Question

If $P(E) = 0.6$, $P(F) = 0.3$ and $P(E \cap F) = 0.2$, find $P(E|F)$ and $P(F|E)$.

Hint:

Remember: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

Answer 1

Concept: Conditional probability is given by: \[ P(E|F) = \frac{P(E \cap F)}{P(F)}, \quad P(F|E) = \frac{P(E \cap F)}{P(E)} \] Step 1: {Calculate $P(E|F)$.}
\[ P(E|F) = \frac{0.2}{0.3} = \frac{2}{3} \]
Step 2: {Calculate $P(F|E)$.}
\[ P(F|E) = \frac{0.2}{0.6} = \frac{1}{3} \]
Step 3: {Conclusion.}
\[ \boxed{P(E|F) = \frac{2}{3}, \quad P(F|E) = \frac{1}{3}} \]