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

We know that,
P(exactly one of A or B occurs)
$=P\left(A\right)+P\left(B\right)-2P(A\cap B)\therefore P\left(A\right)+P\left(B\right)-2P(A\cap B)=p\cdots \left(i\right)P\left(B\right)+P\left(C\right)-2P(B\cap C)=p\cdots \left(ii\right)P\left(C\right)+P\left(A\right)-2P(C\cap A)=p\cdots \left(iii\right)$
On adding Eqs. (i), (ii) and (iii), we get
$2\left[P\right(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A\left)\right]=3p\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)=\frac{3p}{2}\cdots \left(iv\right)$
$GiventhatP(A\cap B\cap C)=p^2\cdots \left(v\right)\therefore$ ~P(at least one of the events A, B, and C occurs)
$=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$
$=\frac{3p}{2}+p^2$ [from Eqs. (iv) and (v)]
$=\frac{3p+2p^2}{2}$