Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $11 / 25$ and the probability of none of them occurring is $2 / 25$. If $P(T)$ denotes the probability of occurrence of the event $T$, then

Question

Aicharya · Accepted Answer

$ P(E \cup F)-P(E \cap F)=\frac{11}{25} $ (i.e. only $E$ or only $F$ )

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

Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $11 / 25$ and…

Solution