Question

Assume that each child who is born is equally likely to be a boy or a girl. If a family has two children, what is the probability that both are girls given that (a) the eldest is a girl, (b) at least one is a girl?

Short answer

The probabilities are as follows: (a) The probability that both children are girls given the eldest is a girl is 0.5. (b) The probability that both children are girls given at least one is a girl is approximately 0.333 or 1/3.

Step-by-step solution

(Case 1: Probability that both children are girls given the eldest is a girl)

: To find the probability that both children are girls given the eldest is a girl, we can define two events: Event A: Both children are girls Event B: The eldest is a girl Now, using the conditional probability formula: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) We know the following probabilities: P(A ∩ B) = Probability that both children are girls and the eldest is a girl. Since the gender of each child is independent, this probability is equal to the product of the probabilities that each child is a girl. So, \(P(A \cap B) = 0.5 * 0.5 = 0.25\). P(B) = Probability that the eldest is a girl, which is 0.5, as both genders are equally likely. Now, we can find the conditional probability: \(P(A|B) = \frac{0.25}{0.5} = 0.5\) So, the probability that both children are girls given the eldest is a girl is 0.5.

(Case 2: Probability that both children are girls given at least one is a girl)

: To find the probability that both children are girls given at least one is a girl, we can redefine the events: Event A: Both children are girls Event B: At least one child is a girl Now, using the conditional probability formula: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) We already know the probability P(A ∩ B) from above, which is 0.25. So, we need to find P(B). P(B) = Probability of at least one child being a girl, which is the complement of both children being boys (the only situation in which there is no girl). So, P(B') = Probability of both children being boys = 0.5 * 0.5 = 0.25 P(B) = 1 - P(B') = 1 - 0.25 = 0.75 Now, we can find the conditional probability: \(P(A|B) = \frac{0.25}{0.75} = \frac{1}{3} \approx 0.333\) So, the probability that both children are girls given at least one is a girl is approximately 0.333 or 1/3.