2006 AMC 10B Problem 21

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

\tfrac{k}{21}

Sum

P(i)P(7-i)

over the six ordered pairs with

i+j=7

Each die shows

k

with probability

\tfrac{k}{1+2+\cdots+6}=\tfrac{k}{21}.

For a total of

7,

the ordered pairs

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

contribute

\begin{aligned} &\scriptsize\dfrac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}\\ &=\dfrac{56}{441}=\dfrac{8}{63}. \end{aligned}

Thus, the correct answer is C.

Problem 21 in other years