Question

in a class of 17 students, there are 10 men and 7 women. four students are selected to present a demonstration on the use of graphing calculators. find the probability that the group consists of
part 1 of 3
(a) all men. round your answer to five decimal places.
the probability that the group consists of all men is 0.08824
part 2 of 3
(b) all women. round your answer to five decimal places.
the probability that the group consists of all women is 0.01471
part 2 / 3
part 3 of 3
(c) three men and one woman. round your answer to five decimal places.
the probability that the group consists of three men and one woman is

Explanation:

Step1: Calculate the combination values

The formula for combination is \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n\) is the total number of items, and \(r\) is the number of items to be chosen.
The total number of ways to choose \(4\) students out of \(17\) is \(C(17,4)=\frac{17!}{4!(17 - 4)!}=\frac{17\times16\times15\times14}{4\times3\times2\times1}=2380\).
The number of ways to choose \(3\) men out of \(10\) is \(C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1} = 120\).
The number of ways to choose \(1\) woman out of \(7\) is \(C(7,1)=\frac{7!}{1!(7 - 1)!}=7\).

Step2: Calculate the probability

The probability \(P\) that the group consists of three men and one woman is given by the formula \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).
The number of favorable outcomes is \(C(10,3)\times C(7,1)=120\times7 = 840\).
So \(P=\frac{840}{2380}\approx0.35294\)

Answer:

\(0.35294\)