Personnel selection. Suppose that 6 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 11 finalists, what is the probability of selecting (A) 3 females and 2​ males? ​(B) 4 females and 1​ male? ​(C) 5​ females?

Personnel selection. Suppose that 6 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 11 finalists, what is the probability of selecting (A) 3 females and 2​ males? ​(B) 4 females and 1​ male? ​(C) 5​ females?

Question
Decimals
asked 2021-03-07
Personnel selection. Suppose that 6 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 11 finalists, what is the probability of selecting
(A) 3 females and 2​ males?
​(B) 4 females and 1​ male?
​(C) 5​ females?

Answers (1)

2021-03-08
Data analysis
Note : As per authoring guidelines, when multiple subparts are posted, then only first three subparts are to be answered.
Given there is selection for a Personnel.
And there are 11 finalists in which 6 are female and 5 are female.
And there are only 5 posts vacant.
To find the probability of selecting as below.
Total ways
Total number of ways of selecting 5 people from 11 finalists is given by,
\(11P5\)
\(= 11\ \times\ 10\ \times\ 9\ \times\ 8\ \times 7\)
\(= 55.440\)
Probability is given by the ratio of number of favourable ways to the total number of ways.
Note :
\(nPr=n\ \frac{!}{n\ -\ r}!\)
A) 3 females and 2​ males
3 females selected from 6 female finalists in \(6P3\) ways
\(= 6\ \times\ 5\ \times 4\)
\(= 120\)
2 males selected from 5 male finalists in \(5P2\) ways
\(= 5\ \times\ 4\)
\(= 20\)
Favourable ways \(= 120\ \times\ 20 = 2400\)
Probability \(= 2400/55.440\)
\(= 0.0432900433\)
Hence probability of selecting 3 females and 2 males is 0.04329
(Rounded to 5 decimals)
B) 4 females and 1 males
4 females selected from 6 female finalists in \(6P4\) ways
\(= 6\ \times\ 5\ \times\ 4\ \times 3\)
\(= 360\)
1 male selected from 5 male finalists in \(5P1\) ways
\(= 5\)
Favourable ways \(= 360\ \times\ 5 = 18000\)
Probability \(= 1800/55.440\)
\(= 0.0324675325\)
Hence probability of selecting 4 females and 1 male is 0.0324675
(Rounded to 7 decimals)
C) 5 females
5 females selected from 6 female finalists in \(6P5\) ways
\(= 6\ \times\ 5\ \times\ 4\ \times\ 3\ \times\ 2\)
\(= 720\)
Favourable ways = 720
Probability \(= 720/55.440\)
\(= 0.012987013\)
Hence probability of selecting 5 females is 0.012987
(Rounded to 6 decimals)
0

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