Question

For the three events A, B and C, P (exactly one of the events A or B occurs) = P(exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) $=p$ and P (all the three events occur simultaneously) = $p^2$, where $0<p<\frac{1}{2}$. Then the probability of at least one of the three events A, B and C occurring is

• A. $\frac{(3p+2p^2)}{2}$
• B. $\frac{(p+3p^2)}{4}$
• C. $\frac{(p+3p^2)}{2}$
• D. $\frac{(3p+2p^2)}{4}$

Answer 1

The correct option is A

$\frac{(3p+2p^2)}{2}$

Explanation for the correct options:

Step1: Probability that exactly one of events :

P (exactly one of the events A or B occurs)$=P\left(A\right)+P\left(B\right)–2P(A\cap B)=p\dots \left(1\right)$

Likewise, $P\left(B\right)+P\left(C\right)–2P(B\cap C)=p\dots \left(2\right)$

$P\left(C\right)+P\left(A\right)–2P(C\cap A)=p\dots \left(3\right)$

Adding the above three equations, we get

$P\left(A\right)+P\left(B\right)+P\left(C\right)–P(A\cap B)-P(B\cap C)–P(C\cap A)=\frac{3p}{2}\dots .\left(4\right)$

Step2:Simplifying and substituting the equations:

Also given that$P(A\cap B\cap C)=p^2$

So, the probability that at least one of the event occurs $=P\left(A\right)+P\left(B\right)+P\left(C\right)–P(A\cap B)-P(B\cap C)–P(C\cap A)+P(A\cap B\cap C)$

$=\frac{3p}{2}+p^2$

$=\frac{(3p+2p^2)}{2}$

Hence, Option(A) is correct.