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Question 9 of 20
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Assume that when human resource managers are randomly selected, $51 \%$ say job applicants should follow up within two weeks. If 7 human resource managers are randomly selected, find the probability that exactly 3 of them say job applicants should follow up within two weeks.
The probability is $\square$.
(Round to four decimal places as needed.)
Answer
So, the probability that exactly 3 out of 7 randomly selected human resource managers say job applicants should follow up within two weeks is approximately 0.2678, which can be written as \(\boxed{0.2678}\).
Step 1 ：First, we calculate the combination of 7 items taken 3 at a time, denoted as \(C(7, 3)\). This is calculated as \(\frac{7!}{3!(7-3)!}\), which equals 35.
Step 2 ：Next, we calculate \(0.51^3\), which equals 0.132651.
Step 3 ：Then, we calculate \((1-0.51)^{7-3}\), which equals \(0.49^4\), and this equals 0.057646.
Step 4 ：Finally, we multiply these three values together to get the probability. So, \(P(X=3) = 35 * 0.132651 * 0.057646\), which equals 0.2678.
Step 5 ：So, the probability that exactly 3 out of 7 randomly selected human resource managers say job applicants should follow up within two weeks is approximately 0.2678, which can be written as \(\boxed{0.2678}\).
