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GO Classes DPP | GATE DA | Probability | Conditional Probability

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1 1 vote

In a probabilistic model, suppose $P(A)=a$, $P(B)=b$, and $P(B^c \mid A^c)=e$, where $0<a<1$ and $0<b<1$. Determine $P(A \mid B)$ in terms of $a$, $b$, and $e$.

  1. $\frac{b-(1-e)(1-a)}{a}$
     
  2. $\frac{b-(1-e)(1-a)}{b}$
     
  3. $\frac{b+(1-e)(1-a)}{b}$
     
  4. $\frac{a-(1-e)(1-b)}{b}$

2 Answers

2 2 votes

We are given $P(B^c \mid A^c)=e$. By the definition of conditional probability,

$P(B^c \cap A^c)=P(B^c \mid A^c)P(A^c)=e(1-a)$.

Now inside $A^c$, the event $B$ is the complement of $B^c$, so

$P(B \cap A^c)=P(A^c)-P(B^c \cap A^c)$

$=(1-a)-e(1-a)$

$=(1-e)(1-a)$.

Next, split $B$ into the disjoint events $(A \cap B)$ and $(A^c \cap B)$. Then

$P(B)=P(A \cap B)+P(A^c \cap B)$,

so

$P(A \cap B)=b-(1-e)(1-a)$.

Therefore,

$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{b-(1-e)(1-a)}{b}$.

So the correct answer is $\boxed{\frac{b-(1-e)(1-a)}{b}}$.

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