I have this assignment that I can not figure out. The book doesn't provide enough inside for me to tackle it. The professor provides only theory, and no examples of how to apply it to real problems. I think, I figured the part a, but not sure. Please, help. I would really appreciate the explanations on how to break this kind of problems in steps and how to work it through.
Rodney is rolling a rather lob-sided die. The probability that the roll results in $i$, $i = 1, 2, 3, 4, 5, 6$ is $p_i = i/21$. Suppose Rodney rolls the die $25$ times, and let $N_i$ be the number of times the roll results in $i$.
(a) What is the distribution of the $6$-variate random variable $(N_1,\ldots , N_6)$?
(b) Rodney wins the a dollar each time the roll results in an even number and loses a dollar each time the results in an odd number. Find Rodney’s expected winnings.
(c) Find the expected number of rolls that result in $2$ given the number of rolls that result in and even number is $14$.
For part c, I have $14$ rolls that result in even number. Let N_2 be the event that the result is 2.
E[N_2] = 14(P(X=2)-P(X=4)-P(X=6)), which results in negative number. It is obviously wrong. Or I have $14$ rolls and only $3$ events that result in an even number, and the result is 14/3?
For (a), let $n_1,\ldots,n_6$ be nonnegative integers such that $\sum_{i=1}^6 n_i=25$. Then $$ \mathbb P((N_1,\ldots, N_6) = (n_1,\ldots,n_6)) = \prod_{i=1}^6 p_i^{n_i}. $$
For (b), let $W_i = \mathsf 1_{\{2,4,6\}}(X_i) - \mathsf 1_{\{1,3,5\}}(X_i)$ where $X_i$ is the result of the $i^{\mathrm{th}}$ roll. Then his total winnings are $W=\sum_{i=1}^{25} W_i$, and because the $X_i$ are i.i.d. so are the $W_i$. Therefore \begin{align} \mathbb E[W] &= \mathbb E\left[\sum_{i=1}^{25} W_i\right]\\ &=\sum_{i=1}^{25}\mathbb E[W_i]\\ &= 25\mathbb E[W_1]\\ &= 25(\mathbb P(X_1\in\{2,4,6\}) - \mathbb P(X_2\in\{1,3,5\})\\ &= 25(\mathbb P(X_1=2)+\mathbb P(X_1=4)+\mathbb P(X_1=6) - (\mathbb P(X_1=1)+\mathbb P(X_1=3)+\mathbb P(X_1=5))\\ &= 25 (2/21 + 4/21 + 6/21 - (1/21+3/21+5/21))\\ &= \frac{25}7. \end{align}
For (c), I will let you figure it out. Don't forget that $2$ is itself an even number!