Answer to Question #303353 in Statistics and Probability for BxJ

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.