Q Alex decides to play the casino game of roulette and to be

Q) Alex decides to play the casino game of roulette and to bet $5 on “red” on each play. If the roulette ball lands on red, then he wins a payout of $5; otherwise, his $5 bet is taken by the roulette dealer. The roulette wheel has 38 slots, of which 18 are red and 20 are not red (black or green). Let X be the number of times Alex plays until he wins for the third time. Find:

1)The probability, to 6 decimal places, that Alex wins for the third time on his sixth play.

2)E( X ) and V ( X ) as simple fractions or to six decimal places.

Solution

1) if he wins third time on 6th play , he wins two times in the first 5 plays. 6th play is fixed he wins in 6th play

Probability = 5 choose 2 * 18/38 * 18/38 * (20/38)^3 * 18/38 = 0.15495511711

2) E(X) = expected value of X

= 3*E(X_1)

where E(X_1) is the expected number of time he plays to win first time.

X_1 is the negative binomial distribution with expected value 1/p p is the prob of success , in our case which is 18/38

So E(X_1) = 38/18 and E(X) = 38/6

what is V???

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