Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?

(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?

In (a)

Apply the formula

[tex]\frac{n!}{(n-r)!(r!)}[/tex]

where n=12 and r=5

substitute values

[tex]=\frac{12!}{(12-5)!(5!)} \\\\\\=\frac{12!}{(7!)(5!)} \\\\\\=\frac{12*11*10*9*8}{5*4*3*2*1} \\\\\\=\frac{95040}{120} \\\\\\=792[/tex]

In (b)

If Jerry did only 5 problems of one assignment then the probability will be

[tex]\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}[/tex]

joaobezerra joaobezerra · Answer 2 · 2022-03-08T11:41:23+01:00

Using the hypergeometric distribution, it is found that there is a 0.0013 = 0.13% probability that the problems he did comprised the group that was selected to be graded.

The problems are chosen without replacement, hence the hypergeometric distribution is used to solve this question.

What is the hypergeometric distribution formula?

The formula is:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Researching the problem on the internet, it is found that:

There are 12 problems, hence N = 12.

5 will be graded, hence k = 5.

Jerry chooses five of them, hence n = 5.

The probability is P(X = 5), then:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 5) = h(5,12,5,5) = \frac{C_{5,5}C_{7,0}}{C_{12,5}} = 0.0013[/tex]

0.0013 = 0.13% probability that the problems he did comprised the group that was selected to be graded.

More can be learned about the hypergeometric distribution at https://brainly.com/question/24826394