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.

• 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.

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).

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.

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.

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)`.

Event points, Even Space:

• 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.

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

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).

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.

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

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.

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

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)

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.

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.

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,
(i) getting three heads
P(getting three heads) = P(E₁)
=Number of times three heads appeared/ Total number of trials
= 70/250
= 0.28

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

(iii) getting one head
P(getting one head) = P(E₃)
=Number of times three heads appeared/ Total number of trials
= 75/250
= 0.30

(iv) getting no head
P(getting no head) = P(E₄)
=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.
(i) getting all heads
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.

(ii) getting two heads
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.

(iii) getting one head
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.

(vi) getting atmost 2 heads
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'.

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.

• 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.

(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.

• 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.

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

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.

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.

• 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.

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