Solution: 2017 Fall Final - 20
Author: Michiel SmidQuestion
You choose a uniformly random element, say $a$, from the set $\{1,2,\dots,100\}$, and you choose a
uniformly random element, say $b$, from the same set $\{1,2,\dots,100\}$. ($a$ and $b$ are chosen
independently of each other.) Define the random variable $X$ to be
- X = $\max(a,b)$.
(a)
${\sum_{k=1}^{100}} k \cdot \left( \frac{1+2(k-1)}{100^{2}} \right)$
(b)
${\sum_{k=1}^{100}} k \cdot \frac{k^{2}}{100^{2}}$
(c)
${\sum_{k=1}^{100}} k \cdot \frac{k(k-1)}{100^{2}}$
(d)
${\sum_{k=1}^{100}} k \cdot \frac{2k}{100^{2}}$
Solution
Let’s explain why some are wrong and one is right
- ${\sum_{k=1}^{100}} k \cdot \frac{k^{2}}{100^{2}}$
This equation says, \enquote{The probability that k is the max value and each value has 1 to k values to be}
This doesn't work because it's saying that the max value is k if $a$ is 1 and $b$ is 1 - ${\sum_{k=1}^{100}} k \cdot \frac{k(k-1)}{100^{2}}$
This equation says, \enquote{a can be any value from 1 to k and b can be any value from 1 to k-1}
This doesn't work because it's saying that the max value is k if $a$ is $k-1$ and $b$ is $k-2$ - ${\sum_{k=1}^{100}} k \cdot \frac{2k}{100^{2}}$
This equation says, \enquote{a can be any value from 1 to k and b can be any value from 1 to k}
This doesn't work because it's saying that the max value is k if $a$ is 1 and $b$ is 1 - ${\sum_{k=1}^{100}} k \cdot bigg( \frac{1+2(k-1)}{100^{2}}bigg)$
This one actually has multiple possibilities
- $a = k$ and $b = k$: 1 possibility
- $a = k$ and $b$ is any value less than k: $k(k-1)$ possibilities
- $a$ is any value less than k and $b = k$: $(k-1)k$ possibilities