Calculating $\mathbb{P}$ and $\mathbb{E}$ for the amount of vodka on a party

Question

You want to give a party and have to estimate the amount of vodka you need. $G$ is the amount of guests that will come to the party. $35$ are invited, $10$ will come for sure, $5$ won't come and the other $20$ are not sure about it yet. You estimate $G = 10 + X$, with $X \sim \text{Bin}(20, \frac{2}{3})$.\

1. $Y$ is the amount of bottles of vodka. You estimate that for all $n \in \{10, 11, \dots, 30\}$ and all $k \in \{0, 1, 2, \dots, 4n\}$: $$ \mathbb{P}(Y = k \mid G = n) = \binom{4n}{k}\left(\frac{3}{4}\right)^k\left(\frac{1}{4}\right)^{4n-k}. $$ Give a formula for $\mathbb{P}(Y = k, G = n), n \in \{10, 11, \dots, 30\}$ en $k \in \{0, 1, 2,\dots ,4n\}$.
2. Calculate $\mathbb{E}(Y \mid G = n), n \in \{10, 11, \dots, 30\}$.
3. Calculate $\mathbb{E}(Y)$ with the Partition Theorem.

My ideas:

1. In general we have $$ \mathbb{P}(Y=k\mid G=n) = \dfrac{\mathbb{P}(Y=k, G=n)}{\mathbb{P}(G = n)} $$ with $\mathbb{P}(G = n) > 0$. We can rewrite it as $$ \mathbb{P}(Y=k, G=n) = \mathbb{P}(Y = k \mid G = n) \cdot \mathbb{P}(G=n). $$ Now we use that $\mathbb{P}(Y = k \mid G = n) = \binom{4n}{k}(\frac{3}{4})^k(\frac{1}{4})^{4n-k}$ in the formula above to find $$ \begin{aligned} \mathbb{P}(Y=k, G=n) &= \binom{4n}{k}\left(\frac{3}{4}\right)^k\left(\frac{1}{4}\right)^{4n-k}\cdot \mathbb{P}(G=n) \\ &= \binom{4n}{k}\left(\frac{3}{4}\right)^k\left(\frac{1}{4}\right)^{4n-k}\cdot \mathbb{P}(10+X=n) \\ &= \binom{4n}{k}\left(\frac{3}{4}\right)^k\left(\frac{1}{4}\right)^{4n-k}\cdot \mathbb{P}(X=n-10) \\ &= \binom{4n}{k}\left(\frac{3}{4}\right)^k\left(\frac{1}{4}\right)^{4n-k}\cdot \binom{20}{n-10}\left(\frac{2}{3}\right)^{n-10}\left(\frac{1}{3}\right)^{30-n}, \end{aligned} $$

2. The conditional expectation of $Y$ given $G = n$ is: $$ \mathbb{E}(Y \mid G = n) = \sum_{k=0}^{4n} k \cdot \mathbb{P}(Y=k \mid G=n). $$ We find: \begin{align*} \mathbb{E}(Y \mid G=n) &= \sum_{k=0}^{4n} k \cdot \binom{4n}{k}\left(\frac{3}{4}\right)^k\left(\frac{1}{4}\right)^{4n-k}\\ \end{align*} Now i don't know anymore?

3. The Partition theorem says $\{B_1, B_2, \dots, B_n\}$ is a partition of $\Omega$ such that $\mathbb{P}(B_i) > 0$ for all $i \in \{1, \dots, n \}$, $$ \mathbb{E}[X] = \sum_{i=1}^n \mathbb{E}[X \mid B_i] \cdot \mathbb{P}(B_i)$$ I believe I have to understand (2) before I can answer (3).

Answer 1

For 2, note that $Y|(G=n) \sim Bin(4n,3/4)$ so $E[Y|G=n]=4n\times 3/4=3n.$