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(C) Let $A$ be the event of getting a road contract and $B$ be the event of getting a building contract. Given: $P(A) = \frac{2}{9}$,$P(B) = \frac{5}{

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$\frac{7}{18}$

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(C) Let $A$ be the event of getting a road contract and $B$ be the event of getting a building contract. Given: $P(A) = \frac{2}{9}$,$P(B) = \frac{5}{9}$,and $P(A \cap B) = \frac{1}{6}$. We need to find the probability of getting neither contract,which is $P(A' \cap B') = P((A \cup B)')$. First,calculate the probability of getting at least one contract: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $P(A \cup B) = \frac{2}{9} + \frac{5}{9} - \frac{1}{6} = \frac{7}{9} - \frac{1}{6}$ Finding a common denominator $(18)$: $P(A \cup B) = \frac{14}{18} - \frac{3}{18} = \frac{11}{18}$. Now,the probability of getting neither contract is: $P((A \cup B)') = 1 - P(A \cup B) = 1 - \frac{11}{18} = \frac{7}{18}$. Thus,the probability is $\frac{7}{18}$.

The probability for a contractor to get a road contract is $\frac{2}{9}$ and to get a building contract is $\frac{5}{9}$. If the probability to get both the contracts is $\frac{1}{6}$,then what is the probability to get neither of these two contracts?

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