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25 Jun 2004, 13:51
Kudos
Bookmarks
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
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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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25 Jun 2004, 14:56
Kudos
Bookmarks
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.
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.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
