Question

Two dice are thrown once. Find the probability of getting an even number on the 1st die or a total of 8.

Answer 1

Given: Two fair dice are thrown once. Find P(even on 1st die OR total = 8).

Step-wise calculation:

1. Total outcomes = 6 × 6 = 36.

2. Let A = “1st die is even”.

Number of outcomes with A = 3 × 6 = 18.

So, P(A) = `18/36`

= `1/2`

3. Let B = “Sum = 8”. 

Favourable outcomes for B are (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) → 5 outcomes.

So, P(B) = `5/36`.

4. A ∩ B = Outcomes with first die even and sum 8:

(2, 6), (4, 4), (6, 2) → 3 outcomes.

So, P(A ∩ B) = `3/36`

= `1/12`

5. Use inclusion–exclusion:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= `18/36 + 5/36 - 3/36`

= `20/36`

= `5/9`

The required probability is `5/9`.