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PS, tough probability

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by ukr.net » Fri Dec 07, 2007 7:35 am
Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

8/28
9/28
10/28
10/18
11/18
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by jayhawk2001 » Sat Dec 08, 2007 9:42 am
3C2 * 3C2 * 2C1 / 8C5
= 9/28

Is it B?
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by hemanth28 » Sat Dec 08, 2007 1:43 pm
Different possible combinations of colors is

G R B
3 2 0
3 0 2
3 1 1
2 1 2
2 2 1
2 0 3
1 2 2
1 1 3
0 2 3


these are the possible colour combinations when 5 balls are picked.

out of this the color combinaton required is 2 1 2
hence the probability is 1/9.
Can someone tell me if i am missing something?
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PS

by ukr.net » Sun Dec 09, 2007 3:00 am
the answer is B 9/28
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by theroadrunnershow » Fri Dec 14, 2007 9:40 am
hemanth28 wrote:Different possible combinations of colors is

G R B
3 2 0
3 0 2
3 1 1
2 1 2
2 2 1
2 0 3
1 2 2
1 1 3
0 2 3


these are the possible colour combinations when 5 balls are picked.

out of this the color combinaton required is 2 1 2
hence the probability is 1/9.
Can someone tell me if i am missing something?



dude ur missing out the various ways the req. trio can occur..always follow combinatorics fr such type of questions..
answer is 9/28
Abhishek sunku
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by II » Tue Dec 25, 2007 5:03 am
jayhawk2001 wrote:3C2 * 3C2 * 2C1 / 8C5
= 9/28

Is it B?
Hi,
I dont understand your equation ... can you please explain.

Thanks.
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by StarDust845 » Thu Dec 27, 2007 1:09 pm
Total number of ways you can pick 5 balls out of 8 is 8C5 (Order is not important).

Out of those combinations the favourable events are where you have 1 red, 2g and 3 blue.

And you can pick up 1r, 2g and 2b in 2C1 * 3C2 * 3C2 ways given (you have 2 red, 3 green and 3 blue).

I think that's how he got the equation.

Hope this helps.

Calista.
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by cris » Thu Dec 27, 2007 4:50 pm
why is this approach wrong????

3/8*2/7*3/6*2/5*2/4

I dont get it

Thanks!
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by StarDust845 » Thu Dec 27, 2007 6:36 pm
That's not the only way to pick up the balls now, is it?

For e.g. you are assuming you get a blue ball first, then another blue, then a green, another green and then a red (i.e. b,b,g,g,r).

But what about the other possibilities? like g,g,b,b,r or b,g,r,b,g etc?

If you add all of those then you MUST get the same probability.

Obviously, this isn't a faster approach.

Calista.
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re

by stock2007 » Sat Jan 05, 2008 5:55 am
SO WHATS THE ANSWER TO THIS QUESTION?
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by stock2007 » Sat Jan 05, 2008 3:31 pm
helloo

anybody help>
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Re: PS

by beatthegmat » Sat Jan 05, 2008 4:02 pm
ukr.net wrote:the answer is B 9/28
ukr.net already revealed the OA as B earlier in this thread...
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by cris » Sat Jan 05, 2008 9:39 pm
The statement says "...He randomly picks 5 from the bag without replacement" and I understand how the answer to this question is calculated.

But I am wondering what would be the answer if the statement said instead "WITH REPLACEMENT after piking each ball"??? :?:
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by cris » Sat Jan 05, 2008 9:39 pm
The statement says "...He randomly picks 5 from the bag without replacement" and I understand how the answer to this question is calculated.

But I am wondering what would be the answer if the statement said instead "WITH REPLACEMENT after piking each ball"??? :?:
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ps

by ukr.net » Wed Feb 27, 2008 4:30 am
OA is B
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