> For the complete documentation index, see [llms.txt](https://justinthezhu.gitbook.io/stat110/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://justinthezhu.gitbook.io/stat110/untitled-2/questions.md).

# Questions

## Recruiting

Suppose the Harvard Consulting, Investment, and Tech Group (HCITG) currently consists of two freshmen and some number of upperclassmen. A new student joins the group, but she forgot to indicate what year she was in! At the next club meeting, a recruiter from Bayes Inc. comes in and plucks a lucky student to join their ranks.

{% tabs %}
{% tab title="Q" %}
Given that the student is a freshman, what is the probability that the student that just joined was a freshman? Suppose that freshmen and upperclassmen are equally likely to join HCITG.
{% endtab %}

{% tab title="A" %}
Let there be $$u$$ upperclassmen in the group, $$A$$ be the event that the new student was a freshman, and $$B$$ be the probability that the selected student was the freshman. We are clearly interested in $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ Here, $$P(B|A) = \frac{3}{3+u}$$, as we are conditioning that a freshman joined the group. Next, we can calculate $$P(B)$$ using LOTP.

$$
P(B) = P(B|A)P(A) + P\left(B|A^c\right)P\left(A^c\right) = \frac{1}{2}\cdot\frac{3}{3+u} + \frac{1}{2}\cdot  \frac{2}{3+u}
$$

Plugging everything in, we find that $$P(A|B) = \frac{3}{5}$$.
{% endtab %}
{% endtabs %}

## Russian Roulette

In a game of Russian Roulette, you find a revolver with **six** chambers containing two real bullets **side-by-side** and four empty chambers. You spin the chamber and point the gun at yourself... Click. No bullet.

{% tabs %}
{% tab title="Q" %}
It is your turn again, but do you want to spin the barrel again or just pull the trigger?
{% endtab %}

{% tab title="A" %}
The probability that you are shot if you spin the barrel again is simply $$\frac{1}{3}$$, as there are still two bullets and four empty chambers. To calculate the probability of being shot if the interviewer pulls the trigger right away, we can use conditional probability! Let $$B$$ be the event that the previous shot was not a bullet, and let $$A$$ be the event that the next adjacent shot in the chamber is a bullet. We know that $$P(B) = \frac{2}{3}$$ and $$P(A \cap B) = \frac{1}{6}$$, as there is only one way for the previous shot to not be a bullet while the next shot is a bullet. Therefore,

$$
P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{4}
$$

So you're better off taking the next shot without spinning the chamber!
{% endtab %}
{% endtabs %}

## Independence

Strategic practice problems to test your understanding of independence.

{% tabs %}
{% tab title="Q" %}
Is it possible for an event to be independent of itself?
{% endtab %}

{% tab title="A" %}
Let $$A$$ be an event. If $$A$$ is independent of itself, then $$P(A)=P(A \cap A)=P(A)^{2},$$ so $$P(A)$$ must be either 0 or 1 . So this is only possible in the extreme cases that the event has constant probability 0 or 1 .
{% endtab %}
{% endtabs %}

{% tabs %}
{% tab title="Q" %}
Is it always true that if $$A$$ and $$B$$ are independent events, then $$A^{c}$$ and $$B^{c}$$ are independent events? Show that it is, or give a counterexample.
{% endtab %}

{% tab title="A" %}
Yes, because we have

$$
P\left(A^{c} \cap B^{c}\right)=1-P(A \cup B)=1-(P(A)+P(B)-P(A \cap B))
$$

and since $$A$$ and $$B$$ are independent,

$$
1-P(A)-P(B)+P(A) P(B)=(1-P(A))(1-P(B))=P\left(A^{c}\right) P\left(B^{c}\right)
$$
{% endtab %}
{% endtabs %}