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

Explanation:

Step1: Calculate the total number of executives

There are \(4 + 3=7\) executives in total.

Step2: Use the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\)

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

Step3: Calculate the probability

The probability \(P=\frac{C(3,3)}{C(7,3)}=\frac{1}{35}\approx0.02857\)

Answer:

\(0.02857\)