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 5
(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 5
(b) all women. round your answer to five decimal places.
the probability that the group consists of all women is 0.01471.
part 3 of 5
(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 0.35294.
part 4 of 5
(d) one man and three women. round your answer to five decimal places.
the probability that the group consists of one man and three women is 0.14706.
part 5 of 5
(a) two men and two woman. round your answer to five decimal places.
the probability that the group consists of two men and two woman is □.

Explanation:

Step1: Calculate the combination values

The formula for combination is \(C(n,k)=\frac{n!}{k!(n - k)!}\), where \(n\) is the total number of items, and \(k\) is the number of items to be chosen.
The total number of students is \(n=10 + 7=17\), and we are choosing \(r = 4\) students. So, \(C(17,4)=\frac{17!}{4!(17-4)!}=\frac{17\times16\times15\times14}{4\times3\times2\times1}=2380\)
The number of ways to choose \(2\) men from \(10\) men is \(C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1} = 45\)
The number of ways to choose \(2\) women from \(7\) women is \(C(7,2)=\frac{7!}{2!(7-2)!}=\frac{7\times6}{2\times1}=21\)

Step2: Calculate the probability

The probability \(P\) that the group consists of two men and two women is given by the formula \(P=\frac{C(10,2)\times C(7,2)}{C(17,4)}\)
Substitute the combination values: \(P=\frac{45\times21}{2380}=\frac{945}{2380}\approx0.39706\)

Answer:

\(0.39706\)