1.

In a single throw of two dice, find the probability of(i) getting a sum less than 6(ii) getting a doublet of odd numbers(iii) getting the sum as a prime number

Answer»

Probability of occurrence of an event = (Total number of favorable outcomes) / (Total number of outcomes)

Possible outcomes are as follow:

(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), (5, 6) ,

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total number of outcomes = 36

(i) getting a sum less than 6

Pick entries having sum less than 6:

(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)

Total number of favorable outcomes = 10

Probability (getting a sum less than 6) = 10/36 or 5/18

(ii) getting a doublet of odd numbers

Pick entries having doublet of odd numbers:

(1, 1), (3, 3), (5, 5)

Total number of favorable outcomes = 3

Probability (getting a doublet of odd numbers) = 3/36 or 1/12

(iii) getting the sum as a prime number

Pick entries having sum as a prime number:

(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)

Total number of favorable outcomes = 15

Probability (getting the sum as a prime number) = 15/36 or 5/12



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