Question

Let $A,B,C$ be three events. If the probability of occurring exactly one event out of $A$ and $B$ is $1-a$, out of $B$ and $C$ and $A$ is $1-a$ and that of occurring three events simultaneously is $a^2$, then the probability that at least one out of $A$, $B$, $C$ will occur is

• A. $1/2$
• B. Greater than $1/2$
• C. Less than $1/2$
• D. Greater than $3/4$

Answer 1

Answer: (B)

$P$ (exactly one event out of $A$ and $B$ occurs)
$=P\left[A\cap B^{\prime }\right)\cup \left(A^{\prime }\cap B\right)]$

$=P(A\cup B)-P(A\cap B)=P(A)+P(B)-2P(A\cap B)$
$\therefore P(A)+P(B)-2P(A\cap B)=1-a...(1)$
Similarly, $P(B)+P(C)-2P(B\cap C)=1-2a...(2)$
$P(C)+P(A)-2P(C\cap A)=1-a...(3)$
$P(A\cap B\cap C)=a^2...(4)$
Now $P(A\cup B\cup C)$
$=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)$
$-P(C\cap A)+P(A\cap B\cap C)$
$=\frac{1}{2}[P(A)+P(B)-2P(B\cap C)+P(B)+P(C)$
$-2P(B\cap C)+P(C)+P(A)-2P(C\cap A)]+P(A\cap B\cap C)$
$=\frac{1}{2}[1-a+1-2a+1-a]+a^2$[using (1), (2), (3) and (4)]
$=\frac{3}{2}-2a+a^2=\frac{1}{2}+(a-1{)}^2>\frac{1}{2}$