Question

Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 47

Concepts:basic probability combinations casework

Difficulty rating: 2390

Small Hint:

Compute separately the probability that all three math books land in the $3$ -box, the $4$ -box, and the $5$ -box

Big Hint:

For the box holding $k$ books, that probability is $\frac{\binom{9}{k-3}}{\binom{12}{k}};$ add the three fractions

Solution:

Focus on one box at a time. The box of $k$ books receives a uniformly random $k$ -subset of the $12$ books, so the probability that it contains all three math books is $\frac{\binom{9}{k-3}}{\binom{12}{k}}.$ For $k = 3,$ $4,$ and $5$ this gives $\frac{1}{220},$ $\frac{9}{495} = \frac{1}{55},$ and $\frac{36}{792} = \frac{1}{22}.$

The events are disjoint, so the total probability is $\begin{aligned} \frac{1}{220} + \frac{1}{55} + \frac{1}{22} &= \frac{1 + 4 + 10}{220} \\ &= \frac{15}{220} = \frac{3}{44}, \end{aligned}$ and $m + n = 3 + 44 = 47.$

2013 AIME I Problem 6

Answer: 47

Difficulty rating: 2390

Small Hint:

Big Hint:

Solution:

Problem 6 in Other Years