SOLUTION: Please help! The question my book says is this There are 40 people in a chemistry class. The teacher wants to have two projects going where one project deals with water quality

Question 1132130: Please help! The question my book says is this
There are 40 people in a chemistry class. The teacher wants to have two projects going where one project deals with water quality and the other with pH level. Which of the following is best method to make sure each student has a fair chance at being picked for either station? Hint: Each side should be equally likely
A) For each student, select a random number using a computer. If the number is positive, then the student goes to the water quality station; if the number is negative, then the student goes to the pH station.
B) There is no way to assure that each student has a fair shot at either station
C) For each student, select a random number using a computer. If the number is even, then the student goes to the water quality station; if the number is odd, then the student goes to the pH station.
D) Assign the students the numbers 1 through 40. Place 40 pingpong balls into a bin. Label them 1 through 40. Select 20 balls at random (shuffle them around first). The first 20 numbers selected will go to the water quality station. The rest go to the pH station.
Ok so that's the problem. Sorry it's a bit long. Anyway, to me it seems like A, C and D are all the same. I could be missing something? I would guess that the chances of picking an even number is 50 50 right? So is picking a positive number? And D pretty much says the same thing as A and C? Please help this question makes no sense at all. Thanks so much.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i would think D because the probability of getting picked for either project is the same and it ensure that 20 go to one project and 20 go to another.

with A, there is no guarantee that there will be exactly 20 positive numbers and 20 negative numbers.

with C, there is no guarantee that there will be exactly 20 even numbers and 20 odd numbers.

it adds a tcomplexity to the decision.

with D, you have 40 slots and are randomly picking 20 for one or the other project.
everybody has the same chance of being picked or not picked out of the 20 draws.
it's simpler with no complications of what to do if more than 20 are picked for one project or the other, as can happen with A and C.

i'd go with D.

for example, with A, it's possible that 40 students all get positive numbers (not likely, but possible).

since only 20 can go to one project, then you would take the first 20 that has positive numbers.

but which 20?

another decision has to be made to decide which 20 are selected that may or may not be randomly executed.

option D avoids that complexity.

as long as the balls were randomly selected, there is no possibility of bias and you only get 20 for one project with the rest for the other project.

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