In a single throw of two dice, find the probability of getting a doublet of odd numbers. (a) 1/9 (b) 1/18 (c) 1/36 (d) 1/12

Question

In a single throw of two dice, find the probability of getting a doublet of odd numbers.

(a) \(\frac{1}{9}\)

(b) \(\frac{1}{18}\)

(c) \(\frac{1}{36}\)

(d) \(\frac{1}{12}\)

Answer 1

(d) \(\frac{1}{12}\)

Total number of exhaustive cases in a single throw of two dice = 6 × 6 = 36

Doublets are obtained as (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) 

Number of doublets of odd numbers = 3

\(\therefore\) Required probability = \(\frac{3}{36}\) = \(\frac{1}{12}\)