Question

A is interviewed for $3$ posts. There are $3$ candidates for post $1,4$ for second post and $2$ for post No. three. The probability of A's being selected for at least one post is:

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{3}{4}$

Let $P\left(A\right)=$ Probability of getting first company by the candidates.

Similarly, $P\left(B\right)$ $\xi$ $P\left(C\right)$ be the probability of getting into second and third company respectively.

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

$\Rightarrow P\left(A\right)=\frac{1}{3}$ $P\left(B\right)=\frac{1}{4}$ $P\left(C\right)=\frac{1}{2}$

$\Rightarrow P(A\cup B\cup C)=\frac{1}{3}+\frac{1}{4}+\frac{1}{2}-(\frac{1}{3}\times \frac{1}{4})-(\frac{1}{3}\times \frac{1}{2})-(\frac{1}{4}\times \frac{1}{2})+(\frac{1}{3}\times \frac{1}{4}\times \frac{1}{2})$

$=\frac{1}{3}+\frac{1}{4}+\frac{1}{2}-\frac{1}{12}-\frac{1}{6}-\frac{1}{8}+\frac{1}{24}$

$=\frac{8+6+12-2-4-3+1}{24}$

$=\frac{3}{4}.$

Hence, the answer is $\frac{3}{4}.$