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GO Classes DPP | GATE CS | Probability | Conditional Probability

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4 4 votes

In a probability space, it is given that $P(A)=a$, $P(B)=b$, and $P(B \mid A^c)=e$. Also, $P(A \mid B^c)=f$. Determine $P(A \mid B)$ and $P(B \mid A)$ in terms of $a$, $b$, $e$, and $f$.

  1. $P(A \mid B)=\frac{a-(1-b)f}{b}, \quad P(B \mid A)=\frac{b-(1-a)e}{a}$
     
  2. $P(A \mid B)=\frac{a+(1-b)f}{b}, \quad P(B \mid A)=\frac{b-(1-a)e}{a}$
     
  3. $P(A \mid B)=\frac{a-(1-b)f}{b}, \quad P(B \mid A)=\frac{b+(1-a)e}{a}$
     
  4. $P(A \mid B)=\frac{a+(1-b)f}{b}, \quad P(B \mid A)=\frac{b+(1-a)e}{a}$

1 Answer

1 1 vote

First find $P(B \mid A)$:

Using total probability with respect to $A$ and $A^c$,
$P(B)=P(B \mid A)P(A)+P(B \mid A^c)P(A^c)$.

$\Rightarrow b=P(B \mid A)\cdot a+e(1-a)$.

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


Now find $P(A \mid B)$:

Using total probability with respect to $B$ and $B^c$,
$P(A)=P(A \mid B)P(B)+P(A \mid B^c)P(B^c)$.

$\Rightarrow a=P(A \mid B)\cdot b+f(1-b)$.

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


$\boxed{\therefore P(A \mid B)=\frac{a-f(1-b)}{b} \qquad \& \qquad P(B \mid A)=\frac{b-e(1-a)}{a}}$

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