Completing a probability table and finding expected value.

**amul28** · May 5th 2010, 05:37 AM

16. The manager of a casino plans the following game. Each player will be given a box containing five coins of which one is “gold” (valued Rs 100), one is “silver” (valued Rs 25) and three are “ordinary” (valued Re 1 each). The player draws two coins at random and without replacement. Let A and B be the values (in rupees) of the two coins so drawn. Define X = |A  B|. The player receives a payoff (X + 25) if none of the two selected coins is ordinary; the payoff is zero otherwise. Complete the following table showing your answers.
P(X = 0) =
P(X = 1) =
P(X =20) =
P(X = 24) =
P(X = 75) =
P(X = 99) =
E(X) =
Expected payoff

**undefined** · May 5th 2010, 12:59 PM

Originally Posted by amul28

Define X = |A  B|.

The symbol between A and B does not render properly on my computer. I see a gray-bordered box with F0 2D inside of it.

I'll assume it's X = |A - B|.

Originally Posted by amul28

The player receives a payoff (X + 25) if none of the two selected coins is ordinary; the payoff is zero otherwise.

If I understand this correctly, the only way to get a nonzero payoff is to pick the gold and the silver coin, in which case the payoff is 100. That means that we only really care about P(X = 75) and the other probabilities are irrelevant.

$P(X = 75) = \frac{1}{C(5,2)}$

The other values are below.

$P(X = 0) = \frac{C(3,2)}{C(5,2)}$
$P(X = 1) = 0$
$P(X = 20) = 0$
$P(X = 24) = \frac{3}{C(5,2)}$
$P(X = 99) = \frac{3}{C(5,2)}$

$E(X)=0\cdot P(X=0)+24\cdot P(X = 24) + 75\cdot P(X=75) + 99\cdot P(X=99)$

Expected payoff will be $100\cdot P(X=75) + 0\cdot P(X \neq 75)$ .

Strange problem..

**amul28** · May 6th 2010, 05:54 AM

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