Question

For three events $A,B$ and $C,P$ (exactly one of the events $A$ occur) $=$ P (exactly one of the events B and C occur) $=$ P (exactly one of the events C or A occurs) $=$ p and P (all the three events occur simultaneously) $=p^2$, where $0<p<1/2$. If the probability of at least one of the three events $A,B$ and $C$ occurs is $11/18$, the value of $p$ is

• A. $1/6$
• B. $1/4$
• C. $1/5$
• D. $1/3$

Answer 1

The correct option is D $1/3$

P (exactly one of A or B occurs)
$=P\left(A\right)+P\left(B\right)-2P(A\cap B)$
Therefore, $P\left(A\right)+P\left(B\right)-2P(A\cap B)=p$ (1)
Similarly, $P\left(B\right)+P\left(C\right)-2P(B\cap C)=p$ (2)
and $P\left(C\right)+P\left(A\right)-2P(C\cap A)=p$ (3)
Adding (1), (2) and (3) and dividing by 2
$\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)=3p/2$ (4)
Now, P (at least one of A, B and C)
$=\frac{3p}{2}+p^2=\frac{3p+2p^2}{2}=\frac{11}{18}$ (given)
$\Rightarrow 2p^2+3p-\frac{11}{9}=0\therefore p=\frac{-3+\sqrt{9+\frac{88}{9}}}{4}=\frac{1}{3}$