Question

Jenn randomly chooses a number $J$ from $1,$ $2,$ $3,$ $\ldots,$ $19,$ $20.$ Bela then randomly chooses a number $B$ from $1,$ $2,$ $3,$ $\ldots,$ $19,$ $20$ distinct from $J.$ The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 29

Concepts:basic probability counting pairs

Difficulty rating: 1950

Small Hint:

All $20 \cdot 19$ ordered pairs $(J, B)$ with $B \neq J$ are equally likely

Big Hint:

For each $J$ there are $19 - J$ values of $B$ with $B \ge J + 2,$ so the favorable count is $1 + 2 + \cdots + 18$

Solution:

There are $20 \cdot 19 = 380$ equally likely ordered pairs $(J, B)$ with $B \neq J.$ The condition $B \ge J + 2$ allows $19 - J$ choices of $B$ for each $J \le 18,$ so the number of favorable pairs is $\begin{aligned} &\sum_{J=1}^{18} (19 - J) \\ &= 18 + 17 + \cdots + 1 = 171. \end{aligned}$

The probability is $\frac{171}{380} = \frac{9}{20},$ so $m + n = 9 + 20 = 29.$

2019 AIME I Problem 2

Answer: 29

Difficulty rating: 1950

Small Hint:

Big Hint:

Solution:

Problem 2 in Other Years