2006 AMC 12B Problem 17

Question

For a particular peculiar pair of dice, the probabilities of rolling

1,

2,

3,

4,

5,

and

6

on each die are in the ratio

1 : 2 : 3 : 4 : 5 : 6.

What is the probability of rolling a total of

7

on the two dice?

\dfrac{4}{63}

\dfrac{1}{8}

\dfrac{8}{63}

\dfrac{1}{6}

\dfrac{2}{7}

The probability of rolling

k

is

\dfrac{k}{1+2+\cdots+6} = \dfrac{k}{21}

Sum

\dfrac{k}{21} \cdot \dfrac{7-k}{21}

over the pairs that total

7

Since the weights sum to

21,

the probability of rolling

k

is

\dfrac{k}{21}.

A total of

7

comes from

(1,6), (2,5), \ldots, (6,1),

so the 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, the correct answer is C.

Problem 17 in other years