Question

question 13
4 pts
a math teacher chooses 4 students at random from a class of 17 to solve a problem at the board. if 10 students in the entire class know how to solve the problem, what is the probability that of the 4 students who were picked, only 1 know(s) how to solve the problem?
note: for any final answer that has up to four decimal places, enter your answer without rounding the number. for any answers with more than four decimal values, round your final answer to four decimal places.
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Explanation:

Step1: Calculate the number of ways to choose 1 student who knows the problem

The number of ways to choose 1 student out of 10 who know the problem is given by the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\), where \(n = 10\) and \(k=1\).

$$C(10,1)=\frac{10!}{1!(10 - 1)!}=\frac{10!}{1!9!}=10$$

Step2: Calculate the number of ways to choose 3 students who do not know the problem

The number of students who do not know the problem is \(17-10 = 7\). The number of ways to choose 3 students out of 7 is \(C(7,3)=\frac{7!}{3!(7 - 3)!}\)

$$C(7,3)=\frac{7!}{3!4!}=\frac{7\times6\times5}{3\times2\times1}=35$$

Step3: Calculate the number of ways to choose 4 students out of 17

Using the combination formula \(C(17,4)=\frac{17!}{4!(17 - 4)!}\)

$$C(17,4)=\frac{17!}{4!13!}=\frac{17\times16\times15\times14}{4\times3\times2\times1}=2380$$

Step4: Calculate the probability

The probability \(P\) is the product of the number of ways to choose 1 student who knows and 3 who do not, divided by the number of ways to choose 4 students.

$$P=\frac{C(10,1)\times C(7,3)}{C(17,4)}=\frac{10\times35}{2380}=\frac{350}{2380}\approx0.1471$$

Answer:

\(0.1471\)