Question

The probability that a person will get an electrification contract is $\left(\frac{2}{5}\right)$ and the probability that he will not get a plumbing contract is $\frac{4}{7}.$ If the probability of getting at least one contract is $\left(\frac{2}{3}\right)$, what is the probability that he will get both ?

Answer 1

$p\left(E_1\right)=\frac{2}{5},P\left(not{E}_2\right)=\frac{4}{7}\Rightarrow p\left(E_2\right)=(1-\frac{4}{7})=\frac{3}{7}$

And, $p(E_1\cup E_2)=p\left(E_{1}or{E}_2\right)=\frac{2}{3}$

Thus required probability = $p(E_1\cap E_2)=P\left(E_1\right)+P\left(E_2\right)-P(E_1\cup E_2)=\frac{2}{5}+\frac{3}{7}-\frac{2}{3}$

= $\frac{17}{105}$