Question

If \(E\) and \(F\) are events such that \[ P(\overline{F})=0.7 \] and \[ P(E\cap F)=0.2, \] then \[ P(E|F)= \] is

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

Hint:

Remember: \[ P(A|B)=\frac{P(A\cap B)}{P(B)} \] and \[ P(B)=1-P(\overline{B}) \] Always convert complement probabilities first before applying conditional probability formulas.

Answer 1

The Correct Option is A

Step 1: Find $ P(F) $ from its complement.
We are given $ P(\overline{F}) = 0.7 $. By the complement rule, $ P(F) = 1 - P(\overline{F}) = 1 - 0.7 = 0.3 $. This is the probability that event $ F $ actually occurs.
Step 2: Write the conditional probability formula.
By definition of conditional probability: \[ P(E|F) = \frac{P(E \cap F)}{P(F)} \] We use this formula whenever we want the probability of $ E $ given that $ F $ has already occurred.
Step 3: Substitute the known values.
We have $ P(E \cap F) = 0.2 $ and $ P(F) = 0.3 $. So: \[ P(E|F) = \frac{0.2}{0.3} \]
Step 4: Simplify the fraction.
\[ \frac{0.2}{0.3} = \frac{2}{3} \] We can see this by writing both as fractions over 10: $ \dfrac{2/10}{3/10} = \dfrac{2}{3} $.
Step 5: Verify the result makes sense.
Since $ P(E \cap F) < P(F) $, we expect $ P(E|F) < 1 $, and $ \dfrac{2}{3} \approx 0.67 < 1 $. This is consistent.
Step 6: State the answer.
\[ \boxed{\frac{2}{3}} \]