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One of your examples in your probabilty guide

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Joined: 16 Jan 2004
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One of your examples in your probabilty guide [#permalink] New post 25 Jun 2004, 12:51
Hi

I went through the guide and really is fantastic, great work!!!

I do have some doubt on one of the questions and the solution posted

Q.
There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue? (Hypergeometric Distribution)

Answer Posted:

The HD formula is:

p = aCa' * bCb' * cCc' * ... * zCz' / (a+b+c+..+z)C(a'+b'+c'+...+z')
where a, b, ... z are show how many times the outcome was obtained in each sub-event, and a', b', ... z' show how many times the corresponding sub-event was tested. nCk stands for combinations formula.

Let us again apply it to the example given in the text above: we have 3 ball colors, or 3 sub-events: g for green, r for red, b for blue. We know that:

g' = 2
r' = 3
b' = 2
g = 1
r = 2
b = 1
Now, let's do the calculations:

p = 2C1 * 3C2 * 2C1 / (2+3+2)C(1+2+1) = 2C1 * 3C2 * 2C1 / 7C4
7C4 = 7! / (4! * 3!) = 5 * 6 * 7 / 6 = 35
2C1 = 2
3C2 = 3! / (2! * 1!) = 3
p = 2 * 2 * 3 / 35 = 12/35
So, the answer is 12/35.


My answer comes out to be 1/35. This problem states "without replacement". The way i did was followed simple f/t

Prob of 2 red marbles = 3/7* 2/6
Prob of 1 green marble = 2/5
Prob of 1 Blue marble = 2/4

TotalProb (3/7*1/3)*2/5*1/2 =1/35


Where did I go wrong. Had it been "with replacement" then I would have gotten the same answer as yours. Please advise
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 [#permalink] New post 25 Jun 2004, 13:56
Hi Pakoo, The anwer you got 1/35 is partly right.

You have to get RRGB balls out of a 7 total balls.

Your explanation for 1/35 is perfectly correct. But you have considered only one possible arrangement For eg: RRGB

But it could even be RGRB RGBR etc.. ie 4!/2! = 12 ways to arrange RRGB.

Therefore the answer is 12/35.
  [#permalink] 25 Jun 2004, 13:56
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One of your examples in your probabilty guide

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