Question

in a company there are seven executives: four women and three men. three are selected at random to attend a management seminar. find these probabilities.
part 1 of 4
(a) all three selected will be women. round your answer to five decimal places.
the probability that all three people selected will be women is 0.11429.
part 2 of 4
(b) all three selected will be men. round your answer to five decimal places.
the probability that all three selected people will be men is 0.02857.
part: 2 / 4
part 3 of 4
(c) two men and one woman will be selected. round your answer to five decimal places.
the probability that two men and one woman will be selected is

Explanation:

Step1: Calculate the combination of selecting 2 men and 1 woman

The number of ways to choose 2 men out of 3 is \(C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=3\).
The number of ways to choose 1 woman out of 4 is \(C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4\).
The number of ways to choose 3 people out of 7 is \(C(7,3)=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1}=35\).

Step2: Calculate the probability

The probability \(P=\frac{C(3,2)\times C(4,1)}{C(7,3)}=\frac{3\times4}{35}=\frac{12}{35}\approx0.34286\)

Answer:

\(0.34286\)