Question

Let A, B & C be $3$ arbitrary events defined on a sample space 'S' and if, $P\left(A\right)+P\left(B\right)+P\left(C\right)=p_1,P(A\cap B)+P(B\cap C)+P(C\cap A)=p_2$ & $P(A\cap B\cap C)=p_3$, then the probability that exactly one of the three events occurs is given by?

• A. $p_1-p_2+p_3$
• B. $p_1-p_2+2p_3$
• C. $p_1-2p_2+p_3$
• D. $p_1-2p_2+3p_3$

Answer 1

The correct option is C $p_1-2p_2+p_3$

Probability that exactly one of event occurs=P(onlyA)+P(onlyB)+P(onlyC)

P(onlyA)=P(A)-P(A∩B)-P(A∩C)+P(A∩B∩C)

P(onlyB)=P(B)-P(A∩B)-P(B∩C)+P(A∩B∩C)

P(onlyC)=P(C)-P(A∩C)-P(B∩C)+P(A∩B∩C)

Probability that exactly one of event occurs=P(onlyA)+P(onlyB)+P(onlyC)

=(P(A)-P(A∩B)-P(A∩C)+P(A∩B∩C))+(P(B)-P(A∩B)-P(B∩C)+P(A∩B∩C))+(P(C)-P(A∩C)-P(B∩C)+P(A∩B∩C))

=P1-2P2+3P3