2016 AIME I Problem 2

Question

Two dice appear to be standard dice with their faces numbered from

1

to

6,

but each die is weighted so that the probability of rolling the number

k

is directly proportional to

k.

The probability of rolling a

7

with this pair of dice is

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

On each die the probability of rolling

k

is

\frac{k}{21},

since

1 + 2 + \cdots + 6 = 21

The probability of a total of

7

is

\sum_{k=1}^{6} \frac{k}{21} \cdot \frac{7 - k}{21}

Since

1 + 2 + \cdots + 6 = 21,

each die rolls

k

with probability

\frac{k}{21}.

A total of

7

arises from the pairs

(k, 7-k)

for

k = 1, \ldots, 6,

so its probability is

\begin{aligned} &\scriptsize \frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} \\ &= \frac{56}{441} = \frac{8}{63}. \end{aligned}

Thus

m + n = 8 + 63 = 71.

Problem 2 in other years