Probability Theory – Problem 42: find the probability that both children are girls

A family has two children whose genders and birth order are unknown.

(1) Given that the older child is a girl, find the probability that both children are girls. $\underline{\qquad\qquad}$

(2) Given that one of the children is a girl, find the probability that both children are girls. $\underline{\qquad\qquad}$

(3) Upon observing one child who turns out to be a girl, find the probability that both children are girls. $\underline{\qquad\qquad}$

(4) Given that at least one child is a girl, find the probability of observing a girl when one child is selected at random. $\underline{\qquad\qquad}$

Let B denote a boy and G denote a girl. For a family with two children, disregarding gender and birth order, the sample space is Ω = {(B, B), (B, G), (G, B), (G, G)}, where the first letter represents the older child and the second letter represents the younger child. Each outcome is equally likely with probability $\frac{1}{4}$. (1) Let event A be "the older child is a girl." Then the sample points in A are {(G, B), (G, G)}. $P(A) = \frac{2}{4} = \frac{1}{2}$ Let event B be "both children are girls." Then the sample points in B are {(G, G)}. $P(B) = \frac{1}{4}$ The desired probability is the conditional probability $P(B|A)$. $A \cap B = \{(G, G)\}$ $P(A \cap B) = \frac{1}{4}$ $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$

(2) Let event C be "one of the children is a girl," meaning at least one child is a girl. Then the sample points in C are {(B, G), (G, B), (G, G)}. $P(C) = \frac{3}{4}$ Let event B be "both children are girls." Then the sample points in B are {(G, G)}. $B \cap C = \{(G, G)\}$ $P(B \cap C) = \frac{1}{4}$ The desired probability is the conditional probability $P(B|C)$. $P(B|C) = \frac{P(B \cap C)}{P(C)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$

(3) "Observing one child who turns out to be a girl" is an observation event; denote it as event S. Let the events corresponding to family compositions GG, GB, BG, BB be $E_{GG}, E_{GB}, E_{BG}, E_{BB}$, respectively. $P(E_{GG}) = P(E_{GB}) = P(E_{BG}) = P(E_{BB}) = \frac{1}{4}$ Under each family composition, the probability of observing a girl is as follows: $P(S|E_{GG}) = 1$ $P(S|E_{GB}) = \frac{1}{2}$ $P(S|E_{BG}) = \frac{1}{2}$ $P(S|E_{BB}) = 0$ By the law of total probability, the probability of observing a girl $P(S)$ is: $P(S) = P(S|E_{GG})P(E_{GG}) + P(S|E_{GB})P(E_{GB}) + P(S|E_{BG})P(E_{BG}) + P(S|E_{BB})P(E_{BB})$ $P(S) = (1 \times \frac{1}{4}) + (\frac{1}{2} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{1}{4}) + (0 \times \frac{1}{4}) = \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}$ The desired probability is $P(E_{GG}|S)$. By Bayes' theorem: $P(E_{GG}|S) = \frac{P(S|E_{GG})P(E_{GG})}{P(S)} = \frac{1 \times \frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$

(4) This problem asks for the probability of "observing a girl" (event S) given that "at least one child is a girl" (event C), i.e., $P(S|C)$. By the definition of conditional probability: $P(S|C) = \frac{P(S \cap C)}{P(C)}$ If event S occurs (a girl is observed), then event C (at least one child is a girl) necessarily occurs. Therefore $S \cap C = S$. $P(S|C) = \frac{P(S)}{P(C)}$ From the computations in (2) and (3), $P(C) = \frac{3}{4}$ and $P(S) = \frac{1}{2}$. $P(S|C) = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}$

(1) $\frac{1}{2}$ (2) $\frac{1}{3}$ (3) $\frac{1}{2}$ (4) $\frac{2}{3}$