Re: OT - simple (not for me) probability question

David (Ross) confirms:
Jack got this one correct (twice!).

He wasn't alone ... Eyes glaze over when the indistinguishable urns
and balls appear, but I may be able to illuminate the problem a
little.

It's usually easier to compute simpler, related results first, and
then build upon them. In this case, one of the simplest results is the
all-red probability, 1/1024 (by inspection). Indistinguishability
plays a role with the next result. The first ball is not red 3/4 of
the time, and the rest are red 1/256 of the time, but this case is
indistinguishable from the one where the only non-red ball comes from
the second urn. In other words, there are five ways to achieve this
result so the probability is 5 x 3/4 x 1/256 = 15/1024.

The more complicated cases are best addressed by studying the effect
of indistinguishability, but those of us who've given up on acquiring
that intuition may find it easier to look at the corresponding four-
urn problem instead. Here, the simple results indirectly produce a
complete solution: none-red = 81/256, one-red = 108/256, all-red =
1/256, three-red = 12/256, and two-red must be the fraction needed to
get these to sum to 1, i.e. 54/256.

Restoring the fifth urn ... if the first ball is red, we need two
more, so the probability is 1/4 x 54/256 = 54/1024. If the first ball
is not red we need three more, so the probability is 3/4 x 12/256 =
36/1024. Note that indistinguishability is already factored into these
results. Thus, the total three-red probability is 90/1024 (as before).


Felix

.

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