Determine the necessary number of games so that $\overline{X}_n$ is on the given interval

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A roulette player sets $n$ times to $0$; the random variable $X_i$ denotes the result of the $i$-th game.

a) Specify the distribution of each $X_i$.

b) Argue why the relative frequency of successes converges stochastically to the theoretical probability of success. What does this result mean in practice?

c) Determine the necessary number $n$ of games, so that the relative frequency of the achievements $\overline{X}_n=\frac{1}{n}\sum_{i=1}^nX_i$ is with a probability of at least $90\%$ in the interval $\frac{1}{37}-\frac{1}{74}<\overline{X}_n<\frac{1}{37}+\frac{1}{74}$.

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I have done the following:

a) Each $X_i$ describes succes or faillure, therefore each $X_i$ is bernoulli distributed.

b) Does this have a relation with the law of large numbers?

c) According to the central limit theorem we have that $\overline{X}$ approximates the normal distribution for large $n$ with the parameters \begin{equation*}E(\overline{X})=\frac{1}{2}=0.5 \ \text{ and } \ V(\overline{X})=\frac{\sigma_X^2}{n}=\frac{\left (\frac{1}{4}\right )^2}{n}=\frac{1}{16n}\Rightarrow \sigma_{\overline{X}}=\frac{1}{4\sqrt{n}}\end{equation*}

The relative frequency of achievements $\overline{X}_n$ should be with probability of at least $90\%$ in the bounds $\frac{1}{37}-\frac{1}{74}<\overline{X}_n<\frac{1}{37}+\frac{1}{74} \Rightarrow \frac{1}{74}<\overline{X}_n<\frac{3}{74}$.

So we have the following:
\begin{align*}P\left (\frac{1}{74}<\overline{X}_n<\frac{3}{74}\right )\geq 90\% &\Rightarrow \Phi \left (\frac{\frac{3}{74}-0.5}{\frac{1}{4\sqrt{n}}}\right )-\Phi \left (\frac{\frac{1}{74}-0.5}{\frac{1}{4\sqrt{n}}}\right )\geq 0.9 \\ & \Rightarrow \Phi \left (-1.83784 \sqrt{n}\right )-\Phi \left (-1.94595 \sqrt{n}\right )\geq 0.9 \\ & \Rightarrow 1-\Phi \left (1.83784 \sqrt{n}\right )-1+\Phi \left (1.94595 \sqrt{n}\right )\geq 0.9 \\ & \Rightarrow -\Phi \left (1.83784 \sqrt{n}\right )+\Phi \left (1.94595 \sqrt{n}\right )\geq 0.9\end{align*}

Is everything so far? If yes, how could we calculate the last expression where we have two values of the distribution function?

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For a) you should be more specific with the distribution of $X_i$. It is the result of one spin of the wheel and is either $0$ or $1$. What is the probability of $0$? of $1$?

For b) yes. You should look at the normal approximation and claim that the fraction of successes converges to $\frac 1{37}$ assuming you are playing with only a single $0$.

For c) $E(\overline X) \neq \frac 12$. If I could find a roulette wheel where my probability of success was $\frac 12$ I would be betting a lot there an be rich. Your $\sigma_X$ is also badly wrong. You have a discrete distribution for $X_i$ with $P(X_i=0)=\frac {36}{37}$