Question

If A and B are two events associated with a random experiment, then
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
If A, B, C are three events associated with a random experiment, then
$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$

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 $\frac{2}{3}$, what is the probability that he will get both?

Answer 1

Probability that the contractor will get an electric contract, $P\left(A\right)=\frac{2}{5}\dots .\left(1\right)$

Probability that the contractor will get a plumbing contract, $P\left(B\right)=\frac{4}{7}\dots .\left(2\right)$

Probability of getting at least one contract is $=\frac{2}{3}$

$P(A\cup B)=\frac{2}{3}$ $i.e.,P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{2}{3}\dots .\left(3\right)$

$P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)$

$\therefore P(A\cap B)=\frac{2}{5}+\frac{4}{7}-\frac{2}{3}$

$=\frac{32}{105}$