Question

For 3 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 & P (all the three events occurs simultaneously) = $p^2$. Where $o<p<\frac{1}{2}$. then the probability of at lest one of three events, A, B & C occurring is

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

Answer 1

The correct option is A $\frac{3p+2p^2}{2}$

$P\left(A\right)-P\left(B\right)-2P(A\cap B)=p$ ...(i)
$P\left(B\right)+P\left(C\right)-2P(B\cap C)=p$ ...(ii)
$P\left(C\right)+P\left(A\right)-2P(C\cap A)=p$ ...(iii)
and, $P(A\cap B\cap C)=p^2$ ...(iv)

Adding (i), (ii) and (iii), we get
$2\left[P\right(A)+P(B)+P(C)+P(A\cap B)-P(B\cap C)-P(A\cap C\left)\right]=3p$
$\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)=\frac{3p}{2}$

$\therefore$ Required probability

$=P(A\cup B\cup C)$
$=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$
$=\frac{3p}{2}+p^2=\frac{3p+2p^2}{2}$