(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.

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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 E_{1} = event of getting six as a product. The number whose product is six will be

E_{1} = [(1, 6), (2, 3), (3, 2), (6, 1)] = 4

Therefore, probability of getting 'six as a product'

P(E_{1}) = Number of Favourable outcomes / Total number of possible outcome

= 4/36

= 1/9

**(ii)** getting sum <= 3:

Let E_{2} = event of getting sum <= 3. The number whose sum <= 3 will be E_{2} = [(1, 1), (1, 2), (2, 1)] = 3

Therefore, probability of getting 'sum <= 3'

P(E_{2}) = Number of Favourable outcomes/ Total number of possible outcome

= 3/36

= 1/12

**(iii)** getting sum <= 10:

Let E_{3} = event of getting sum <= 10.

The number whose sum <= 10 will be

E_{3} =
[(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(E_{3}) = Number of Favourable outcomes /Total number of possible outcome

= 33/36

= 11/12

**(iv)** getting a doublet: Let E_{4} = event of getting a doublet. The number which doublet will be E_{4} = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6

Therefore, probability of getting 'a doublet'

P(E_{4}) = Number of Favourable outcomes /Total number of possible outcome

= 6/36

= 1/6

**(v)** getting a sum of 8

Let E_{5} = event of getting a sum of 8. The number which is a sum of 8 will be

E_{5} = [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)] = 5

Therefore, probability of getting 'a sum of 8'

P(E_{5}) = Number of Favourable outcomes / Total number of possible outcome

= 5/36

**(vi)** getting sum divisible by 5

Let E

_{6} = event of getting sum divisible by 5.

The number whose sum divisible by 5 will be

E

_{6} = [(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)] = 7

Therefore, probability of getting 'sum divisible by 5'

P(E

_{6}) = Number of Favourable outcomes / Total number of possible outcome

= 7/36

**(vii)** getting sum of atleast 11:

Let E_{7} = event of getting sum of atleast 11.

The events of the sum of atleast 11 will be

E_{7}= [(5, 6), (6, 5), (6, 6)] = 3

Therefore, probability of getting 'sum of atleast 11'

P(E_{7}) = Number of Favourable outcomes /Total number of possible outcome

= 3/36

= 1/12

**(viii)** getting a multiple of 3 as the sum

Let E_{8} = event of getting a multiple of 3 as the sum.

The events of a multiple of 3 as the sum will be

E_{8} = [(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(E_{8}) = Number of Favourable outcomes / Total number of possible outcome

= 12/36

= 1/3

**(ix)** getting a total of atleast 10:

Let E_{9} = event of getting a total of atleast 10.

The events of a total of atleast 10 will be

E_{9}= [(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)] = 6

Therefore, probability of getting 'a total of atleast 10'

P(E_{9}) = Number of Favourable outcomes / Total number of possible outcome

= 6/36

= 1/6

**(x)** getting an even number as the sum:
Let E_{10} = event of getting an even number as the sum.

The events of an even number as the sum will be

E_{10} = [(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(E_{1}0) = Number of Favourable outcomes/ Total number of possible outcome

= 18/36
= 1/2

**(xi)** getting a prime number as the sum:

Let E_{11} = event of getting a prime number as the sum. The events of a prime number as the sum will be

E_{11} = [(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(E_{11}) = Number of Favourable outcomes / Total number of possible outcome

= 15/36

= 5/12

**(xii)** getting a doublet of even numbers:

Let E_{12} = event of getting a doublet of even numbers.

The events of a doublet of even numbers will be

E_{12} = [(2, 2), (4, 4), (6, 6)] = 3

Therefore, probability of getting 'a doublet of even numbers'

P(E_{12}) = 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 E_{13} = 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

E_{13} = [(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(E_{13}) =Number of Favourable outcomes / Total number of possible outcome

= 11/36