Question

$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $1/6$ and probability that neither of them occurs is $1/3.$ Then, the probability of the two events are respectively:

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

Answer 1

The correct option is A $\frac{1}{2},\frac{1}{3}.$

Suppose $P\left(A\right)=a,P\left(B\right)=b$ then

Given that:

$a\times b=\frac{1}{6}$ .....(1)

Or

$b=\frac{1}{6a}$

$(1-a)\times (1-b)=\frac{1}{3}$ .....(2)

$(1-a-b+ab)=\frac{1}{3}$

$(1-a-b+\frac{1}{6})=\frac{1}{3}$

$(-a-b+\frac{7}{6})=\frac{1}{3}$

$a+b=\frac{7}{6}-\frac{1}{3}$

$a+\frac{1}{6a}-\frac{5}{6}=0$ (After substituting value of $b$)

$6a^2-5a+1=0$ (After taking $LCM$)

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

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

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

Hence, $P\left(A\right)=a=\frac{1}{3}or\frac{1}{2}$ and $P\left(B\right)=b=\frac{1}{2}or\frac{1}{3}.$