If a student may not do more than one​ job, then what is the probability that the instructor will choose at most one senior to do the​ jobs?

Question

Refer to the demographic information shown to the right that has been collected by an instructor after surveying her class.

The instructor randomly chooses one student to pass out​ papers, one student to clean the​ chalkboard, and one student to empty the trash. If a student may not do more than one​ job, then what is the probability that the instructor will choose at most one senior to do the​ jobs?

Total number of students in the class 50

Total number of students in the class	50		Number of boys	17
Number of girls	33
Number of Freshman	8
Number of Sophomores	16
Number of Juniors	11
Number of Seniors 15

There is a __​% probability that the instructor will choose at most one senior to do the jobs.

​(Round the final answer to one decimal place as needed. Round all intermediate values to three decimal places as​ needed.)

Answer 1

This again is a binomial density problem with p=.3 (15/50) and q=.7.

The question means no seniors (all non-seniors) or one senior.

This is given by (.7)³ + 3*.3*(.7)², where the 3 comes from the binomial coefficient of 3 choose 1.

Apparently you didn't follow my method correctly...which should have given you .784.

However, when I re-read the problem, I missed a requirement.

The requirement that no student my have 2 jobs means this is NOT a binomial but a hypergeometric problem, as follows.

There are 50 choose 3 ways of picking 3 students from 50.

There are 35 pick 3 ways of choosing 3 non-seniors from 35 and

there are 15*35 choose 2 ways of choosing 1 senior and 2 non-seniors from the group.

If you do the required multiplications, you will get .789 for this probability.

Please not that 35 choose 2 is "shorthand" for the binomial coefficient of 35 and 2, that is ₃₅C₂.

I'm sorry for the error, but this will give you a chance to go back and think over the problem.

Please make another comment if I can help further.