Question

Let $A,B$ and $C$ be three events such that the probability that exactly one of $A$ and $B$ occurs is $(1–k),$ the probability that exactly one of $B$ and $C$ occurs is $(1–2k),$ the probability that exactly one of $C$ and $A$ occurs is $(1–k)$ and the probability of all $A,B$ and $C$ occur simultaneously is $k^2,$ where $0<k<1.$ Then the probability that at least one of $A,B$ and $C$ occur is

• A. Greater than $\frac{1}{2}$
• B. Exactly equal to $\frac{1}{2}$
• C. Greater than $\frac{1}{8}$ but less than $\frac{1}{4}$
• D. Greater than $\frac{1}{4}$ but less than $\frac{1}{2}$

Answer 1

The correct option is A Greater than $\frac{1}{2}$

$P\left(A\right)+P\left(B\right)-2P(A\cap B)=1-k\cdots \left(i\right)$
$P\left(B\right)+P\left(C\right)-2P(B\cap C)=1-2k\cdots \left(ii\right)$
$P\left(C\right)+P\left(A\right)-2P(C\cap A)=1-k\cdots \left(iii\right)$
$\left(i\right)+\left(ii\right)+\left(iii\right)$
$\Rightarrow \sum P\left(A\right)-\sum P(A\cap B)=\frac{3-4k}{2}$
$P(A\cup B\cup C)=\sum P\left(A\right)-\sum P(A\cap B)+P(A\cap B\cap C)$
$=\frac{3-4k}{2}+k^2$
$=(k-1{)}^2+\frac{1}{2}>\frac{1}{2}$