Julia plays the following game: There is a bag full of 10 chips. There are 3 win

Question

Julia plays the following game: There is a bag full of 10 chips. There are 3 winning chips and 7 losing chips. Julia is allowed to draw 2 chips (without replacement). For each winning chip she pulls out, she wins $2. However, the game costs $1.50 to play. (So her actual winnings if she pulls out a single winning chip is only $0.50.) Let X be her winnings.

(a) Find the expected value and variance for the random variable X

.(b) Based on your answer to part (a), should she play the game?

(c) What is a fair price for this game? That is, how much should Julia be charged for this game for the expected value to be 0?

Explanation / Answer

here probabilty P( X=-1.50) =P( both are losing chips)=(7/10)*(6/9)=42/90=7/15

P(X=0.5)=P( first chip winning and second losing +first chip losing and second winning)

=(3/10)*(7/9)+(7/10)*(3/9)=7/15

P(X=2.5)=P( both chips are winning) =(3/10)*(2/9)=1/15

hence from above :

a) mean E(X) =-0.30

variance Var(X)=1.4933

b) as expected return is negative therefore she should not play the game.

c)for above to be fair E(X) =0

therefore Julia must be charged =1.5-0.3 =$1.2 for game to be fair

x p(x) xP(x) x2P(X) (x-)2 (x-)2P(x) -1.5 7/15 -0.700 1.050 1.440 0.672 0.5 7/15 0.233 0.117 0.640 0.299 2.5 1/15 0.167 0.417 7.840 0.523 total 1 = -0.30 1.583 9.920 2= 1.4933

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