Question

If $A$ and $B$ are two events such that $P(A) = 0.4$, $P(B) = 0.8$ and $P(B|A) = 0.6$, then $P(\bar{A} \cap B) =$

• A. 0.56
• B. 0.24
• C. 0.16
• D. 0.32

Hint:

$P(\bar{A} \cap B)$ represents the probability of "only B". Visualizing this on a Venn diagram makes the relation $P(B) - P(A \cap B)$ immediately clear.

Answer 1

The Correct Option is A

Step 1: Tools we need.
Conditional probability: $P(B|A) = \frac{P(A\cap B)}{P(A)}$. Also the set fact $P(\bar A\cap B) = P(B) - P(A\cap B)$.

Step 2: Write the known values.
$P(A) = 0.4$, $P(B) = 0.8$, $P(B|A) = 0.6$.

Step 3: Find the intersection.
Rearranging the conditional formula: \[ P(A\cap B) = P(B|A)\,P(A) = 0.6 \times 0.4 = 0.24 \]

Step 4: Set up the target.
$\bar A\cap B$ is the part of $B$ outside $A$, so \[ P(\bar A\cap B) = P(B) - P(A\cap B) \]

Step 5: Substitute.
\[ = 0.8 - 0.24 \]

Step 6: Answer.
\[ = 0.56 \] \[ \boxed{ 0.56 } \]