A die is thrown 3 times. Events A and B stated below:
4 on the 3rd throw
6 on the 1st and 5 on the 2nd throw
What will be the probability of A knowing that B has already taken place?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 1)

Answer 1

Given:

A die is thrown 3 times.

Events A and B stated:-

4 on the 3rd throw.

6 on the 1st and 5 on the 2nd throw.

Formula Used:

P(A|B) = P(A ∩ B)/P(B)

Calculation:

A die is thrown 3 times.

Total cases = 6 × 6 × 6 =216

A : 4 on the third throw.

B : 6 on the first and 5 on the second throw.

A = {(1, 1, 4), (1, 2, 4),....(1, 6, 4), (2, 1, 4), (2, 2, 4),....(2, 6, 4),

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

(5, 1, 4), (5, 2, 4),.....(5, 6, 4), (6, 1, 4), (6, 2, 4),....(6, 6, 4)}

⇒ P(A) =

B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

⇒ P(B) =

Thus A ∩ B = {(6, 5, 4)}

So, P(A ∩ B) =

Now, P(A|B) =

⇒ P(A|B) =

∴ The probability of A knowing that B has already taken place is 1/6