Probability - Partition TheoremWatch

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Thread starter 14 years ago
#1
A bag contains three balls - one red, one white, one orange.

A ball is chosen at random and replaced.

If the ball is white, the player receives two counters.

If the ball is orange, the player receives one counter.

If the ball is red, the player receives no counters.

What is the expected number of counters that the player will gain before choosing the red ball for the first time?
0
14 years ago
#2
Let X be the number of balls chosen up to and including the first red one. Since X ~ Geometric(1/3), E(X) = 3.

Let C be the number of counters gained from the first X draws.

Then

E(C | X = x) = (3/2)(x - 1)

because, conditional on X = x, each of the first x - 1 balls is independently white (with probability 1/2) or orange (with probability 1/2).

E(C)
= (sum over x) E(C | X = x)P(X = x)
= (sum over x) (3/2)(x - 1)P(X = x)
= (3/2)(E(X) - 1)
= 3
0
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