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
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
Show 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.
Updated On: Jun 25, 2026
- \(\dfrac{2}{3}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{3}{4}\)
- \(\dfrac{1}{4}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Find \(P(F)\).
Given \[ P(\overline{F})=0.7 \] Using the complement rule, \[ P(F)=1-P(\overline{F}) \] Therefore, \[ P(F)=1-0.7 \] Hence, \[ P(F)=0.3 \]
Step 2: Use the formula of conditional probability.
We know that \[ P(E|F)=\frac{P(E\cap F)}{P(F)} \] Given \[ P(E\cap F)=0.2 \] and \[ P(F)=0.3 \] Therefore, \[ P(E|F)=\frac{0.2}{0.3} \] \[ =\frac{2/10}{3/10} \] \[ =\frac{2}{3} \]
Step 3: Final conclusion.
Hence, \[ \boxed{\frac{2}{3}} \]
Given \[ P(\overline{F})=0.7 \] Using the complement rule, \[ P(F)=1-P(\overline{F}) \] Therefore, \[ P(F)=1-0.7 \] Hence, \[ P(F)=0.3 \]
Step 2: Use the formula of conditional probability.
We know that \[ P(E|F)=\frac{P(E\cap F)}{P(F)} \] Given \[ P(E\cap F)=0.2 \] and \[ P(F)=0.3 \] Therefore, \[ P(E|F)=\frac{0.2}{0.3} \] \[ =\frac{2/10}{3/10} \] \[ =\frac{2}{3} \]
Step 3: Final conclusion.
Hence, \[ \boxed{\frac{2}{3}} \]
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