Consider 10 boxes and 10 balls. We throw each ball, uniformly, in a random box. What is the probability that, after we have thrown all 10 balls, none of the 10 boxes is empty?
a) $\frac{10!}{10^{10}}$
b) $\frac{10^{10}}{10!}$
c) $1 - \frac{10 \cdot (9/10)^{10}}{10^{10}}$
d) None of the above.
Solution: 2014 Fall Midterm - 16
Author: Michiel SmidQuestion
Solution
We can write it out $…$
The first ball has a $10/10$ possibility of getting into an empty box
The second ball has $9/10$ possibility of getting into an empty box
Assuming the previous balls are in different boxes, the third ball has 8/10 possibility of getting into an empty box
We can write this down as $…$
$=\frac{10}{10} \cdot \frac{9}{10} \cdot \frac{8}{10} \cdot \frac{7}{10} \cdot \frac{6}{10} \cdot \frac{5}{10} \cdot \frac{4}{10} \cdot \frac{3}{10} \cdot \frac{2}{10} \cdot \frac{1}{10}$
$=\frac{10!}{{10}^{10}}$