Question

Let $A,B,C$ be $3$ arbitrary events defined on a sample space $S$ and if, $P\left(A\right)+P\left(B\right)+P\left(C\right)=p_1,P(A\cap B)+P(B\cap C)+P(C\cap A)=p_2$ and $P(A\cap B\cap C)=p_3$, then 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 D $p_1-2p_2+3p_3$

Given $P\left(A\right)+P\left(B\right)+P\left(C\right)=p_1$

$\Rightarrow P(A\cap B)+P(B\cap C)+P(C\cap A)=p_2$

$\Rightarrow P(A\cap B\cap C)=p_3$

$\Rightarrow P$ ( Exactly are event $A$ occurs between $A,BandC$ ) $=p_1$

$P$ ( Exactly are event B occurs between $A,BandC$ ) $=p_2$

$P$ ( Exactly are event C occurs between $A,BandC$ ) $=p_3$

$\Rightarrow P\left(A\right)+P\left(B\right)-2P(A\cap B)=p_1$

$P\left(B\right)+P\left(C\right)-2P(B\cap C)=p_2$

$P\left(C\right)+P\left(A\right)-2P(A\cap C)=p_3$

$\Rightarrow 2p_1+2p_2+2p_3-2p_2=p_1-p_2-p_3$

$2p_1+2p_2+2p_3-2p_2-p_1+p_2+p_3$

$p_1-2p_2+3p_3$

Hence, the answer is $p_1-2p_2+3p_3.$