Question

A and B are two independent events. The probability that A and B happen simultaneously is $1/12$ and neither A nor B happens is $1/2$ then

• A. $P\left(A\right)=1/3,P\left(B\right)=1/4$
• B. $P\left(A\right)=1/2,P\left(B\right)=1/6$
• C. $P\left(A\right)=1/5,P\left(B\right)=1/3$
• D. $P\left(A\right)=1/6,P\left(B\right)=1/2$

Answer 1

The correct option is D $P\left(A\right)=1/6,P\left(B\right)=1/2$

$P(A\cap B)=\frac{1}{12}$

$P(\overline{A}\cap \overline{B})=\frac{1}{2}$

$\Rightarrow P\left(A\right)P\left(B\right)=\frac{1}{12}$

$\Rightarrow P\left(\overline{A\cup B}\right)=\frac{1}{2}$

$\Rightarrow P(A\cup B)=1-\frac{1}{2}=\frac{1}{2}$

$\therefore P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{1}{2}$

$\Rightarrow P\left(A\right)+P\left(B\right)=\frac{1}{2}+P(A\cap B)$

$\Rightarrow P\left(A\right)+P\left(B\right)=\frac{1}{2}+\frac{1}{12}=\frac{6+1}{12}=\frac{7}{12}$

$\Rightarrow P\left(B\right)=\frac{7}{12}-P\left(A\right)$

$\Rightarrow P\left(A\right)(\frac{7}{12}-P\left(A\right))=\frac{1}{12}$

$\Rightarrow -{\left(P\left(A\right)\right)}^2+\frac{7P\left(A\right)}{12}-\frac{1}{12}=0$

Let $\alpha =P\left(A\right)$ then the above equation becomes

$-{\alpha }^2+\frac{7}{12}\alpha -\frac{1}{12}=0$

$\Rightarrow 12{\alpha }^2-7\alpha +2=0$

$\Rightarrow 6{\alpha }^2-7\alpha +1=0$

$\Rightarrow 6\alpha (\alpha -1)-1(\alpha -1)=0$

$\Rightarrow 6\alpha =1,\alpha =1$

$\therefore P\left(A\right)=\frac{1}{6}$ and $P\left(A\right)P\left(B\right)=\frac{1}{12}$

$\Rightarrow P\left(B\right)=\frac{1}{P\left(A\right)}\times \frac{1}{12}$

$P\left(B\right)=\frac{1}{12}\times 6=\frac{1}{2}$