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A company hires management trainees for entry level sales positions. Past experience indicates that only 29% will still be employed at the end of nine months. Assume the company recently hired 9 trainees. Assuming independence, what is the probability that exactly five of the trainees will still be employed at the end of nine months?

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Answer:

0.0656 is the probability that exactly five of the trainees will still be employed at the end of nine months.

Explanation:

We are given the following information:

We treat management trainees still employed at the end of nine months as a success.

P(Employed) = 29% = 0.29

Then the number of trainees follows a binomial distribution, where


P(X=x) = \binom{n}{x}.p^x.(1-p)^(n-x)

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 9

P(exactly five of the trainees will still be employed at the end of nine months)

We have to evaluate:


P(x =5)\\\\= \binom{9}{5}(0.29)^5(1-0.29)^4\\\\= 0.0656

0.0656 is the probability that exactly five of the trainees will still be employed at the end of nine months.

User GJohannes
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