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?
Answer & explanation
Correct answer: option 1
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