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 woman. round your answer to five decimal places.
the probability that the group consists of one man and three woman is

Explanation:

Step1: Calculate the total number of ways to choose 4 students

The total number of students is \(n = 10+7=17\). The number of ways to choose \(r = 4\) students out of \(n=17\) is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\). So, \(C(17,4)=\frac{17!}{4!(17 - 4)!}=\frac{17\times16\times15\times14}{4\times3\times2\times1}=2380\).

Step2: Calculate the number of ways to choose 1 man and 3 women

The number of ways to choose 1 man out of 10 is \(C(10,1)=\frac{10!}{1!(10 - 1)!}=10\). The number of ways to choose 3 women out of 7 is \(C(7,3)=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1}=35\). By the multiplication principle, the number of ways to choose 1 man and 3 women is \(C(10,1)\times C(7,3)=10\times35 = 350\).

Step3: Calculate the probability

The probability \(P\) is the number of favorable outcomes divided by the number of total outcomes. So, \(P=\frac{350}{2380}\approx0.14706\).

Answer:

\(0.14706\)