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 \(P(A \mid L)\)?

Let \(L\) be the event that the child is left-handed.

From the given data, the conditional probabilities of having a left-handed child are:
\(P(L \mid A) = \dfrac{24}{100}\)
\(P(L \mid B) = \dfrac{9}{100}\)
\(P(L \mid C) = \dfrac{17}{100}\)
\(P(L \mid D) = \dfrac{22}{100}\)

We are given the prior probabilities:
\(P(A) = P(B) = P(C) = P(D) = \dfrac{1}{4}\)

We need to find the probability that both parents are left-handed given that the child is left-handed, which is \(P(A \mid L)\).

Using Bayes' theorem:
\(P(A \mid L) = \dfrac{P(A)P(L \mid A)}{P(A)P(L \mid A) + P(B)P(L \mid B) + P(C)P(L \mid C) + P(D)P(L \mid D)}\)

Substituting the values:
\(P(A \mid L) = \dfrac{\dfrac{1}{4} \times \dfrac{24}{100}}{\dfrac{1}{4} \times \dfrac{24}{100} + \dfrac{1}{4} \times \dfrac{9}{100} + \dfrac{1}{4} \times \dfrac{17}{100} + \dfrac{1}{4} \times \dfrac{22}{100}}\)

Canceling the common factor of \(\dfrac{1}{400}\) from the numerator and the denominator:
\(P(A \mid L) = \dfrac{24}{24 + 9 + 17 + 22}\)

\(P(A \mid L) = \dfrac{24}{72} = \dfrac{1}{3}\)

Answer: \(\dfrac{1}{3}\)