Question 534794: A box contains 50 balls numbered from 1 to 50. If 10 balls are drawn with replacement, what is the probability that at least two of them have the same number?
Any help would be appreciated! Thank you!
Found 2 solutions by stanbon, Alyrian:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A box contains 50 balls numbered from 1 to 50. If 10 balls are drawn with replacement, what is the probability that at least two of them have the same number?
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P(at least 2 have same #) = 1 - P(no 2 have the same number)
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Examine P(no 2 have same number)
# of ways to succeed: 50C10
# of possible outcomes: 50^10
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P(no 2 have same number) = 50C10/50^10 = 1.05x10^-7
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So, P(at least 2 have same #) = 1 - (1.05x10^-7)
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Cheers,
Stan H.
Answer by Alyrian(2) (Show Source):
You can put this solution on YOUR website! The previous solution was almost correct with a little flaw in counting.
So you want the number of permutations of size 10 from 50, with replacement.
This is just 50^10, or n^r.
Then you need to find the number of PERMUTATIONS without replacement, as this
will give you the numbers of ways you can rearrange 10 distinct object picked
from 50.
This is 50P10, nPr.
Then the probability of drawing 10 balls and not getting any matches is
50P10/50^10
This is the complement of what you're actually looking for, so subtract this
value from 1
1 - 50P10/50^10
The numerical value is 0.61829331945.
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