Question

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?

Answer 1

Answer:

(b) 1/792

Step-by-step explanation:

The complete question is;

Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.

(a) How many different groups of 5 problems can be chosen from the 12

problems?

(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

$\frac{n!}{(n-r)!(r!)}$

where n=12 and r=5

substitute values

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

In (b)

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

$\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}$