Re: Probability question
- From: Jack Tomsky <jtomsky@xxxxxxxxxxxxx>
- Date: Mon, 26 Jun 2006 16:20:07 EDT
Hi there everyone, I have come across a problem while
trying to learn
a bit more about probability, and was wondering if
anyone could help.
There exist 3 balls (red (R), green (G), blue (B))
and four holes
(1,2,3,4).
All three balls have to go into a hole.
Given that
p(R->1) = p(G->1) = p(B->1) = f(1)
p(R->2) = p(G->2) = p(B->2) = f(2)
p(R->3) = p(G->3) = p(B->3) = f(3)
p(R->4) = p(G->4) = p(B->4) = f(4)
, where p(R->1) means the probability of the red ball
going into hole
no.1
what is the probability that the balls will go, in no
particular
order, in holes 1,2 and 3?
I have playing with numbers a bit, and was thinking
of some kind of
combination/permutation multiplied by the
probability. But it doesn't
work. Any ideas?
Thank you all for your help.
The probability is [3!/(1!1!1!0!)]*f(1)*f(2)*f(3) = 6*f(1)*f(2)*f(3).
Jack
.
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