Question

A and B are two independent events. The probability that both A and B occur is $\frac{1}{6}$ and the probability that neither of them occur is $\frac{1}{3}$.Then P(A) is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{5}{6}$
• D. $\frac{1}{2}$ or $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{2}$ or $\frac{1}{3}$

$P(A\cup B)=\frac{2}{3}$

$P(A\cap B)=\frac{1}{6}=P\left(A\right)P\left(B\right)$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$\frac{2}{3}=P\left(A\right)+P\left(B\right)-\frac{1}{6}$

$P\left(A\right)+P\left(B\right)=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$

$P\left(B\right)=\frac{5}{6}-P\left(A\right)$

$\frac{1}{6}=P\left(A\right)(\frac{5}{6}-P(A\left)\right)$

$5P-6P^2=1$

$6P^2-5P+1=0$

$6P^2-3P-2P+1=0$

$3P(2P-1)-(2P-1)=0$

$P=\frac{1}{2},\frac{1}{3}$