Question

In a single throw of two dice, find the probability of an odd number on one dice and a number less than or equal to 4 on the other dice.   

Answer 1

The number of possible outcomes is 6 × 6 = 36. 

We are looking for outcomes where one die shows an odd number (1, 3, 5) and the other shows a number less than or equal to 4 (1, 2, 3, 4).

Favourable outcomes are:

When die 1 is odd and die 2 is ≤ 4:

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

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

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

When die 2 is odd and die 1 is ≤ 4:

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

Total favourable outcomes = 12 + 4 = 16

So, the required probability is:

`16/36`

`4/9`