Manager
Joined: 16 Jan 2004
Posts: 64
Location: NJ

One of your examples in your probabilty guide [#permalink]
Show Tags
25 Jun 2004, 13:51
This topic is locked. If you want to discuss this question please repost it in the respective forum.
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 subevent, and a', b', ... z' show how many times the corresponding subevent 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 subevents: 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
