Question

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.​

Answer 1

Answer:

Given :-

Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

To find :-

(i) a prime number;

(ii) a number lying between 2 and 6;

(iii) an odd number.

Solution :-

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = ½ = 0.5

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

P (getting a number between 2 and 6) = 3/6 = ½ = 0.5

(iii) Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) = 3/6 = ½ = 0.5

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Answer 2

Answer:

(i) Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = ½ = 0.5

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

P (getting a number between 2 and 6) = 3/6 = ½ = 0.5

(iii) Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) = 3/6 = ½ = 0.5