Q139 / 240 · KCET 2026 · Mathematics

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 $P(A \mid L)$?

  • A
    $\dfrac{17}{100}$
  • B
    $\dfrac{19}{25}$
  • Correct answer
    C
    $\dfrac{1}{3}$
  • D
    $\dfrac{2}{3}$

Solution

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: $\dfrac{1}{3}$

Hence, the correct answer is option: 'C'.

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