Venn Diagram Conditional Probability Questions Solution I was having a go at this question, I got an answer of $\frac{2}{29}$, not sure if this is correct, I would appreciate if someone could explain the solution to it.


You are correct, please include your working next time.
From the definition of conditional probability we have $$P(B\cap A'|C)=\frac{P(B\cap A'\cap C)}{P(C)}$$
$$=\frac{\frac{x}{5}}{x+\frac{x}{5}+\frac{x}{2}+2x^2+x}=\frac{\frac{x}{5}}{\frac{27}{10}x+2x^2}=\frac{\frac{1}{5}}{\frac{27}{10}+2x}.$$
But the sum of the probabilities must be $1$ so we have $$x+x+x^2+2x+\frac{x}{2}+x+\frac{x}{5}+2x^2+x+13x-1=1 $$
$$3x^2+\frac{197}{10}x-2=0$$
and solving gives $x=\frac{1}{10}$ and $x=-\frac{20}{3}$, but $x$ cannot be negative.
Thus we have $$P(B\cap A'|C)=\frac{\frac{1}{5}}{\frac{27}{10}+\frac{1}{5}}=\frac{\frac{1}{5}}{\frac{29}{10}}=\frac{2}{29}.$$
