Question

A set of 5 MP3-players is taken from a group of 50 players, for testing. If two of the original 50 are defective, how many ways can the 5 players be selected

Answer 1

There are C(50, 5) combinations of 5 players taken from 50 players.
\[\large C(50, 5)=\frac{50!}{5!(50-5)!}=\frac{50\times49\times48\times47\times46}{5\times4\times3\times2\times1}=you\ can\ calculate\]

Answer 2

Note that the question has no requirement regarding the two defective players when making the selection.

Answer 3

is that the combination equation?

Answer 4

Yes.

Answer 5

okay, so the 2 defective players is irrelevant then right?

Answer 6

Yes, as I mentioned previously the question has no requirement regarding the two defective players when making the selection.

Answer 7

okay cool, thanks :) just so i understand fully, what would b the difference in equations if the defective players were a factor? like, if we didnt count the defective ones? would i have to subtract 50-2 in the bottom of the equation?

Answer 8

The question is rather unusual. It does not make it clear whether or not the selection of five players is random. The form of question that I would expect would ask for the probability of getting none of the defectives in a random sample of five players, or the probability of getting one or both defective players in a random sample of five players, in each case taking the sample without replacement.

Answer 9

Therefore I can't think of any other form of solution to the question other than the one that I have given you.

Answer 10

so we cant compute how many ways the 5 players can be selected if no player is defective?

Answer 11

wouldnt it just be 48C5?

Answer 12

since were taking out the 2 defective ones?

Answer 13

The number of ways of selecting the sample without getting a defective player could be calculated by finding the number of ways of selecting five good players from the 48 good players, which is C(48, 5).
\[\large C(48, 5)=\frac{48!}{5!(48-5)!}=\frac{48\times47\times46\times45\times44}{5\times4\times3\times2\times1}\]

Answer 14

okay thanks! that helps alot!

Answer 15

You're welcome :)