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)

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)