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 $P(\bar{A} \cap B) =$
If $A$ and $B$ are two events such that $P(A) = 0.4$, $P(B) = 0.8$ and $P(B|A) = 0.6$, then $P(\bar{A} \cap B) =$
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$P(\bar{A} \cap B)$ represents the probability of "only B". Visualizing this on a Venn diagram makes the relation $P(B) - P(A \cap B)$ immediately clear.
Updated On: Jun 3, 2026
- 0.56
- 0.24
- 0.16
- 0.32
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The Correct Option is A
Solution and Explanation
Step 1: Concept
We use the conditional probability formula $P(B|A) = \frac{P(A \cap B)}{P(A)}$ and the set theory identity $P(\bar{A} \cap B) = P(B) - P(A \cap B)$.
Step 2: Meaning
We first calculate the probability of the intersection $P(A \cap B)$ and then subtract it from $P(B)$ to get the probability of $B$ occurring without $A$.
Step 3: Analysis
From the conditional probability formula: \[ P(A \cap B) = P(B|A) \cdot P(A) \] \[ P(A \cap B) = 0.6 \times 0.4 = 0.24 \] Now, compute $P(\bar{A} \cap B)$: \[ P(\bar{A} \cap B) = P(B) - P(A \cap B) \] \[ P(\bar{A} \cap B) = 0.8 - 0.24 = 0.56 \]
Step 4: Conclusion
The value of $P(\bar{A} \cap B)$ is equal to 0.56.
Final Answer: (A)
We use the conditional probability formula $P(B|A) = \frac{P(A \cap B)}{P(A)}$ and the set theory identity $P(\bar{A} \cap B) = P(B) - P(A \cap B)$.
Step 2: Meaning
We first calculate the probability of the intersection $P(A \cap B)$ and then subtract it from $P(B)$ to get the probability of $B$ occurring without $A$.
Step 3: Analysis
From the conditional probability formula: \[ P(A \cap B) = P(B|A) \cdot P(A) \] \[ P(A \cap B) = 0.6 \times 0.4 = 0.24 \] Now, compute $P(\bar{A} \cap B)$: \[ P(\bar{A} \cap B) = P(B) - P(A \cap B) \] \[ P(\bar{A} \cap B) = 0.8 - 0.24 = 0.56 \]
Step 4: Conclusion
The value of $P(\bar{A} \cap B)$ is equal to 0.56.
Final Answer: (A)
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