Question

Six women and five men walk into an

appliance store at the same time. There are five

salesperson available to help them.Find the

probability that a salesperson will first help at

least 2 women.

Answer 1

Number of men = 5

Number of women = 6

Number of salesperson =5

The total number of men and women will be 5+6 =11

Then

$$11C5 =462$$

Let Y be the probability of women being served first.

The probability that a sales person will first help atleast two women will be given as ;

$P (Y ≥ 2) = 1- [P(Y=0)+P(Y=1)]$

Where;

$P(Y=0) =6C0 *5C5/11C5$

$= 1/462$

$P(Y=1) = 6C1 *5C4/11C5$

$= 30/462$

$$\therefore P(Y ≥ 2) =1-[1/462+30/462]$$

$=431/462$

$= 0.932900432$

The probability that a salesperson will first help atleast two women is approximately 0.9329.