Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E∩F) = 0.2 then P(E|F) ?

1. Break Down the Problem

To find $P(E|F)$, we will use the conditional probability formula:

$P(E|F) = \frac{P(E \cap F)}{P(F)}$

2. Relevant Concepts

• Conditional Probability Formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

3. Analysis and Detail

We'll apply the values provided:

1. Given Values:

   • $P(E) = 0.6$
   • $P(F) = 0.3$
   • $P(E \cap F) = 0.2$

2. Substituting into the Formula:

   $P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0.2}{0.3}$

4. Verify and Summarize

Now let’s calculate:

$P(E|F) = \frac{0.2}{0.3} = \frac{2}{3} \approx 0.6667$

Final Answer

Thus, the probability $P(E|F)$ is approximately $0.6667$ or $\frac{2}{3}$.