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 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 0.34286.
part 4 of 4
(a) one man and two women will be selected. round your answer to five decimal places.
the probability that one man and two women will be selected is □.

Explanation:

Step1: Calculate combination

Use combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\), where \(n = 7\) (total executives), \(k = 3\) (number of people selected).
Total number of ways to select 3 people from 7 is \(C(7,3)=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1}=35\).
Number of ways to select 1 man and 2 women: \(C(3,1)\times C(4,2)\).
\(C(3,1)=\frac{3!}{1!(3 - 1)!}=3\), \(C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4\times3}{2\times1}=6\).
So \(C(3,1)\times C(4,2)=3\times6 = 18\).

Step2: Calculate probability

Probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).
\(P=\frac{18}{35}\approx0.51429\).

Answer:

\(0.51429\)