Question

A family has two children. Given that one of the children is a girl and that she was born on a Monday, what is the probability that both children are girls?

(a) \(\frac{13}{27}\)

(b) \(\frac{23}{54}\)

(c) \(\frac{12}{19}\)

(d) \(\frac{43}{58}\)

I have been asked this question in a job interview.

This is a very interesting question from Discrete Probability in chapter Discrete Probability of Discrete Mathematics

Answer 1

The correct option is (a) \(\frac{13}{27}\)

The explanation is: We let Y be the event that the family has one child who is a girl born on Tuesday and X be the event that both children are boys, and apply Bayes’ Theorem. Given that there are 7 days of the week and there are 49 possible combinations for the days of the week the two girls were born on and 13 of these have a girl who was born on a Monday, so P(Y|X) = \(\frac{13}{49}\). P(X) remains unchanged at \(\frac{1}{4}\). To calculate P(Y), there are 142 = 196 possible ways to select the gender and the day of the week the child was born on. There are 132 = 169 ways which do not have a girl born on Monday and which 196 – 169 = 27 which do, so P(Y) = \(\frac{27}{196}\). This gives is that P(X|Y) = \(\frac{\frac{13}{19}*\frac{1}{4}}{\frac{27}{196}} = \frac{13}{27}\).