# Questions

## 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 $$X|N \sim Bin(N,p)$$.

{% tabs %}
{% tab title="Q" %}
Find $$E(X)$$
{% endtab %}

{% tab title="A" %}
Use Adam's Law!

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

{% tabs %}
{% tab title="Q" %}
Find $$Var(X)$$
{% endtab %}

{% tab title="A" %}
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}
$$
{% endtab %}
{% endtabs %}

## 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 $$\sigma^2$$.

{% tabs %}
{% tab title="Q" %}
Find the mean of how much a random customer spends (note that the customer may spend nothing).
{% endtab %}

{% tab title="A" %}
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}
$$
{% endtab %}
{% endtabs %}

{% tabs %}
{% tab title="Q" %}
Find the variance of how much a random customer spends (note that the customer may spend nothing).
{% endtab %}

{% tab title="A" %}
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}
$$
{% endtab %}
{% endtabs %}

## 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.

{% tabs %}
{% tab title="Q" %}
What is the expected amount of time until he reaches safety?
{% endtab %}

{% tab title="A" %}
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

$$
\begin{aligned}
E(W) &= E(W|D\_1)P(D\_1) + E(W|D\_2)P(D\_2) + E(W|D\_3)P(D\_3) \\
&= 3 \frac{1}{3} + (5 + E(W)) \frac{1}{3} + (7 + E(W)) \frac{1}{3} \\
\implies\frac{1}{3} E(W) &= 5 \\
E(W) &= 15\end{aligned}
$$
{% endtab %}
{% endtabs %}


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