Question

A professor has learned that nine students in her class of 35 will cheat on the exam. She decides to focus her attention on ten randomly chosen students during the exam. a. What is the probability that she finds at least one of the students cheating

Answer 1

`(73,331)/(75,516)\approx 97.11\%`

Explanation:

The probability that she will find at least one student cheating is equal to the probability that she finds no students cheating subtracted from 1.

Each time she randomly chooses a student the probability she will catch a cheater is equal to the number of cheaters divided by the number of students.

Therefore, for the first student she chooses, there is a
`(9)/(35)` chance that the student chosen is a cheater and therefore a
`(26)/(35)` chance she does not catch a cheater. For the second student, there are only 34 students to choose from. If we stipulate that the first student chosen was not a cheater, then there is a
`(9)/(34)` chance she will catch a cheater and a
`(25)/(34)` chance she does not catch the cheater.

Therefore, the probability she does not catch a single cheater after randomly choosing ten students is equal to:

`(26)/(35)\cdot (25)/(34)\cdot (24)/(33)\cdot (23)/(32)\cdot (22)/(31)\cdot (21)/(30)\cdot (20)/(29)\cdot (19)/(28)\cdot (18)/(27)\cdot (17)/(26)`

Subtract this from one to get the probability she finds at least one of the students cheating after randomly selecting nine students. Let event A occur when the professor finds at least one student cheating after randomly selecting ten students from a group of 35 students.

`P(A)=1-(26)/(35)\cdot (25)/(34)\cdot (24)/(33)\cdot (23)/(32)\cdot (22)/(31)\cdot (21)/(30)\cdot (20)/(29)\cdot (19)/(28)\cdot (18)/(27)\cdot (17)/(26),\\\\P(A)=1-(2,185)/(75,516),\\\\P(A)=\boxed{(73,331)/(75,516)}\approx 0.97106573441\approx \boxed{97.11\%}`