Question

Of the 30 students in the sixth period math class, 8 are also in the same fourth period science class. Which can be used to determine the probability that if three students are chosen at random from the math class to do a group project, the first student chosen to be in the group is in the fourth period science class but the other two are not?

A. (8/30)(22/29)(21/28)
B. (8/30)(8/30)(8/30)
C. (8/30)(7/29)(6/28)
D. (8/30)(22/30)(21/30)

Answer 1

A. (8/30)(22/29)(21/28)

Explanation:

The key is in reading the question really thoroughly, like, way more than you would expect. The first person is from the group of 8, thus (8/30), but the other two cannot be from the group of 8. 30(students total) minus 8 (the 8 people from science class)=22 (remaining in math). You choose one person who is not from the group of 8, and you are left with 21 other people. The number of students available to be in a group goes down each time: 30 down to 29 down to 28 (the denominators). Kaboom, A. (8/30)(22/29)(21/28).

Answer 2

Answer: A.
`(8)/(30)*(22)/(29)*(21)/(28)`

Explanation:

Given : Total number of students are in the sixth period math class = 30

Number of sixth period math class students are also in the same fourth period science class = 8

Number of students are not in the same fourth class = 30-8=22

Then , the the probability that if three students are chosen at random from the math class to do a group project, the first student chosen to be in the group is in the fourth period science class but the other two are not will be :-

`\frac{\text{ students are also in same fourth period}}{\text{Total students}}*\frac{\text{students are not in fourth period}}{\text{Remaining total students}}*\frac{\text{remaining students are not in fourth period}}{\text{Total remaining students}}\\\\=(8)/(30)*(22)/(29)*(21)/(28)`

Hence, the correct answer is option A.
`(8)/(30)*(22)/(29)*(21)/(28)`