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

2/5

Let e be the event that " number 4 appeared at least once" and F be the event that " the sum of the numbers appearing is 6"

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

and      F = { (1,5), (2, 4), (3, 3), (4,2), (5, 1)}

We have P(E) = 11/36  and P(F) = 5/36

Also    E∩F = {(2, 4), (4, 2)}

Therefore   P(E∩F) = 2/36

Hence, the required probability is P(EIF) =P(E∩F)/ P(F) 

                                                           = (2/36)/(5/36)

                                                           = 2/5