Let $A$ and $B$ be two events such that $P\left(A \cap B^{c}\right) = 0.20, P\left(A^{c} \cap B\right) = 0.15, P\left(A^{c} \cap B^{c}\right) = 0.1$, then $p(A/B)$ is equal to,
Step-by-Step Solution
Key Concept: Partition the sample space into four disjoint regions: A∩B, A∩B^c, A^c∩B, and A^c∩B^c. Use P(A∩B^c) + P(A^c∩B) + P(A^c∩B^c) + P(A∩B) = 1 to find P(A∩B), then apply the conditional probability formula P(A/B) = P(A∩B)/P(B).
Given $P(A \cap B') = P(A) - P(A \cap B) = 0.20$ and $P(A' \cap B) = P(B) - P(A \cap B) = 0.15$. Adding these equations yields $P(A) + P(B) - 2P(A \cap B) = 0.35$. Using $P(A' \cap B') = 1 - P(A \cup B) = 1 - P(A) - P(B) + P(A \cap B) = 0.1$, we get $P(A) + P(B) - P(A \cap B) = 0.9$, so $P(A \cap B) = 0.55$. Given $P(A) = 0.75$ and $P(B) = 0.70$, we find $P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.55}{0.70}$.
Correct Answer: 1