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# Probabilities Question

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Probabilities Question [#permalink]  15 Sep 2005, 13:15
Hi again, I got another question, this time about probabilities:

There are 10 different stamps. Peter's collection includes 3 stamps of the 10. If 2 stamps are drawn from the 10, what is the probability that neither of the 2 stamps are among the 3 held by Peter?

Thanks!

M
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Lets first find probability when BOTH two stamps drawn are from the Peter's collection of 3 stamps. The probability for this would be:

3C2 / 10C2 = 3/45 = 1/15

Probability that ONE of the stamp picked matched that of Peter's collection =

3C1 / 10C2 = 3/45 = 1/15

the probability that NIETHER was from Peter's collection = 1 - 1/15 - 1/15 or
p = 13/15

Last edited by duttsit on 15 Sep 2005, 14:51, edited 1 time in total.
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7/10*6/9 = 7/15
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ranga41 wrote:
7/10*6/9 = 7/15

hows the second draw probability 6/9 here? wont it depend on first draw. I mean if:
A) first draw resulted into one of the stamp in peter's collection, this probability would be : 7/9

B) first draw did not result into perter's collection stamps, second draw probability would be:
6/9

As first draw probability remains same, the total probability would be:

7/10 * 7/9 + 7/10 * 6/9 or 91/90

Guess ans to this question depends/changes on if:
- both stamps are drawn "together"
- stamps are drawn one after another.

The ans will vary based on the two.

Last edited by duttsit on 15 Sep 2005, 15:00, edited 2 times in total.
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First pick: 7/10 stamps are NOT included in Peter's collection
Second pick: 6/9 stamps are NOT included in Peter's collection, since there are 9 stamps left and 3 of those 9 are within his collection

7/10 * 6/9 = 7/15
Intern
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Can we use binomial distribution here ?????

10C2*(7/10)^2*(3/10)^8
Intern
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You draw first time and it is 7/10 to miss
Then you draw second time and it 6/9 to miss
So - 6*7/90
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silentell wrote:
First pick: 7/10 stamps are NOT included in Peter's collection
Second pick: 6/9 stamps are NOT included in Peter's collection, since there are 9 stamps left and 3 of those 9 are within his collection

7/10 * 6/9 = 7/15

7/15 looks good to me.

I think that the suggestion of 91/90 is improbable, indeed impossible !

0<= P <= 1 for all probabilities.
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don't overthink it. It is as simple as 7c2/10c2= 7/15
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