Question

An $MBM$ applies for a job in two firms $X$ and $Y$. The probability of his being selected in firm $X$ is $0.7$ and being rejected at $Y$ is $0.5$. The probability of at least one of his applications being rejected is $0.6$. The probability that he will be selected in one of the firms, is

• A. $0.6$
• B. $0.4$
• C. $0.8$
• D. None of these

Answer 1

The correct option is C $0.8$

Let $A$ and $B$ denote the events that the person is selected in firms $X$ and $Y$ respectively.
Then in the usual notations, we are given:
$P\left(A\right)=0.7\Rightarrow P\left(\overline{A}\right)=1-0.7=0.3P\left(\overline{B}\right)=0.5\Rightarrow P\left(B\right)=1-0.5=0.5$
and $P(\overline{A}\cup \overline{B})=0.6$
The probability that the persons will be selected in one of the two firms $X$ or $Y$ is given by:
$P(A\cup B)=1-P(\overline{A}\cap \overline{B})=1-(P\left(\overline{A}\right)+P\left(\overline{B}\right)-P(\overline{A}\cup \overline{B}))=1-(0.3+0.5-0.6)=0.8$