Roulette. An American roulette wheel contains 38 numbers: 18 are red, 18 are black, and 2 are green. When the roulette wheel is spun, the ball is equally likely to land on any of the 38 number. You win \(1: otherwise you lose your \)1. Let X be the amount you win on your \(1 bet. Then x is a random variable whose probability distribution is as follows.

Part (a) Verify that the probability distribution is correct.

Part (b) Find the expected value of the random variable X.

Part (c) On average, how much will lose per play?

Part (d) Approximately how much would you expect to lose if you bet \)1 on red 100 times? 1000 times?

Part (e) Is roulette a profitable game to pay ?Explain

Let's call 'X' the random variable that represents the winning amount on a $1 bet. The following is the probability distribution of X:

The following table demonstrates that,

$$\begin{gathered} \sum _{}P(X=x)=0.474+0.526 \\ =1 \end{gathered}$$

The variable 'X' is a finite population variable, as is the case for discrete random variables.

$\sum _{}P(X=x)=1$

As a result, the probability distribution given is correct.

The random variable X's expected value will be

$$\begin{gathered} E\left(X\right)=\sum _{}xP(X=x) \\ E\left(X\right)=\left[\right(1\times 0.474)+(-1\times 0.526\left)\right] \\ E\left(X\right)=(0.474-0.526) \\ E\left(X\right)=-0.052 \end{gathered}$$

The random variable X is expected to have a value of -0.052. As a result, on average, 5.2 cents will be lost per play.

If $1 is bet 100 times on red,

$$\begin{gathered} 5.2cent\times 100 \\ \$5.2 \end{gathered}$$

If $1 is bet 1000 times on red,

$$\begin{gathered} 5.2cent\times 1000 \\ \$52 \end{gathered}$$

Roulette is not a profitable game because the chances of winning are extremely low.