Solution — 2023 Question 8 Probability

$$\begin{array}{cc}\mathrm{P}(B|A)& =\frac{\mathrm{P}(A\cap B)}{\mathrm{P}\left(A\right)}\\ \frac{3}{5}& =\frac{\mathrm{P}(A\cap B)}{x}\\ \mathrm{P}(A\cap B)& =\frac{3}{5}x\end{array}$$

$$\begin{array}{cc}\mathrm{P}(A\cup B)& =\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}(A\cap B)\\ \frac{13}{24}& =x+\mathrm{P}\left(B\right)-\frac{3}{5}x\\ & =\frac{2}{5}x+\mathrm{P}\left(B\right)\\ \mathrm{P}\left(B\right)& =\frac{13}{24}-\frac{2}{5}x∎\end{array}$$

$$\begin{array}{cc}\mathrm{P}(A^{\prime }\cap B)& =\mathrm{P}\left(B\right)-\mathrm{P}(A\cap B)\\ & =\frac{13}{24}-\frac{2}{5}x-\frac{3}{5}x\\ & =\frac{13}{24}-x\end{array}$$

$$\begin{array}{cc}\frac{3}{5}x& =2(\frac{13}{24}-x)\\ & =\frac{13}{12}-2x\\ \frac{13}{5}x& =\frac{13}{12}\\ x& =\frac{5}{12}∎\end{array}$$