Understanding Conditional Probability for Children's Gender

This problem involves conditional probability. We need to find the probability of an event occurring given that another event has already occurred.

Let B denote a boy and G denote a girl. Assuming the probability of having a boy or a girl is equal ($\frac{1}{2}$), the possible combinations for two children, listed from older to younger, are:

• BB (Boy, Boy)
• BG (Boy, Girl)
• GB (Girl, Boy)
• GG (Girl, Girl)

Each combination has an equal probability of $\frac{1}{4}$.

Calculating the Probability

We are given the condition that the older child is a boy. Let this event be 'A'. The outcomes where the older child is a boy are:

• BB
• BG

So, the reduced sample space, given event A, consists of these two equally likely outcomes.

We want to find the probability that both children are boys (event 'B') given that the older one is a boy (event 'A'). The outcome where both children are boys is BB.

Within the reduced sample space {BB, BG}, the outcome 'BB' occurs once.

Therefore, the conditional probability is calculated as:

$ P(\text{Both Boys} | \text{Older is Boy}) = \frac{\text{Number of outcomes where both are boys and older is a boy}}{\text{Number of outcomes where older is a boy}} $

$ P(\text{Both Boys} | \text{Older is Boy}) = \frac{1}{2} $

Alternatively, using the formula for conditional probability $P(B|A) = \frac{P(A \cap B)}{P(A)}$:

• Event A: Older child is a boy {BB, BG}. $P(A) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
• Event B: Both children are boys {BB}.
• Event $A \cap B$: Older child is a boy AND both are boys {BB}. $P(A \cap B) = \frac{1}{4}$.
• $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1/4}{1/2} = \frac{1}{4} \times 2 = \frac{1}{2}$.