Recent studies suggest that $12\%$ of the world population is left handed. Depending on parents' hand usage, the chances of having left handed children are as follows:
A: Both parents are left handed, chances of having left handed children $= 24\%$
B: Both parents are right handed, chances of having left handed children $= 9\%$
C: Father left handed and mother right handed, chances of having left handed children $= 17\%$
D: Father right handed and mother left handed, chances of having left handed children $= 22\%$
Given $P(A) = P(B) = P(C) = P(D) = 1/4$ and L denotes child is left handed. What is the probability that $P(A \mid L)$?

The probability that a man and his wife live after $20$ years are $\dfrac{1}{4}$ and $\dfrac{1}{3}$ respectively. The probability that neither the man nor his wife live after $20$ years is

Probability of obtaining an even prime number on each die when a pair of dice is rolled is

Probability of occurrence of an event A is $\dfrac{1}{2}$ and that of B is $\dfrac{3}{10}$. If A and B are mutually exclusive, then the probability of occurrence of neither A nor B is