Question

Suppose $E_1,E_2,E_3$ be three mutually exclusive events such that $P\left(E_i\right)=p_i$, for $i=1,2,3$,

then $P(E_1\cap E_2^{\prime })+P(E_2\cap E_3^{\prime })+P(E_3\cap E_1^{\prime })$ equals,

• A. $p_1(1-p_2)+p_2(1-p_3)+p_3(1-p_1)$
• B. $p_1{p}_2+p_2{p}_3+p_3{p}_1$
• C. $p_1+p_2+p_3$
• D. None of these

Answer 1

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

we have

$P(E_1\cap E_2^{\prime })+P(E_2\cap E_3^{\prime })+P(E_3\cap E_1^{\prime })$

since events are mutually exclusive so

we use $E_1\cap E_2^{\prime }=E_1$

$E_3\cap E_1^{\prime }=E_3$

$E_2\cap E_3^{\prime }=E_2$

so equation becomes

$E_1+E_2+E_3=(E_1\cap E_2^{\prime })+(E_3\cap E_1^{\prime })+(E_2\cap E_3^{\prime })$

so $p_1+p_2+p_3$

answer is option $C$.