# Chicken and Egg

Recall in the Chicken-Egg problem that we have a chicken that lays eggs, where the number of eggs $N$ follows the distribution $N\sim Pois(\lambda)$. Given that the chicken lays $N$ eggs, the number of eggs that hatch $X$ follows the distribution $X|N \sim Bin(N,p)$.

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Find $E(X)$

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\begin{aligned} E(X) &= E[E(X|N)] = E[Np] = p E[N] =\lambda p \end{aligned}
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Find $Var(X)$

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Use Eve's Law!

\begin{aligned} Var(X) &= E[Var(X|N)] + Var[E(X|N)] \\ &= E[Np(1-p)] + Var[Np] \\ &= \lambda p(1-p) + p^2 \lambda \\ &= \lambda p \end{aligned}

# Consumerism

When customers enter a particular store, each makes a purchase with probability $p$, independently. Given that a customer makes a purchase, the amount spent has mean $\mu$ (in dollars) and variance $\sigma^2$.

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Find the mean of how much a random customer spends (note that the customer may spend nothing).

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Let $X$ be the amount a random customer spends at the store, and let $I$ be the indicator that a random customer makes a purchase. We then apply the Law of Total Expectation:

\begin{aligned} E[X] &= E[X | I= 0]P(I=0) \\&+ E[X | I=1]P(I=1)\\ &= \mu p \end{aligned}
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Find the variance of how much a random customer spends (note that the customer may spend nothing).

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Let $X$ be the amount a random customer spends at the store, and let $I$ be the indicator that a random customer makes a purchase. We then apply the Law of Total Expectation:

\begin{aligned} E[X^2] &= E[X^2 | I=0]P(I=0) \\ &+ E[X^2 | I=1]P(I=1) \\ &= E[X^2 | I=1]P(I=1) \\ &= (\mu^2 + \sigma^2)p \\ Var[X] &= E[X^2] - E[X]^2 \\ &= \sigma^2 p +\mu^2 p (1-p) \end{aligned}

# Trapped Miners

A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 minutes. The other two doors lead to tunnels that will return him to the mine after 5 and 7 minutes of travel each. The miner is equally likely to choose any of the doors at any time.

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What is the expected amount of time until he reaches safety?

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Let $W$ be the total amount of time taken until he reaches safety, and $D_1$, $D_2$, and $D_3$ be the events of going through doors 1, 2, and 3 respectively.

Using LoTE, we have

\begin{aligned} E(W) &= E(W|D_1)P(D_1) + E(W|D_2)P(D_2) + E(W|D_3)P(D_3) \\ &= 3 \frac{1}{3} + (5 + E(W)) \frac{1}{3} + (7 + E(W)) \frac{1}{3} \\ \implies\frac{1}{3} E(W) &= 5 \\ E(W) &= 15\end{aligned}