The probability that a person will get an electrification contract ia (2/5) and the probability that he will not get a plumbing contract is (4/7).
The probability that a person will get an electrification contract ia (2/5) and the probability that he will not get a 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?
Let A denote the event that a person will get electrification contract and B denote the event that the person will get a plumbing contract
Given : $\mathrm{P}(\mathrm{A})=\frac{2}{5}, \mathrm{P}($ not $\mathrm{B})=\mathrm{P}\left({ }^{\bar{B}}\right)=\frac{4}{7}, \mathrm{P}(\mathrm{A}$ or $\mathrm{B})=\frac{2}{3}$
To find: Probability that he will get both electrification and plumbing contract $=\mathrm{P}(\mathrm{A}$ and B)
Formula used : $\mathrm{P}(\mathrm{B})=1-\mathrm{P}\left({ }^{\bar{B}}\right)$
P(A or B) = P(A) + P(B) - P(A and B)
$P(B)=1-\frac{4}{7}=\frac{3}{7}$
$P(B)=\frac{3}{7}$
Probability of getting at least one contract $=\frac{2}{3}$
$\frac{2}{3}=\frac{2}{5}+\frac{3}{7}-\mathrm{P}(\mathrm{A}$ and $\mathrm{B})$
$\frac{2}{3}=\frac{14+15}{35}-\mathrm{P}(\mathrm{A}$ and $\mathrm{B})$
$\mathrm{P}(\mathrm{A}$ and $\mathrm{B})=\frac{29}{35}-\frac{2}{3}=\frac{87-70}{105}=\frac{17}{105}$
$\mathrm{P}(\mathrm{A}$ and $\mathrm{B})=\frac{17}{105}$
The probability that he will get both electrification and plumbing contract $=\frac{17}{105}$