roulette number repeat...two workings dif answers..1wrong-why?
Hi,
Suppose a roulette (with 38 numbers) is spun 12 times, and the probability of any number repeated is required.
So:
Working 1: 38C1*12C2*(1/38)^2*(37/38)^10.......gives a value greater than 1 so obviously wrong. (success occuring twice for a single number, then accounting for 38 separate no.s)
Working 2: No repeat= 38*37*36*..*27/(38)^12. So at least one repeat=1-no repeat=1-.143=.857 probability...looks better, and sufficient for my task.
Problem is...how do I find out the probability of any number occuring thrice exactly, or at least probability of any number occuring twice exactly, as the working 1 doesn't work, and working 2 is only applicable for the special case and no moreWhy just when I think I have learnt something about probability it eludes me.
Heeeelp!
Originally Posted by ssadi
There aretotal of outcomes.
There areways to have all different outcomes.
The probability of all different outcomes is.
So the probability of at least one repeated outcome is.
Now I am not sure that is actually what you mean by "any number repeated".
no, I don't think I meant what you think I meant
You are saying the probability of a particular set of number of appearing
I am saying that say..2 appears at least twice, or 15 appears exactly 3 times... in the 12 trials
How to get that probability?
for example
12 trials
any number in the set (00,0...36) repeats itself
within the 12 trials
I don't know how i can be more specific
similarly, say
any number in the set (00, 0...36) appears exactly 3 times in the 12 trials
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