What kind of random variable is this? Expected Value of Binomial

Question

A new roulette wheel has 21 slots with 6 symbols. 6 with an A, 5 with a C, 4 with a G, 3 with B, 2 with R, and 1 with D. The wheel is spun 10 times.

What's the expected number $X$ of different symbols seen?

Let $Y_i=1$ if $i$ spin is an A and 0 otherwise. Find the Expected Value and Variance of $Y$?

Let $Z_i$ indicate if $i$ spin is C. find $Cov(Y,Z)$?

I believe I am close on the answer to these questions, but I would like some confirmation.

My first question is what kind of variable is $X$? How does one find this part. I think the $E(Y)=np=10*6/21$ and the $Var(Y)=np(1-p)=100/49$.
Also, the $

\begin{equation*} Cov(Y,Z)=E(Y,Z)-E(Y)E(Z) = n ( p_B - p_Z p_Y ) \end{equation*}

such that $p_b$ is the probability that they both happen at the same time, which is clearly 0. I got the above formula from wikipedia Binomial Distribution.

Am I at all close on this?

Answer 1

Q1. Define event $A=$ "Symbol $A$ appears at least once in the $10$ spins". Define events $C,G,B,R,D$ similarly for the other symbols. Then, with $I_A$ being the indicator variable for event $A$, etc., meaning $I_A=1$ if $A$ occurs and $0$ otherwise, we have $X=I_A+I_C+I_G+I_B+I_R+I_D$, so

\begin{eqnarray*} E(X) &=& E(I_A+I_C+I_G+I_B+I_R+I_D) \\ && \\ &=& E(I_A)+E(I_C)+E(I_G)+E(I_B)+E(I_R)+E(I_D) \qquad\text{by linearity of expectation} \\ && \\ &=& P(A)+P(C)+P(G)+P(B)+P(R)+P(D) \\ && \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $E(I_A)=1\cdot P(A)+0\cdot P(A^c)$, etc.} \\ && \\ &=& (1-P(A^c))+(1-P(C^c))+(1-P(G^c))+(1-P(B^c))+(1-P(R^c))+(1-P(D^c)) \\ && \\ &=& 1-\left(\frac{15}{21}\right)^{10} + 1-\left(\dfrac{16}{21}\right)^{10} + 1-\left(\dfrac{17}{21}\right)^{10} + 1-\left(\dfrac{18}{21}\right)^{10} + 1-\left(\dfrac{19}{21}\right)^{10} \\ &&\qquad + 1-\left(\dfrac{20}{21}\right)^{10} \qquad\qquad\text{since $A^c$ requires all $10$ spins to be not $A$, etc.} \\ && \\ &=& 6-\left(\frac{15}{21}\right)^{10} -\left(\dfrac{16}{21}\right)^{10} -\left(\dfrac{17}{21}\right)^{10} -\left(\dfrac{18}{21}\right)^{10} -\left(\dfrac{19}{21}\right)^{10} -\left(\dfrac{20}{21}\right)^{10}. \end{eqnarray*}

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Q2. I agree with your answers for $E(Y)$ and $Var(Y)$. (I assume that $Y=Y_1+\ldots+Y_{10}$ and $Z=Z_1+\ldots+Z_{10}$.)

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Q3. Since $Y=\sum_{i=1}^{10}{Y_i}\quad $ and $\quad Z=\sum_{i=1}^{10}{Z_i}$,

\begin{eqnarray*} E(YZ) &=& E\left[\left(\sum_{i=1}^{10}{Y_i}\right) \times \left(\sum_{i=1}^{10}{Z_i}\right)\right] \\ && \\ &=& E\left[\sum_{i=1}^{10}{Y_i Z_i} + \sum_{i\neq j}^{10}{Y_i Z_j}\right] \qquad\qquad\text{by expanding the product} \\ && \\ &=& E\left[\sum_{i=1}^{10}{Y_i Z_i}\right] + E\left[\sum_{i\neq j}^{10}{Y_i Z_j}\right] \qquad\qquad\text{by linearity of expectation} \\ && \\ &=& 0 + 90 E\left[Y_1 Z_2\right] \\ && \\ &=& 90 E\left[Y_1\right] E\left[Z_2\right] \\ && \\ &=& 90 \dfrac{6}{21}\dfrac{5}{21} \\ && \\ &=& \dfrac{300}{49} \\ && \\ \therefore\quad Cov(Y,Z) &=& E(YZ)-E(Y)E(Z) \\ && \\ &=& \dfrac{300}{49} - \dfrac{20}{7}\dfrac{50}{21} \\ && \\ &=& - \dfrac{100}{147}. \end{eqnarray*}