The probability for a contractor to get a road contract is $\frac{2}{9}$ and to get a building contract is…
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 contract is $\frac{1}{6}$, then what is the probability to get neither of these two contracts?
$\frac{7}{9}$
$\frac{4}{9}$
$\frac{7}{18}$
$\frac{4}{18}$
Solution
Let $A$ be the event of getting a road contract and $B$ be the event of getting a building contract.
$\therefore \quad P(A)=\frac{2}{9}, P(B)=\frac{5}{9}$
$P(A \cap B)=\frac{1}{6}$
$\therefore \quad P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\begin{aligned} & =\frac{2}{9}+\frac{5}{9}-\frac{1}{6} \\ & =\frac{7}{9}-\frac{1}{6}=\frac{14-3}{18}=\frac{11}{18}\end{aligned}$
$\therefore \quad P\left((A \cup B)^{\prime}\right)=1-P(A \cup B)$
$=1-\frac{11}{18}=\frac{7}{18}$
Hence, the probability to get neither of these two contracts is $7 / 18$.