JAMB Mathematics 2023 Paper

Concept:Use the relation between the probabilities to find the number of white balls, then apply the multiplication rule for dependent events without replacement.Explanation:Let the number of white balls be $n$.Total balls in the bag $= n + 8$.So $P(\text{white}) = \frac{n}{n+8}$ and $P(\text{red}) = \frac{8}{n+8}$.Given: $P(\text{white}) = \frac{1}{2}P(\text{red})$.Therefore, $\frac{n}{n+8} = \frac{1}{2} \times \frac{8}{n+8} = \frac{4}{n+8}$.This gives $n = 4$, so there are 4 white balls and $4 + 8 = 12$ balls in all.For drawing a red ball first without replacement: $P(\text{red}) = \frac{8}{12}$.After one red ball is removed, 11 balls remain, including 4 white balls.So $P(\text{white after red}) = \frac{4}{11}$.Thus, $P(\text{red and white}) = \frac{8}{12} \times \frac{4}{11} = \frac{32}{132} = \frac{8}{33}$.Answer:$\frac{8}{33}$