Question

The mailing list of an agency that markets scuba-diving trips to the Florida Keys contains 78% males and 22% females. The agency calls 29 people chosen at random from its list.
(a) What is the probability that 17 of the 29 people are men?
(b) What is the probability that the first woman is reached on the 8th call?

Answer 1

Answer:

a) 0.98% probability that 17 of the 29 people are men.

b) 3.86% probability that the first woman is reached on the 8th call

Step-by-step explanation:

For each person chosen by the agency, there are only two possible outcomes. Either it is a man, or it is a woman. The probability of selecting a man or a women in each trial is independent from other trials. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

(a) What is the probability that 17 of the 29 people are men?

This is P(X = 17) when $n = 29, p = 0.78$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 17) = C_{29,17}.(0.78)^{17}.(0.22)^{12} = 0.0098$

0.98% probability that 17 of the 29 people are men.

(b) What is the probability that the first woman is reached on the 8th call?

On the first 7 trials, all men, each with a 78% probability.

On the 8th trial, a women, with a 22% probability. So

$P = (0.78)^{7}*0.22 = 0.0386$

3.86% probability that the first woman is reached on the 8th call