Mathematics - Probability Question with Solution | TestHub

$$P(E\cup F)-P(E\cap F)=\frac{11}{25}$$ (i.e. only $E$ or only $F$ )
Neither of them occurs $=\frac{2}{25}$ $\Rightarrow P(\bar{E}\cap \bar{F})=\frac{2}{25}$ From Eq. (i), we get $$P(E)+P(F)-2P(E\cap F)=\frac{11}{25}$$ From Eq. (ii), we get $$\begin{array}{c}(1-P(E))(1-P(F))=\frac{2}{25}\\ \Rightarrow 1-P(E)-P(F)+P(E)\cdot P(F)=\frac{2}{25}\end{array}$$ (iv) From Eqs. (iii) and (iv), we get $$\begin{array}{rl}& P(E)+P(F)=\frac{7}{5}\text{ and }P(E)\cdot P(F)=\frac{12}{25}\\ & \therefore P(E)\cdot \left\{\frac{7}{5}-P(E)\right\}=\frac{12}{25}\end{array}$$ $$\begin{array}{rl}& \Rightarrow (P(E){)}^2-\frac{7}{5}P(E)+\frac{12}{25}=0\\ & \Rightarrow \left(P(E)-\frac{3}{5}\right)\left(P(E)-\frac{4}{5}\right)=0\\ & \therefore P(E)=\frac{3}{4}\text{ or }\frac{4}{5}\Rightarrow P(F)=\frac{4}{5}\text{ or }\frac{3}{5}\end{array}$$