Question

A die is thrown twice and the sum of the number appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Answer 1

When a die is thrown twice, the sample is

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

n(S) = 36

Let A be the event of getting the sum of the numbers is 6.

Let B be the event of the number 4 has appeared at least once.

We have to find `"P"("B"/"A")`

A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

n(A) = 5

∴ P(A) = `5/36`

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

A ∩ B = {(2, 4), (4, 2)}

n(A ∩ B) = 2

P(A ∩ B) = `2/36`

∴ `"P"("B"/"A") = ("P"("A" ∩ "B"))/("P"("A")) = (2/36)/(5/36) = 2/5`