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Probability: sampling without replacement

  • Thread starter bennyska
  • Start date
1. The problem statement, all variables and given/known data
there are 30 red balls, 30 white balls, and 30 blue balls in a container. you draw 10 of them, without replacement. what is the probability that at least one color won't be selected.


2. Relevant equations



3. The attempt at a solution
i've been working on this one for too long. someone gave me a solution, but after i thought about it i realized it didn't work if you don't replace. the best solution i could come up with is the following:
for the total number of possibilities, since i can draw 90 for my first, 89 for my second and so on, i have 90!/80! ways of drawing balls.
for the number which exclude one color, i have 90*89 (and here i have to reduce the pool by 30 so those don't get selected)*58*57*...51.
but then someone pointed out that order doesn't matter, and i wasn't sure if there was something i could multiply this number by to make order irrelevant.

as always, i'm looking for hints more than a solution. also, variations on this problem would be cool too. i'm here to learn, people!
 

HallsofIvy

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The probability that the first ball selected is NOT red is 60/90. The probability the second ball is not red is then 59/89, the probability that the third ball is not red is 58/88, etc. until the probability that the tenth ball selected is not red is 51/81. The probability that a red ball is not selected in 10 tries is the product of those.

Do the same with the probability that the balls selected are not white and then blue. The probability that, in 10 tries, one of the colors will not be selected is the sum of those.
 
so (60/90 * 59/89 * ... *51/81) * 3 = .0395. does that sound right?
 
You have to additionally get rid of the double counts that occur, for example, when you pick ten reds in a row. That's going to show up in both the "no white" and "no blue" categories.
 

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