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There are six balls in a black sack. Among those six are [#permalink]
27 Jun 2003, 01:56

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There are six balls in a black sack. Among those six are black, white, and yellow balls. Three balls are taken at random without repetition. What is the probability that the white and yellow balls ARE taken, but the black one IS NOT?

1C1 + 1C1 + 1C1/ 6C3 = 1/20 The 3 grey balls are identical. Therefore, 1 of the 3 grey balls can be chosen in only 1 way. And since the balls have been chosen all at once WITHOUT REPETITION, the ordering does not matter. Therefore, IMO, it should be 1/20.

1C1 + 1C1 + 1C1/ 6C3 = 1/20 The 3 grey balls are identical. Therefore, 1 of the 3 grey balls can be chosen in only 1 way. And since the balls have been chosen all at once WITHOUT REPETITION, the ordering does not matter. Therefore, IMO, it should be 1/20.

I don't agree with you.

It should be 1C1 *1C1*3C1 / 6C3 = 3/20.
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There are six balls in a black sack. Among those six are 1 black, 1 white, 1 yellow and rest grey balls. Three balls are taken at random without repetition. What is the probability that the white and yellow balls ARE taken, but the black one IS NOT?

its pretty straight:

total possibilitites = 6c3 = 20 possibilities for white and yellow but a black = 3

Actually I meant to edit my response, but somehow it did not go through. Absolutely, as we have always done in the past, the answer should be 3/20. The one question I did have was, if Repetition vs. No Repetition should make a difference in our approach. In the same question, if balls are drawn one by one, the approach would be like this: 1/6 * 1/5 * 3/4 * 3! = 3/20. However, if all balls are drawn together, how do we assign 1st, 2nd or 3rd position to any of the balls? The approach should then be: 1C1 * 1C1 * 1C1/6C3 = 1/20. Even if we numbered the 3 grey balls for the sake of our clarification, W-Y-G1 is the same as W-Y-G2 or W-Y-G3. And there is only one chance that the person takes to draw the 3 balls. But we do know that in a total of 6C3 = 20 combinations, there are 3 combinations of W-Y-G. Hence, the prob of getting that combination is 1/20 * 3 = 3/20. All I am saying is that 3C1 sounds like" choosing 1 thing out of 3, which can be done in 3 ways", whereas there is only 1 way to choose the color.

there are 6*5*4 = 120 possible arrangements of the selected balls. Of those, 6 include white and yellow but not black..so 6/120

Black balls ARE part of that 6.

Here is what I think: From 1black, 1 white, 1 yellow, and 3 gray balls ---> With 3 draws (no repetition)

(1/6)(1/5)(3/4)= 3/120 = 1/40

where 1/6= prob of 1 white where 1/5 = prob of 1 yellow (after white is drawn) where 3/4= prob of selecting not black on final draw

am i right? or where did I go wrong?

here you have to multiply by3c2 , as there are 6 ways of picking . i.e the white can be first or second or third. similarly for the other colours. so the requirement is wyg, this can be picked wyg wgy gyw gwy ywg ygw. here order is important because the balls are not being taken out together, rather one by one. if they were to be taken out together , then the order would not matter.