Question

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P \left( E _{1} \mid E _{2}\right)=\frac{1}{2}$, $P \left( E _{2} \mid E _{1}\right)=\frac{3}{4}$ and $P \left( E _{1} \cap E _{2}\right)=\frac{1}{8}$. Then:

• A. $P \left( E _{1} \cap E _{2}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)$
• B. $P \left( E _{1}^{\prime} \cap E _{2}^{\prime}\right)= P \left( E _{1}^{\prime}\right) \cdot P \left( E _{2}\right)$
• C. $P \left( E _{1} \cap E _{2}^{\prime}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)$
• D. $P \left( E _{1}^{\prime} \cap E _{2}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)$

Answer 1

Answer: (C)

(A) $P \left( E _{1}\right) \cdot P \left( E _{2}\right)=\frac{1}{6} \cdot \frac{1}{4}=\frac{1}{24} \neq P \left( E _{1} \cap E _{2}\right)$
(B) $P \left( E _{1}^{\prime} \cap E _{2}^{\prime}\right) =1- P \left( E _{1} \cup E _{2}\right) $
$\left.=1-\left( P \left( E _{1}\right)+ P \left( E _{2}\right)- P _{1} E _{1} \cap E _{2}\right)\right)$
$=1-\left(\frac{1}{6}+\frac{1}{4}-\frac{1}{8}\right)=\frac{17}{24}$
$P \left( E _{1}^{\prime}\right) P \left( E _{2}\right)=\frac{5}{6} \times \frac{1}{4}=\frac{5}{24}$
(C) $P \left( E _{1} \cap E _{2}^{\prime}\right)= P \left( E _{1}\right)- P \left( E _{1} \cap E _{2}\right)$
$=\frac{1}{6}-\frac{1}{8}=\frac{1}{24}$
(D) $P \left( E _{1}^{\prime} \cap E _{2}\right)= P \left( E _{2}\right)- P \left( E _{1} \cap E _{2}\right)$
$=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$