Question

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 $P(A|L)$?

• A. $\frac{17}{80}$
• B. $\frac{24}{75}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Hint:

In Bayes' theorem problems, it's common to find distractor information (like the 12% overall population statistic here). Focus on the specific events defined and the specific probabilities given for them to construct your calculation.

Answer 1

The Correct Option is C

To solve this problem, we need to determine the probability that both parents are left-handed given that the child is left-handed, denoted as \(P(A|L)\). This is a classic application of Bayes' Theorem, which relates conditional probabilities. The theorem is given by:

\(P(A|L) = \frac{P(L|A) \cdot P(A)}{P(L)}\)

Let's identify each term:

1. \(P(L|A) = 0.24\): The probability that a child is left-handed given that both parents are left-handed.
2. \(P(A) = \frac{1}{4}\): The probability that both parents are left-handed.
3. \(P(L)\): The total probability that a child is left-handed. We calculate it using the law of total probability:

\(P(L) = P(L|A) \cdot P(A) + P(L|B) \cdot P(B) + P(L|C) \cdot P(C) + P(L|D) \cdot P(D)\)

1. Substituting the given values:
   • \(P(L|A) = 0.24\)
   • \(P(L|B) = 0.09\)
   • \(P(L|C) = 0.17\)
   • \(P(L|D) = 0.22\)
   • Each parent condition has equal probability: \(P(A) = P(B) = P(C) = P(D) = \frac{1}{4}\)

Now, applying Bayes' theorem:

\(P(A|L) = \frac{P(L|A) \cdot P(A)}{P(L)} = \frac{0.24 \times \frac{1}{4}}{0.18} = \frac{0.24 \times 0.25}{0.18} = \frac{0.06}{0.18} = \frac{1}{3}\)

Therefore, the probability that both parents are left-handed given that the child is left-handed is \(\frac{1}{3}\).