The probability that a person will get an electric contact is `2/5` and the probability that he will not get plumbing contract is `4/7`. If the probability of getting at least one contract is `2/3`, what is the probability that he will get both.

Explanation

`p(E_{1})=frac{2}{5}, P(text { not } E_{2})=frac{4}{7} Rightarrow p(E_{2})=(1-frac{4}{7})=frac{3}{7}`

And, `p(E_{1} cup E_{2})=p(E_{1}.` or `.E_{2})=frac{2}{3}`

Thus required probability `=p(E_{1} cap E_{2})=P(E_{1})+P(E_{2})-P(E_{1} cup E_{2})=frac{2}{5}+frac{3}{7}-frac{2}{3}=frac{17}{105}`