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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).

Question:

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?

Solution:

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 $=\frac{17}{105}$

 

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