Question

Let A and B be two events such that $P(A\cap B^c)=0.20,P(A^c\cap B)=0.15,P(A^c\cap B^c)=0.1$, then $P\left(A/B\right)$ is equal to

• A. $11/14$
• B. $2/11$
• C. $2/7$
• D. $1/7$

Answer 1

The correct option is A $11/14$

$\because P(A\cup B)=1-P(A^c\cap B^c)\Rightarrow P(A\cup B)=1-0.1=0.9$
We know that

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

Adding the above two equations
$P(A\cap B^c)+P(A^c\cap B)=P\left(A\right)-P(A\cap B)+P\left(B\right)-P(A\cap B)\Rightarrow 0.2+0.15=P\left(A\right)+P\left(B\right)-2P(A\cap B)\Rightarrow P\left(A\right)+P\left(B\right)-P(A\cap B)-P(A\cap B)=0.35\Rightarrow P(A\cup B)-P(A\cap B)=0.35$

putting the value of $P(A\cup B)$obtained earlier in the above result
$\Rightarrow P(A\cap B)=0.9-0.35=0.55AlsoP\left(B\right)=P(A^c\cap B)+P(A\cap B)=0.15+0.55=0.70\Rightarrow P\left(B\right)=0.70\because P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}$
hence putting the values
$P\left(A/B\right)=\frac{0.55}{0.70}=\frac{11}{14}$