with/without replacement

waxydock

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Aug 22, 2006
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Balls numbered 1, 2, 3, 4, and 5 are in an urn.

a) If three are drawn with replacement, what is the probability of getting the set {1,2,3} - unordered.

b) If three are drawn without replacement, what is the probability of getting the set {1,2,3} - unordered.

I'm not 100% sure how to do part (a). Is it 5! +5! + 5! ? For part (b), I think the answer is 1 / 5C3 = 1/10

Please help, thanks
 
waxydock said:
Balls numbered 1,2,3,4,5 are in an urn.
a) If three are drawn with replacement, what is the probability of
getting the set {1,2,3} - unorded.
b) If three are drawn without replacement, what is the probability of getting the
set {1,2,3} - unorded.

Im not 100% sure how to do a), is it 5! +5! + 5! ?
but for b) i think the answer is 1 / 5C3 = 1/10

Please help, thanks
I agree on (b).

For (a), the probability is 3!/5^3, which is (the probability of drawing 1,2,3 in a particular order) times (the number of ways of ordering 1,2,3). By the way, this probability comes out of the multinomial distribution if you are learning that.
 
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