In a Las Vegas casino, gamblers lose a mean of on each roulette game, with a sta

Question

In a Las Vegas casino, gamblers lose a mean of on each roulette game, with a standard deviation of . (Since the standard deviation is quite high with respect to the mean, roulette players often win money on a bet.) Suppose a gambler plays roulette for bets straight. What is the probability that he will be winning after bets? (Hint: He is winning when the amount of his total earnings is positive.)
Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.

Explanation / Answer

here expected value after 150 bets =150*(-0.2)=-30

and std devation =2.41*(150)1/2 =29.516

hence probability of winning =P(X>0)=1-P(X<0)=1-P(Z<(0-(-30)/29.516)=1-P(Z<1.0164)=1-0.8453=0.1547 ~ 0.155

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