The conditional probability formula is:
[tex]\begin{gathered} P(A|B)=\frac{P(A\cap B)}{P(B)} \\ \text{ Where} \\ P(A|B)\text{ is the probability of A given B,} \\ P(A\cap B)\text{ is the probability of A and B, and} \\ P(B)\text{ is the probability of B} \end{gathered}[/tex]
On the other hand, the probability formula is:
[tex]P(A)=\frac{\text{ Number of favourable outcomes to A}}{\text{ Total number of possible outcomes}}[/tex]
To solve the exercise, we first complete the table:
So, in this case, we have:
[tex]\begin{gathered} P(\text{ Student has a brother and they have a sister})=\frac{2}{23} \\ P(\text{They have a sister})=\frac{8}{23} \\ P(\text{ Student has a brother | They have a sister})=\frac{\frac{2}{23}}{\frac{8}{23}} \\ P(\text{ Student has a brother | They have a sister})=\frac{2\cdot23}{8\cdot23} \\ P(\text{ Student has a brother | They have a sister})=\frac{2}{8} \\ \text{ Simplify} \\ P(\text{ Student has a brother | They have a sister})=\frac{1\cdot2}{4\cdot2} \\ P(\text{ Student has a brother | They have a sister})=\frac{1}{4} \end{gathered}[/tex]
Therefore, the probability that a student has a brother given that they have a sister is 1/4.
