Question

If E and F are two independent events, such that $P(E\cap F)=\frac{1}{6},P(E^C\cap F^C)=\frac{1}{3}$ and (P(E)-P(F))(1-P(F))>0, then

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

Answer 1

Answer: (A), (C)

The correct options are
A $P\left(E\right)=\frac{1}{2}$

C $P\left(F\right)=\frac{1}{3}$


$P(E\cap F)=P\left(E\right)P\left(F\right)=\frac{1}{6}.....\left(i\right)P(E^c\cap F^c)=(1-P(E\left)\right)(1-P(F\left)\right)=\frac{1}{3}\Rightarrow P\left(E\right)+P\left(F\right)=\frac{5}{6}.......\left(ii\right)\Rightarrow |P\left(E\right)-P\left(F\right)|=\frac{1}{6}$
As (P(E)-P(F))(1-P(F))>0
$\Rightarrow P\left(E\right)>P\left(F\right)\Rightarrow P\left(E\right)-P\left(F\right)=\frac{1}{6}...\left(iii\right)$
Solving (ii)and (iii) $\Rightarrow P\left(E\right)=\frac{1}{2},P\left(F\right)=\frac{1}{3}$