The probability that a person will get an electric contract is $$25$$ and the probability that he will not get plumbingcontract is $$47$$.If the probability of getting at least one contract is $$23$$, the probability that he will get both the contracts is k , then $$105k$$ is

$$P(E)=25$$;$$P(F)=P($$ plumbing $$)=1$$−$$47=37P(E$$∪$$F)=P(E)+P(F)$$−$$P(E$$∩$$F)23=25+37$$−k⇒$$k=17105$$

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The probability that a person will get an electric contract is $\frac{2}{5}$ and the probability that he will not get plumbing
contract is $\frac{4}{7}$ .If the probability of getting at least one contract is $\frac{2}{3}$ , the probability that he will get both the contracts is k , then 105k is

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Solution

$\begin{array}{l} P (E) = \frac{2}{5}; P (F) = P (plumbing) = 1 - \frac{4}{7} = \frac{3}{7} \\ P (E \cup F) = P (E) + P (F) - P (E \cap F) \\ \frac{2}{3} = \frac{2}{5} + \frac{3}{7} - k \Rightarrow k = \frac{17}{105} \end{array}$

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