Question

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happens is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{12}$. Then

• A. $P\left(E\right)=\frac{1}{3},P\left(F\right)=\frac{1}{4}$ or $P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{3}$
• B. $P\left(E\right)=\frac{1}{2},P\left(F\right)=\frac{1}{6}$
• C. $P\left(E\right)=\frac{1}{6},P\left(F\right)=\frac{1}{2}$
• D. $P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{3}$

Answer 1

The correct option is A $P\left(E\right)=\frac{1}{3},P\left(F\right)=\frac{1}{4}$ or $P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{3}$

Since $E,F$ and independent.

$P(E\cap F)=P\left(E\right)P\left(F\right)\Rightarrow \frac{1}{2}=P\left(E\right)P\left(F\right)$

Again,

$\therefore E,F$ are independent $\Rightarrow \overline{E},\overline{F}$ are independent

$\therefore P(\overline{E}\cap \overline{F})=P\left(\overline{E}\right)P\left(\overline{F}\right)$

$\Rightarrow \frac{1}{2}=(1-P\left(E\right))(1-P\left(F\right))=1-P\left(E\right)-P\left(F\right)+P\left(E\right)P\left(F\right)=1-P\left(E\right)-P\left(F\right)+\frac{1}{12}$

$\Rightarrow P\left(E\right)+P\left(F\right)=\frac{13}{12}-\frac{1}{2}=\frac{(13-6)}{12}=\frac{7}{12}$

$\Rightarrow x+y=\frac{7}{12},$

where $x=P\left(E\right),y=P\left(F\right)$ and $xy=\frac{1}{12}$

$\Rightarrow x(\frac{7}{12}-x)=\frac{1}{12}$

$\Rightarrow x^2-\left(\frac{7}{12}\right)x+\frac{1}{12}=0$

$\Rightarrow (x-\frac{1}{4})(x-\frac{1}{3})=0$

$\Rightarrow x=\frac{1}{4}$ or $\frac{1}{3}$

$\therefore P\left(E\right)=\frac{1}{4}$ or $P\left(E\right)=\frac{1}{3}$

$\therefore P\left(F\right)=\frac{1}{3}$ or $P\left(F\right)=\frac{1}{4}$

Hence $P\left(E\right)=\frac{1}{4},P\left(F\right)=\frac{1}{3}$ or $P\left(E\right)=\frac{1}{3},P\left(F\right)=\frac{1}{4}.$