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

Step 1: Understanding the Concept:
This problem requires the application of Bayes' Theorem. We need to find the conditional probability of event A given that event L has occurred, i.e., $P(A|L)$. The "12" statistic is general background information and is superseded by the specific probabilities provided for this scenario.

Step 2: Key Formula or Approach:
Use Bayes' Theorem: \[ P(A|L) = \frac{P(L|A) \cdot P(A)}{P(L)} \] where the total probability $P(L)$ is found using the Law of Total Probability: \[ P(L) = P(L|A)P(A) + P(L|B)P(B) + P(L|C)P(C) + P(L|D)P(D) \]
Step 3: Detailed Explanation:
From the problem statement, we have the following probabilities: Prior probabilities for the parent configurations: \[ P(A) = P(B) = P(C) = P(D) = \frac{1}{4} = 0.25 \] Conditional probabilities of having a left-handed child given the parent configuration: \[ P(L|A) = 24% = 0.24 \] \[ P(L|B) = 9% = 0.09 \] \[ P(L|C) = 17% = 0.17 \] \[ P(L|D) = 22% = 0.22 \] First, calculate the total probability of a child being left-handed, $P(L)$, using the Law of Total Probability: \[ P(L) = P(L|A)P(A) + P(L|B)P(B) + P(L|C)P(C) + P(L|D)P(D) \] \[ P(L) = \left(0.24 \times \frac{1}{4}\right) + \left(0.09 \times \frac{1}{4}\right) + \left(0.17 \times \frac{1}{4}\right) + \left(0.22 \times \frac{1}{4}\right) \] \[ P(L) = \frac{1}{4} (0.24 + 0.09 + 0.17 + 0.22) \] \[ P(L) = \frac{1}{4} (0.72) = 0.18 \] Now, apply Bayes' Theorem to find $P(A|L)$: \[ P(A|L) = \frac{P(L|A) \cdot P(A)}{P(L)} \] \[ P(A|L) = \frac{0.24 \times \frac{1}{4}}{0.18} \] \[ P(A|L) = \frac{0.06}{0.18} = \frac{6}{18} = \frac{1}{3} \]
Step 4: Final Answer:
The probability $P(A|L)$ is $\frac{1}{3}$.