), (2, 3), (4, 1), (3, 2)}, and
B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1,3), (2, 3), (4, 3), (5, 3), (6, 3)}
(i) A ⋂ B = {(2, 3), (3, 2)} != O/.
Hence, A and B are not mutually exclusive.
(ii) Also, A ⋃ B != S.
Therefore, A and B are not exhaustive events.

#### 3. Two dice are thrown simultaneously. Find the probability of (i) getting six as a product (ii) getting sum <= 3 (iii) getting sum <= 10 (iv) getting a doublet (v) getting a sum of 8 (vi) getting sum divisible by 5 (vii) getting sum of at least 11 (viii) getting a multiple of 3 as the sum (ix) getting a total of atleast 10 (x) getting an even number as the sum (xi) getting a prime number as the sum (xii) getting a doublet of even numbers (xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die

Solution:
Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces.
We know that in a single thrown of two different dice, the total number of possible outcomes is (6 * 6) = 36.
(i) getting six as a product
Let E1 = event of getting six as a product. The number whose product is six will be
E1 = [(1, 6), (2, 3), (3, 2), (6, 1)] = 4
Therefore, probability of getting 'six as a product'
P(E1) = Number of Favourable outcomes / Total number of possible outcome
= 4/36
= 1/9

(ii) getting sum <= 3:
Let E2 = event of getting sum <= 3. The number whose sum <= 3 will be E2 = [(1, 1), (1, 2), (2, 1)] = 3
Therefore, probability of getting 'sum <= 3'
P(E2) = Number of Favourable outcomes/ Total number of possible outcome
= 3/36
= 1/12

(iii) getting sum <= 10:
Let E3 = event of getting sum <= 10.
The number whose sum <= 10 will be
E3 = [(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),
(6, 1), (6, 2), (6, 3), (6, 4)] = 33
Therefore, probability of getting 'sum <= 10' P(E3) = Number of Favourable outcomes /Total number of possible outcome
= 33/36
= 11/12

(iv) getting a doublet: Let E4 = event of getting a doublet. The number which doublet will be E4 = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6
Therefore, probability of getting 'a doublet'
P(E4) = Number of Favourable outcomes /Total number of possible outcome
= 6/36
= 1/6

(v) getting a sum of 8
Let E5 = event of getting a sum of 8. The number which is a sum of 8 will be
E5 = [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)] = 5
Therefore, probability of getting 'a sum of 8'
P(E5) = Number of Favourable outcomes / Total number of possible outcome
= 5/36

(vi) getting sum divisible by 5
Let E6 = event of getting sum divisible by 5.
The number whose sum divisible by 5 will be
E6 = [(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)] = 7
Therefore, probability of getting 'sum divisible by 5'
P(E6) = Number of Favourable outcomes / Total number of possible outcome
= 7/36

(vii) getting sum of atleast 11:
Let E7 = event of getting sum of atleast 11.
The events of the sum of atleast 11 will be
E7= [(5, 6), (6, 5), (6, 6)] = 3
Therefore, probability of getting 'sum of atleast 11'
P(E7) = Number of Favourable outcomes /Total number of possible outcome
= 3/36
= 1/12

(viii) getting a multiple of 3 as the sum
Let E8 = event of getting a multiple of 3 as the sum.
The events of a multiple of 3 as the sum will be
E8 = [(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3) (6, 6)] = 12 Therefore, probability of getting 'a multiple of 3 as the
P(E8) = Number of Favourable outcomes / Total number of possible outcome
= 12/36
= 1/3

(ix) getting a total of atleast 10:
Let E9 = event of getting a total of atleast 10.
The events of a total of atleast 10 will be
E9= [(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)] = 6
Therefore, probability of getting 'a total of atleast 10'
P(E9) = Number of Favourable outcomes / Total number of possible outcome
= 6/36
= 1/6

(x) getting an even number as the sum: Let E10 = event of getting an even number as the sum.
The events of an even number as the sum will be
E10 = [(1, 1), (1, 3), (1, 5),
(2, 2), (2, 4), (2, 6),
(3, 3), (3, 1), (3, 5),
(4, 4), (4, 2), (4, 6),
(5, 1), (5, 3), (5, 5),
(6, 2), (6, 4), (6, 6)] = 18
Therefore, probability of getting 'an even number as the sum
P(E10) = Number of Favourable outcomes/ Total number of possible outcome
= 18/36 = 1/2

(xi) getting a prime number as the sum:
Let E11 = event of getting a prime number as the sum. The events of a prime number as the sum will be
E11 = [(1, 1), (1, 2), (1, 4), (1, 6),
(2, 1), (2, 3), (2, 5),
(3, 2), (3, 4), (4, 1), (4, 3),
(5, 2), (5, 6), (6, 1), (6, 5)] = 15 Therefore, probability of getting 'a prime number as the sum'
P(E11) = Number of Favourable outcomes / Total number of possible outcome
= 15/36
= 5/12

(xii) getting a doublet of even numbers:
Let E12 = event of getting a doublet of even numbers.
The events of a doublet of even numbers will be
E12 = [(2, 2), (4, 4), (6, 6)] = 3
Therefore, probability of getting 'a doublet of even numbers'
P(E12) = Number of Favourable outcomes/ Total number of possible outcome
= 3/36 = 1/12

(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die
Let E13 = event of getting a multiple of 2 on one die and a multiple of 3 on the other die.
The events of a multiple of 2 on one die and a multiple of 3 on the other die will be
E13 = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11
Therefore, probability of getting 'a multiple of 2 on one die and a multiple of 3 on the other die'
P(E13) =Number of Favourable outcomes / Total number of possible outcome
= 11/36

#### 4. Two dice are thrown. Find (i) the odds in favour of getting the sum 5, and (ii) the odds against getting the sum 6.

Solution:
We know that in a single thrown of two die,
the total number of possible outcomes is (6 * 6) = 36.
Let S be the sample space. Then, n(S) = 36.
(i) the odds in favour of getting the sum 5:
Let E1 be the event of getting the sum 5. Then,
E1 = {(1, 4), (2, 3), (3, 2), (4, 1)}
P(E1) = 4
Therefore, P(E1) = n(E1)/n(S)
= 4/36 = 1/9
odds in favour of E1 = P(E1)/[1 - P(E1)]
= (1/9)/(1 - 1/9) = 1/8.

(ii) the odds against getting the sum 6:
Let E2 be the event of getting the sum 6. Then,
E2 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
P(E2) = 5
Therefore, P(E2) = n(E2)/n(S) = 5/36
odds against E2 = [1 - P(E2)]/P(E2)
= (1 - 5/36)/(5/36)
= 31/5.

Probability for Rolling Three Dice

• Probability for rolling three dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each (three) dies.
• When three dice are thrown simultaneously/randomly, thus number of event can be
6^3 = (6 * 6 * 6) = 216
because each die has 1 to 6 number on its faces.

Worked out problems involving probability for rolling three dice

#### 1. Three dice are thrown together. Find the probability of: (i) getting a total of 5 (ii) getting a total of atmost 5 (iii) getting a total of at least 5. (iv) getting a total of 6. (v) getting a total of atmost 6. (vi) getting a total of at least 6.

Solution:
Three different dice are thrown at the same time.
Therefore, total number of possible outcomes will be 6^3 = (6 * 6 * 6) = 216.
(i) getting a total of 5:
Number of events of getting a total of 5 = 6
i.e. (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2)
Therefore, probability of getting a total of 5
P(E1) =Number of Favourable outcomes / Total number of possible outcome
= 6/216
= 1/36

(ii) getting a total of atmost 5:
Number of events of getting a total of atmost 5 = 10 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2).
Therefore, probability of getting a total of atmost 5
P(E2) = Number of Favourable outcomes / Total number of possible outcome
= 10/216
= 5/108

(iii) getting a total of at least 5:
Number of events of getting a total of less than 5 = 4
i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1).
Therefore, probability of getting a total of less than 5
P(E3) =Number of Favourable outcomes / Total number of possible outcome
= 4/216
= 1/54 Therefore, probability of getting a total of at least 5 = 1 - P(getting a total of less than 5)
= 1 - 1/54
= (54 - 1)/54
= 53/54

(iv) getting a total of 6:
Number of events of getting a total of 6 = 10
i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).
Therefore, probability of getting a total of 6
P(E4) = Number of Favourable outcomes / Total number of possible outcome
= 10/216
= 5/108

(v) getting a total of atmost 6:
Number of events of getting a total of atmost 6 = 20
i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (1, 2, 2), (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).
Therefore, probability of getting a total of atmost 6
P(E5) = Number of Favourable outcomes / Total number of possible outcome
= 20/216
= 5/54

(vi) getting a total of at least 6:
Number of events of getting a total of less than 6 (event of getting a total of 3, 4 or 5) = 10
i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1) (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1).
Therefore, probability of getting a total of less than 6
P(E6) = Number of Favourable outcomes / Total number of possible outcome
= 10/216
= 5/108
Therefore, probability of getting a total of at least 6 = 1 - P(getting a total of less than 6)
= 1 -5/108
= (108 - 5)/108
= 103/108

Coin toss Probability

Suppose a coin tossed then we get two possible outcomes either a 'head' (H) or a 'tail' (T), and it is impossible to predict whether the result of a toss will be a 'head' or 'tail'.
The probability for equally likely outcomes in an event is:
Number of favourable outcomes / Total number of possible outcomes
Total number of possible outcomes = 2

(i) If the favourable outcome is head (H). Number of favourable outcomes = 1. Therefore, P(getting a head) Number of Favourable outcomes = P(H) = total number of possible outcomes = 1/2.

(ii) If the favourable outcome is tail (T). Number of favourable outcomes = 1. Therefore, P(getting a tail) Number of Favourable outcomes = P(T) = total number of possible outcomes = 1/2.

Word problems on coin toss probability

#### 1. If three fair coins are tossed randomly 195 times and it is found that three heads appeared 22 times, two heads appeared46 times, one head appeared 36 times and zero head appeared 25 times. What is the probability of getting (i) three heads, (ii) two heads, (iii) one head, (iv) 0 head.

Solution: Total number of trials = 195.
Number of times three heads appeared = 22.
Number of times two heads appeared = 46.
Number of times one head appeared = 36.
Number of times zero head appeared = 25.
= P(E1) = Number of times three heads appeared / total number of trials
= 22/195

= P(E2) = Number of times two heads appeared / total number of trials
= 46/195

= P(E3) = Number of times one head appeared / total number of trials
= 36/195
= P(E4) = Number of times zero head appeared / total number of trials
= 25/195
Note: Remember when 3 coins are tossed randomly, the only possible outcomes E1 are E2, E3, E4
and=P(E1) + P(E2) + P(E3) + P(E4)

#### 2. Two coins are tossed randomly 120 times and it is found that two tails appeared 60 times, one tail appeared 48 times and no tail appeared 12 times. If two coins are tossed at random, what is the probability of getting (i) 2 tails, (ii) 1 tail, (iii) 0 tail

Solution:
Total number of trials = 120
Number of times 2 tails appear = 60
Number of times 1 tail appears = 48
Number of times 0 tail appears = 12
Let E1, E2 and E3 be the events of getting 2 tails, 1 tail and 0 tail respectively. (i) P(getting 2 tails)
= P(E1) = Number of times 2 tails appear / total number of trials
= 60/120
= 0.50

(ii) P(getting 1 tail)
= P(E2) = Number of times 1 tail appear / total number of trials
= 48/120
= 0.40

(iii)P(getting 0 tail)
= P(E3) = Number of times no tail appear / total number of trials
= 12/120
= 0.10

Note:
Remember while tossing 2 coins simultaneously, the only possible outcomes are E1, E2, E3and,
P(E1) + P(E2) + P(E3)
= (0.50 + 0.40 + 0.10)
= 1

#### 3. Suppose a fair coin is randomly tossed for 75 times and it is found that head turns up 45 times and tail 30 times. What is the probability of getting (i) a head and (ii) a tail?

Solution:
Total number of trials = 75.
Number of times head turns up = 45
Number of times tail turns up = 30
(i) Let X be the event of getting a head.
= P(X) = Number of times head turns up / total number of trials
= 45/75
= 0.60

(ii) Let Y be the event of getting a tail.
P(getting a tail)
= P(Y) = Number of times tail turns up / total number of trials
= 30/75
= 0.40

Note: Remember when a fair coin is tossed and then X and Y are the only possible outcomes, and P(X) + P(Y)
= (0.60 + 0.40)
=1

4.Match the following events with the corresponding probabilities:
(i) The objects is not a circle       (a) 5/16
(ii) The objects is a triangle        (b) 4/16
(iii) The objects is not a triangle   (c) 7/16
(iv) The objects is not a square      (d) 9/16
(v) The objects is a circle           (e) 12/16
(vi) The objects is a square 	      (f) 11/16


Solution:
Number of blue triangles in a container = 4
Number of green squares = 5
Number of red circles = 7
Total number of objects = 4 + 5 + 7 = 16

(i) The objects is not a circle:
P(the object is a circle) = Number of circles/Total number of objects
= 7/16
P(the object is not a circle)
= 1 - P(the object is a circle)
= 1 - 7/16
= (16 - 7)/16
= 9/16

(ii) The objects is a triangle:
P(the object is a triangle) = Number of triangle/Total number of objects
= 4/16

(iii) The objects is not a triangle:
P(the object is a triangle) = Number of triangles/Total number of objects
= 4/16
P(the object is not a triangle) = 1 - P(the object is a triangle)
= 1 - 4/16
= (16 - 4)/16
= 12/16

(iv) The objects is not a square:
P(the object is a square) = Number of squares/Total number of objects
= 5/16
P(the object is not a square) = 1 - P(the object is a square)
= 1 - 5/16
= (16 - 5)/16 = 11/16

(v) The objects is a circle:
P(the object is a circle) = Number of circles/Total number of objects
= 7/16
(vi) The objects is a square:
P(the object is a square) = Number of squares/Total number of objects
= 5/16

Answers:
(i) The objects is not a circle       (d) 9/16
(ii) The objects is a triangle        (b) 4/16
(iii) The objects is not a triangle   (e) 12/16
(iv) The objects is not a square      (f) 11/16
(v) The objects is a circle           (c) 7/16
(vi) The objects is a square	      (a) 5/16



5. A single card is drawn at random from a standard deck of 52 playing cards.
Match each event with its probability.
Note: fractional probabilities have been reduced to lowest terms.
Consider the ace as the highest card.
(i) The card is a diamond                (a) 1/2
(ii) The card is a red king              (b) 1/13
(iii) The card is a king or queen        (c) 1/26
(iv) The card is either a red or an ace  (d) 12/13
(v) The card is not a king               (e) 2/13
(vi) The card is a five or lower         (f) 1/4
(vii) The card is a king                 (g) 4/13
(viii) The card is black	             (h) 7/13



Solution:
Total number of playing cards = 52

(i) The card is a diamond:
Number of diamonds in a deck of 52 cards = 13
P(the card is a diamond) = Number of diamonds/Total number of playing cards
= 13/52 = 1/4

(ii) The card is a red king:
Number of red king in a deck of 52 cards = 2
P(the card is a red king) = Number of red kings/Total number of playing cards
= 2/52 = 1/26

(iii) The card is a king or queen:
Number of kings in a deck of 52 cards = 4
Number of queens in a deck of 52 cards = 4
Total number of king or queen in a deck of 52 cards = 4 + 4 = 8
P(the card is a king or queen) = Number of king or queen/Total number of playing cards
= 8/52 = 2/13

(iv) The card is either a red card or an ace:
Total number of red card or an ace in a deck of 52 cards = 28
P(the card is either a red card or an ace) = Number of cards which is either a red card or an ace/Total number of playing cards
= 28/52 = 7/13

(v) The card is not a king:
Number of kings in a deck of 52 cards = 4
P(the card is a king) = Number of kings/Total number of playing cards
= 4/52 = 1/13
P(the card is not a king) = 1 -P(the card is a king)
= 1 - 1/13
= (13 - 1)/13
= 12/13

(vi) The card is a five or lower:
Number of cards is a five or lower = 16
P(the card is a five or lower) = Number of card is a five or lower/Total number of playing cards
= 16/52
= 4/13

(vii) The card is a king:
Number of kings in a deck of 52 cards = 4
P(the card is a king) = Number of kings/Total number of playing cards
= 4/52
= 1/13

(viii) The card is black:
Number of black cards in a deck of 52 cards = 26
P(the card is black) = Number of black cards/Total number of playing cards

= 26/52
= 1/2

Match the following events with the corresponding probabilities are shown below:
(i) The card is a diamond               (f) 1/4
(ii) The card is a red king             (c) 1/26
(iii) The card is a king or queen       (e) 2/13
(iv) The card is either a red or an ace (h) 7/13
(v) The card is not a king              (d) 12/13
(vi) The card is a five or lower        (g) 4/13
(vii) The card is a king                (b) 1/13
(viii) The card is black	            (a) 1/2


Theoretical Probability

Definition of Theoretical Probability:
Let a random experiment produce only finite number of mutually exclusive and equally likely outcomes. Then the probability of an event E is defined as
P(E) = Number of Favourable outcomes / Total number of possible outcome
The formula for finding the theoretical probability of an event is
P(E) =Number of Favourable outcomes / Total number of possible outcome
Theoretical probability is also known as Classical or A Priori probability.

Problems based on Theoretical Probability

#### 1. A fair coin is tossed550 times and the outcomes were noted as: Head = 350, Tail = 200 Find the probability of the coin showing up (i) a head (ii) a tail.

Solution:
Number of times coin is tossed = 550
Number of tails = 200
(i) Probability of getting a head
Number of Favourable outcomes P(H) = Total number of possible outcome
= 350/550 = 7/11

(ii) Probability of getting a tail
Number of Favourable outcomes P(T) = Total number of possible outcome
= 200/550 =4/11

#### 2. In a cricket match the Sachin hit a boundary 5 times out of 30 balls he plays. Find the probability that he (i) hit a boundary (ii) do not hit a boundary.

Solution:
Total number of balls Sachin played = 30
Number of boundary hit = 5
Number of times he did not hit a boundary = 30 - 5 = 25
(i) Probability that he hit a boundary
Number of Favourable outcomes P(A) = Total number of possible outcome
= 5/30 =1/6

(ii) Probability that he did not hit a boundary
Number of Favourable outcomes P(B) = Total number of possible outcome
= 25/30
= 5/6

#### 3. The record of weather stations report shows that out of the past 95 consecutive days, its weather forecast was correct 65 times. Find the probability that on a given day: (i) it was correct (ii) it was not correct.

Solution:
Total number of days = 95
Number of correct weather forecast = 65
Number of not correct weather forecast = 95 - 65 = 30
(i) Probability of 'it was correct forecast'
Number of Favourable outcomes P(X) = Total number of possible outcome
= 65/95
= 13/19

(ii) Probability of 'it was not correct forecast'
Number of Favourable outcomes P(Y) = Total number of possible outcome
= 30/95
= 6/19

#### 4. Two fair coins are tossed 225 times simultaneously and their outcomes are noted as: (i) Two tails = 65, (ii) One tail = 110 and (iii) No tail = 50 Find the probability of occurrence of each of these events.

Solution:
Total number of times two fair coins are tossed = 225
Number of times two tails occur = 65
Number of times one tail occur = 110
Number of times no tail occur = 50

(i) Probability of occurrence of 'two tails'
Number of Favourable outcomes P(X) = Total number of possible outcome
= 65/225
= 13/45

(ii) Probability of occurrence of 'one tail'
P(y) = 110/225
= 22/45

(iii) Probability of occurrence of 'no tail'
P(z) = 50/225
= 2/9

Random Experiments

• Random experiments are also known as observations. The word 'Probability' is used very often in our daily life; such as 'probably he is an honest man', 'what is the probability of a double head in a throw of a pair of coin?', 'probably it will rain in the evening' and so on.
• These days, an attempt towards a theory of probability is extensively used in various fields of what we are interested in. The instant answer is obviously an event.
• At first we will discuss about the precise meaning of the term 'event' and how it is used in our mathematical theory. Now our next step will help us to get the idea of what are random experiments or observations.

#### Definition of random experiment

• An experiment for which we know the set of all different results but it is not possible to predict which one of the set will occur at any particular execution of the experiment is called a random experiment.
For example: tossing a fair coin, casting an unbiased die and drawing a card from a pack of 52 cards.
• Let us take the experiment of tossing a coin. If we toss a coin then we get two possible outcomes either a 'head' (H) or a 'tail' (T), and it is impossible to predict whether the result of a toss will be a 'head' or 'tail'.
• Let us consider a similar experiment rolling a die from a box. It we a die then there are only six possible outcomes. The faces of the die are marked as 1, 2, 3, 4, 5 and 6 and these are the only possible outcomes. But here also the outcome of a particular throw is completely unpredictable.
• Again similarly when we are concerned about the measurement of a chemical quantity with the required instrument the outcome of the experiment does not exactly give the true value of the quantity but a value close to it due to that are called experimental errors.
• If repeated observations are taken the measured values are not again the same to the previous one but it will fluctuate in an unpredictable manner. Here we can take, at least for the theoretical considerations that the possible outcomes comprise all the real numbers, but the number given by a single measurement can't be exactly predicted.
• In our mathematical theory we will only consider the experiments or observations, for which we know a priori the set of all different possible outcomes, such that it is impossible to predict which particular outcome will occur at any particular performance of the experiment, are called random experiments.
• As such, if a random experiment is repeated under identical conditions the outcomes or results may vary or fluctuate at random.

Empirical Probability

#### Definition of Empirical Probability

The experimental probability of occurring of an event is the ratio of the number of trials in which the event occurred to the total number of trials.
The empirical probability of the occurrence of an event E is defined as:
P(E) = Number of trials in which event occurred / Total number of trials

Problems on empirical probability

1. Let us take the experiment of tossing a coin.

• When we toss a coin then we know that the results are either a head or a tail.
• Thus, in tossing a coin, all possible outcomes are 'Head' and 'Tail'.
• Suppose, we toss a coin 150 times and we get head, say, 102 times.
• Here we will find the probability of getting

(i) a head and, (ii) a tail

(i) Probability of getting a head
Let E1 be the event of getting a head.
P(E1) = Number of times getting heads / Total number of trials
= 102/150 = 0.68

(ii) Probability of getting a tail:
Total number of times a coin is tossed = 150
Number of times we get head = 102
Therefore, number of times we get tail = 150 -102 = 48
Now, let E2 be the event of getting a tail.
Then, P(getting a tail)
= P(E2) = Number of times getting tails / Total number of trials
= 48/150
= 0.32
Note: Remember, when a coin is tossed, then E1 and E2 are the only possible outcomes,
and P(E1) + P(E2) = (0.68 + 0.32) = 1

#### 2. Consider an experiment of rolling a die.

When we roll a die then the upper face of the die are marked as 1, 2, 3, 4, 5 or 6. These are the only six possible outcomes.
Suppose we throw a die 180 times and suppose we get 5 for 72 times.
Let E = event of getting 5 (dots).
Then, clearly, P(E) = 72/180= 0.40

#### 3. Let us take the case of tossing two coins simultaneously.

When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H,H) or (H,T) or (T,T) respectively.
Let us toss two coins randomly for 100 times.
Suppose the outcomes are:

• Let E1 be the event of getting 2 heads.
Then, P(E1) = 35/100 = 0.35
• Let E2 be the event of getting 1 head.
Then, P(E2) = 30/100 =0.30
• Let E3 be the event of getting 0 head.
Then, P(E3) = 35/100 = 0.35.

Note: Remember, when two coins are tossed randomly, then E1, E2 and E3 are the only possible outcomes, and P(E1) + P(E2) + P(E3)
= (0.35 + 0.30 + 0.35)
= 1

Meritpath...Way to Success!

Meritpath provides well organized smart e-learning study material with balanced passive and participatory teaching methodology. Meritpath is on-line e-learning education portal with dynamic interactive hands on sessions and worksheets.