Let E and F be two events such that \(\displaystyle P(E) =\frac{1}{3}, P(F) =\frac{1}{4}\) and \(\displaystyle P(E\cap F) =\frac{1}{5}\). Then the value of \(P(F|E)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{3}{5}\)
By the definition of conditional probability, the condition E goes in the denominator:
\(\displaystyle P(F|E) = \frac{P(E\cap F)}{P(E)}\)
\(\displaystyle = \frac{1/5}{1/3} = \frac{3}{5}\)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{4}{5}\) — Incorrect. This is \(\frac{1/5}{1/4} = P(E|F)\), with P(F) in the denominator. The question asks for \(P(F|E)\).
Option (2) 1 — Incorrect. \(P(F|E) = 1\) would need \(P(E\cap F) = P(E) = \frac{1}{3}\), but it is \(\frac{1}{5}\).
Option (4) \(\displaystyle\frac{1}{12}\) — Incorrect. This is \(P(E)P(F)\), a product of the two probabilities, not a conditional probability.