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by kksekar » Thu May 10, 2012 9:08 am
A bag has 4 red, 5 blue and 2 green marbles. If 5 marbles ( 2 red, 2 blue & 1 green ) are selected one after the another with out replacement, what is the probability of drawing those 5 marbles?
Please solve for the probability of each marble and whats the probability if the marbles are replaced?
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by mathbyvemuri » Mon May 14, 2012 4:25 am
(2 red, 2 blue and 1 green) from (4 red, 5 blue and 2 green)

Without replacement:
4C2/11C2 * 5C2/9C2 * 2C1/7C1

With replacement:
4C2/11C2 * 5C2/11C2 * 2C1/11C1
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by [email protected] » Tue May 15, 2012 4:30 am
hey,
and how would you solve it, if the order would matter?

Thanks!
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by mathbyvemuri » Tue May 15, 2012 7:13 pm
[email protected] wrote:hey,
and how would you solve it, if the order would matter?

Thanks!
Result does not change even if the order changes. see here:
Let us change the order now.
(1 green,2 red, 2 blue) from (2 green,4 red, 5 blue)

Without replacement:
2C1/11C1 * 4C2/10C2 * 5C2/8C2 = 2/11 * 4*3/10*9 * 5*4/8*7 = (2*4*3*5*4)/(11*10*9*8*7)
The other order considered in my earlier explanation is: 4C2/11C2 * 5C2/9C2 * 2C1/7C1
= (4*3*5*4*2)/(11*10*9*8*7)
Both are equal

With replacement:
2C1/11C1 * 4C2/11C2 * 5C2/11C2 = (2*4*3*5*4)/(11*11*10*11*10)
The other order considered in my earlier explanation is: 4C2/11C2 * 5C2/11C2 * 2C1/11C1
= (2*4*3*5*4)/(11*11*10*11*10)
Both are equal
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by kksekar » Thu May 17, 2012 12:02 am
its differs from the official answer, which is 20/77
got from 4c2*5c2*2c1/11c5 = 20/77

please explain why the answer differs when you consider the probability of each marble separately?
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by mathbyvemuri » Thu May 17, 2012 4:13 am
kksekar wrote:its differs from the official answer, which is 20/77
got from 4c2*5c2*2c1/11c5 = 20/77

please explain why the answer differs when you consider the probability of each marble separately?
What is the OA for the case "with replacement"
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by kksekar » Mon May 28, 2012 4:22 am
with replacement is not part of the Official question and so there is no OA
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