Question

One professor grades homework by randomly choosing 5 out of 12 homework problems to grade. a) Andrew 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:

$Probability = \frac{1}{792}$

Step-by-step explanation:

Given

$Questions = 12$

$Selection = 5$

Required

Probability of grading the selected 5 by Andrew

First, we need to determine the total number of selections.

This is calculated as: $^{12}C_{5}$

Solving further:

$^{12}C_{5} = \frac{12!}{(12-5)!5!}$

$^{12}C_{5} = \frac{12!}{7!5!}$

$^{12}C_{5} = \frac{12*11*10*9*8*7!}{7!*5*4*3*2*1}$

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

$^{12}C_{5} = \frac{95040}{120}$

$^{12}C_{5} =792$

There is only 1 for Andrew's question selection.

So, the required probability is:

$Probability = \frac{1}{792}$