Question

A, B, C are three mutually disjoint exhaustive events with $P(A) \neq 0$, $P(B) \neq 0$, $P(C) \neq 0$. $E$ is any arbitrary event. If $P(A) = \dfrac{4}{9}$, $P(B) = \dfrac{2}{9}$, $P(E|A) = \dfrac{3}{10}$,
$P(E|B) = \dfrac{5}{10}$, $P(E|C) = \dfrac{8}{10}$, and $P(E) = \dfrac{12}{23}$, then $P(C) =$

• A. \(\dfrac{2}{3}\)
• B. \(\dfrac{1}{3}\)
• C. \(\dfrac{5}{6}\)
• D. \(\dfrac{3}{4}\)

Hint:

Apply total probability carefully when dealing with mutually exclusive exhaustive events and conditional probabilities.

Answer 1

The Correct Option is B

Using total probability theorem: \[ P(E) = P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C) \] Substitute known values: \[ \dfrac{12}{23} = \dfrac{4}{9} \cdot \dfrac{3}{10} + \dfrac{2}{9} \cdot \dfrac{5}{10} + P(C) \cdot \dfrac{8}{10} \Rightarrow \dfrac{12}{23} = \dfrac{12}{90} + \dfrac{10}{90} + \dfrac{8}{10}P(C) = \dfrac{22}{90} + \dfrac{8}{10}P(C) \] \[ \Rightarrow \dfrac{12}{23} - \dfrac{11}{45} = \dfrac{8}{10}P(C) \Rightarrow \dfrac{539 - 253}{1035} = \dfrac{8}{10}P(C) \Rightarrow \dfrac{286}{1035} = \dfrac{8}{10}P(C) \Rightarrow P(C) = \dfrac{286 \cdot 10}{1035 \cdot 8} = \dfrac{2860}{8280} = \dfrac{1}{3} \]