# 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.

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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.

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 $P(A|B) = \frac{3}{5}$.

# 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.

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It is your turn again, but do you want to spin the barrel again or just pull the trigger?

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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!

# Independence

Strategic practice problems to test your understanding of independence.

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Is it possible for an event to be independent of itself?

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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 .

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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.

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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,

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