A committee consists of 10 women and 7 men. Three members are chosen as officers. What is the probability that all three officers are women? What is the probability that one of the officers chosen is a woman and the other two are men? What is the probability that at lease one officer is a man?

Using the hypergeometric distribution, it is found that there is a:

0.1765 = 17.65% probability that all three officers are women.

0.3088 = 30.88% probability that one of the officers chosen is a woman and the other two are men.

0.8235 = 82.35% probability that at lease one officer is a man.

The members are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The parameters are:

In this problem:

The probability that all are women is P(X = 3), thus:

0.1765 = 17.65% probability that all three officers are women.

From this, also 1 - 0.1765 = 0.8235 = 82.35% probability that at lease one officer is a man.

The probability that one of the officers chosen is a woman and the other two are men is P(X = 1), thus:

0.3088 = 30.88% probability that one of the officers chosen is a woman and the other two are men.

A similar problem is given at link

Step-by-step explanation:

1.

2.

3.

so possibility can be 1 man and two women,

two men and 1 women,

three men.

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation:

2

step-by-step explanation: