Question

If A and B are mutually exclusive events such that P(A = 0.35 and P(B=0.45, find

(i) $P(A\cup B)$

(ii) $P(A\cap B)$

(iii) $P(A\cap \overline{B})$

(iv) $P(\overline{A}\cap \overline{B})$

Answer 1

Since, it is given that, A and B are mutually exclusive events.

$\therefore P(A\cap B)=0[\because A\cap B=\varphi ]P\left(A\right)=0.35,P\left(B\right)=0.45$

(i) $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

=0.35+0.45-0=0.80

(ii) $P(A\cap B)=0$

(iii) $P(A\cap \overline{B})=P\left(A\right)-P(A\cap B)$ =0.35-0=0.35

(iv) $P(\overline{A}\cap \overline{B})=P(A\cup B{)}^{‘}=1-P(A\cup B)$ =1-0.8=0.2