Question

Suppose that, over the course of several days, a gambler makes 2400 row bets in roulette, betting $1 each time. For a row bet, the gambler selects a row of 3 spaces out of the 38 possible spaces on a roulette table. If the gambler wins on one play, the gambler get his/her dollar back plus eleven more, for a net gain of $11. However, if the gambler loses, he/she loses $1. Find the probability that, after making these 2400 row bets, that the gambler wins at least $0. (Round your answers to four decimal places.)

Answer 1

Here

selected 3 spaces out of 38 possible spaces and if win he will get 11$ more

So winning probability

P(X=11) = 3/38

If lose , he will loses 1$

P(X=-1) = 35/38

E(X) = 11 (3/38) + (-1 (35/38) ) = -2/38 = -0.0526

E(X²) = 11² (3/38) + (-1)² (35/38) = 10.4737

Var (X) = E(X²) - (E(X))² = 10.4737 - (-0.0526)² = 10.4709

Using Central utility theorem, the distribution for the net winnings would be computed as:

X~N ( 2400*(-0.0526) , 2400*10.4709 )

X ~ N (-126.24 , 25130.16)

The probability that, after making these 2400 row bets, that the gambler wins at least $0.

= 0.2129