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 atleast once. 

We have to find P(\(\frac{B}{A}\)) 

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

∴ P(A) = \(\frac{5}{36}\)

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) = \(\frac{2}{36}\)