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.

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step-by-step explanation: