Question:

Step 1: Define the sample space. Let each child be characterized by their gender and day of birth. There are 2 genders (Boy, Girl) and 7 days of the week (Monday, ..., Sunday). Thus, each child has $2 \times 7 = 14$ possible outcomes. For two children, the total number of possible outcomes in the sample space is $14 \times 14 = 196$. Step 2: Determine the number of outcomes where at least one child is a boy born on Sunday. Let $E$ be the event that at least one child is a boy born on Sunday. It is easier to calculate the complement event $E^c$, where neither child is a boy born on Sunday. For a single child, there are 14 possibilities. If a child is not a boy born on Sunday, there are $14 - 1 = 13$ possibilities (i.e., Boy-Monday, ..., Boy-Saturday, Girl-Monday, ..., Girl-Sunday). Thus, the number of outcomes where neither child is a boy born on Sunday is $13 \times 13 = 169$. The number of outcomes where at least one child is a boy born on Sunday is the total number of outcomes minus the number of outcomes in $E^c$: $|E| = 196 - 169 = 27$. Step 3: Determine the number of outcomes where both children are boys AND at least one child is a boy born on Sunday. Let $B$ be the event that both children are boys. We are looking for the number of outcomes in $B \cap E$. If both children are boys, each child can be a boy born on any of the 7 days. The possibilities for two boys are: \begin{itemize} \item Child 1 is a boy born on Sunday, Child 2 is a boy born on any day: $1 \times 7 = 7$ outcomes. \item Child 2 is a boy born on Sunday, Child 1 is a boy born on any day: $7 \times 1 = 7$ outcomes. \end{itemize} The outcome (Boy-Sunday, Boy-Sunday) is included in both of these counts. Using the Principle of Inclusion-Exclusion: $|B \cap E| = (\text{Child 1 is Boy-Sunday, Child 2 is Boy-any day}) + (\text{Child 2 is Boy-Sunday, Child 1 is Boy-any day}) - (\text{Both are Boy-Sunday})$ $|B \cap E| = 7 + 7 - 1 = 13$. The probability that both children are boys, given that at least one child is a boy born on Sunday, is: $P(B | E) = \frac{|B \cap E|}{|E|} = \frac{13}{27}$.