Answer:
Option C
Explanation:
We have , $P(A\cap B)=P(A).P(B),P(B)=\frac{2}{7}, $and $ P(A\cup B^{c})=0.8$
Consider $P(A\cup B^{c})=0.8$
$\Rightarrow$ $P(A^{c}\cap B^{c})=0.8\Rightarrow1-P(A^{c}\cap B)=0.8$
$\Rightarrow P(A^{c}\cap B^{})=0.2\Rightarrow P(A^{c}).P( B)=0.2$
[ $\because$ A and B are independent events , therefore $A^{c}$ and B, are also independent]
$ \Rightarrow P(A^{c}) =0.2\times \frac{7}{2}=0.7\Rightarrow P(A)=0.3$
and So, $P(A\cap B)=P(A).(B)=\frac{3}{10}.\frac{2}{7}=\frac{6}{70}$
Now, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$=\frac{3}{10}+\frac{2}{7}-\frac{6}{70}=\frac{21+20-6}{70}=\frac{35}{70}=\frac{1}{2}$
