We get
(A) The probability that her winnings will exceed zero is 0.50.
(B) The probability that her winnings will be less than zero is 0.25.
(c) Her winnings' estimated worth is 0.
Given that,
A player engages in a game where the odds of winning or losing one dollar are equal. The games are stand-alone. The game will be played until the first win occurs, at which point the player will instantly stop.
We have to find
(A) determine the probability that her winnings will exceed zero.
(B) calculate the probability that her winnings will be less than zero.
(c) calculate her winnings' estimated worth.
We know that,
a) probability that her winnings are greater than zero =P(wins 1st game) =1/2 =
b) probability that her winnings are less than zero =P(lose 1st 2 games) =(1/2)×(1/2 )=1/4 =0.25
c) below is pmf of X (gain from the game)
P(X=1)=P(win 1st game) =1/2
P(X=0)=P(lose 1st and win 2nd) =(1/2)×(1/2) =1/4
P(X=-1)=P(lose 1st two and wins 3rd)=(1/2)×(1/2)×(1/2) =1/8
P(X=-2)=1/16
therefore expected value E(X )=ΣxP(x) =1×1/2+0×1/4-1×1/8-2×1/16........
= (1/2)+0-(1×1/8+2×1/16+3×1/32+4×1/64+...)
=(1/2)-(1/4)×(1/(1/2))
=(1/2)-(1/2)
=0
Therefore,
(A) The probability that her winnings will exceed zero is 0.50.
(B) The probability that her winnings will be less than zero is 0.25.
(c) Her winnings' estimated worth is 0.
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