Question

Fill in the blanks in following table:

	P(A)	P(B)	P(A$\cap$B)	P(A$\cup$B)
(i)	$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	......
(ii)	0.35	.....	0.25	0.6
(iii)	0.5	0.35	....	0.7

Answer 1

(i) Here $P\left(A\right)=\frac{1}{3}$, $P\left(B\right)=\frac{1}{5}$, $P(A\cap B)=\frac{1}{15}$
We know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\therefore P(A\cup B)=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}=\frac{5+3-1}{15}=\frac{7}{15}$
(ii) Here $P\left(A\right)=0.35$, $P(A\cap B)=0.25,P(A\cup B)=0.6$
We know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\therefore 0.6=0.35+P\left(B\right)-0.25$
$\Rightarrow P\left(B\right)=0.6-0.35+0.25$
$\Rightarrow P\left(B\right)=0.5$
(iii) Here $P\left(A\right)=0.5,P\left(B\right)=0.35$, $P(A\cup B)=0.7$
We know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\therefore 0.7=0.5+0.35-P(A\cap B)$
$\Rightarrow P(A\cap B)=0.5+0.35-0.7$
$\Rightarrow P(A\cap B)=0.15$