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Homework answers / question archive / Roulette Probabilities A success, s in Bernoulli trials is often derived from the collection of outcomes

Roulette Probabilities A success, s in Bernoulli trials is often derived from the collection of outcomes

Statistics

Roulette Probabilities

A success, s in Bernoulli trials is often derived from the collection of outcomes. For example, an American roulette wheel consists of 38 numbers, of which 18 are black, 18 are red, and 2 are green. When the roulette wheel is spun, the ball is equally likely to land on any one of the 38 numbers. If you're interested in which number the ball lands on, each play at the roulette wheel has 38 possible outcomes. Suppose however, you're betting on red. Then you're interested only in whether the ball lands on a red number or it doesn't. Hence successive bets on red constitute a sequence of Bernoulli trials with success probability |18 divided by 38.
If four plays at a roulette wheel, what is the probability that the ball lands on red

a) Exactly twice
b) At least once
c) Determine and interpret the average times the ball lands on red

pur-new-sol

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Let C(n,m)=n!/(m!*(n-m)1) be the number of ways to select m items from n items.
Let X be the number of times that the ball lands on red.
a) P(X=2) = C(4,2) * (18/38)^2 * (20/38)^2 = 37.29%
b) P(X>=1) = 1 - P(X=0)
= 1 - C(4,0) * (20/38)^4
= 1 - 7.67%
= 92.33%
c) We still needs to compute
P(X=4) = C(4,4) * (18/38)^4 = 5.03%
P(X=1) = C(4,1) * (18/38) * (20/38)^3 = 27.62%
P(X=3) = C(4,3) * (18/38)^3 * (20/38) = 22.38%
P(X=0) = 7.67%
therefore, the average time is the expected time
E(X) = 0*P(X=0) + 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + 4*P(X=4)
= 1.8946