Random Variables and Expected Values Problem

Question

Let $ b\geq 1$,$c\geq 1$ and $h\geq 1$ be integers, and let $n=b+c+h$.

You have $b$ balls, denoted $B_1, B_2, \dots , B_B$, $c$ cubes $C_1, C_2, \dots , C_c$, and $h$ hexagons $H_1, H_2, \dots , H_h$. Let $m\geq 1$ be an integer with $m\leq b$ and $m\leq n-b$.

All $n$ objects are in a box, from this box you choose a uniformly random subset consisting of $m$ objects.

Define the random variables:

• X = the number of balls in the chosen subset
• Y = the number of cubes in the chosen subset
• Z = the number of hexagons in a chosen subset

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Two questions about this:

1. How would I start, or go about proving that $\Pr(X=k) = \frac{\binom{b}{k} \binom{n-b}{m-k}}{\binom{n}{m}}$
2. How would I prove that $\sum_{k=0}^m \frac{k\binom{b}{k} \binom{n-b}{m-k}}{\binom{n}{m}} = \frac{bm}{n}$

I'm not too sure where to start for these, so any help is appreciated. If there's anything I can do to improve my question I'm happy to get input :)

Answer 1

1. This problem can be solved by a counting argument. $X = k$ is the event that we choose exactly $k$ balls (and therefore, $m-k$ other objects) when picking $m$ objects out of all $n$. You can count the number of ways to do this by multiplying $\binom{b}{k}$ ways to choose $k$ balls and $\binom{n-b}{m-k}$ ways to choose the $m-k$ non-balls. This is divided by the total number of ways to choose $m$ objects out of $n$, $\binom{n}{m}$.
2. Hint: Rewrite the sum as $\sum_{k=0}^m k \Pr(X = k)$. But this is just the expectation of $X$; you can solve for it using linearity of expectation. Namely, let $X_i$ be the indicator random variable for the $i^{th}$ object being a ball; $E[X_i] = \Pr(X_i = 1) = \frac{b}{n}$ (each object is equally likely to be chosen in any spot). Summing over $1 \leq i \leq m$ gives: $$ E[X] = E\left[ \sum_{i=1}^m X_i \right] = \sum_{i=1}^m E[X_i] = \frac{bm}{n} $$