Predict how many times the ball will land on red or black A roulette wheel has 38 slots around the rim. The first 36 slots are numbers from 1 to 36. Half of the 36 slots are red, the other half are black. The remaining 2 slots are numbered 0 and 00 and are green. If the roulette wheel is spun 114 times, predict about how many times the ball will land on red or black?
 A: Under ideal conditions, every $38$ spins the ball will land on red $18$ times, black $18$ and green $2$ times.
So after $114$ spins, we expect there to be $54$ reds and $54$ blacks, with $6$ greens.
A: Let $X_i$ be an indicator random variable denoting the event that the spinner landed on red or black on the $i^{th}$ trial.
I.e. for each $i\in\{1,2,\dots,118\}$ let $X_i=\begin{cases}1~\text{if the roulette landed on red or black on the}~i^{th}~\text{trial}\\
0~\text{otherwise}\end{cases}$
Then, letting $X=\sum\limits_{i=1}^{118}X_i$, you have $X=$ total number of times the spinner landed on red or black out of 118 trials.
We have $E[X]=E[\sum\limits_{i=1}^{118}X_i]=\sum\limits_{i=1}^{118}E[X_i]$ by the linearity of expectation.
We see that $E[X_i]=E[X_j]$ for every $i,j\in\{1,2,\dots,118\}$, so the above simplifies to simply $118E[X_i]$.
In a single trial, the probability that the spinner lands on red or black is $\frac{36}{38}$.
We have then as a final answer $E[X]=118\cdot \frac{36}{38}= 108$
A: @brandi Deriving solution to this problem involves two simple steps.


*

*First find the probability of ball landing on red or black in one trial. it is 1 - {probability of ball landing on green} => 1 - $\frac {2}{36}$ => $\frac {34}{36}$

*Now calculate for 114 trials. it is 114 * $\frac {34}{36}$ => 108
