2006 AMC 10B Problem 21

P21-25~4 minPrint

Basic Probability Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed Exploit symmetry to reduce work or pair up objects

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?

Show solution without answering

Key insight

The weights 1 through 6 add to 21, so face k comes up with chance k/21. Then add up the chances of the six ordered ways to make 7.

Solution

1. Turn the ratio into chances. The chances are in the ratio $1:2:3:4:5:6$ , and $1 + 2 + 3 + 4 + 5 + 6 = 21$ . The six chances must add to $1$ , so a $1$ has chance $\frac{1}{21}$ , a $2$ has chance $\frac{2}{21}$ , and so on up to a $6$ with chance $\frac{6}{21}$ .
2. List the ways to make $7$ . Write (first die, second die):
$(1,6),\ (2,5),\ (3,4),\ (4,3),\ (5,2),\ (6,1).$
These are six different outcomes, because rolling $1$ then $6$ is not the same as rolling $6$ then $1$ .
3. Chance of each way. The dice do not affect each other, so multiply. For example, $(1,6)$ has chance $\frac{1}{21} \cdot \frac{6}{21} = \frac{6}{441}$ . Doing all six:

outcome	$(1,6)$	$(2,5)$	$(3,4)$	$(4,3)$	$(5,2)$	$(6,1)$
chance $\times 441$	$6$	$10$	$12$	$12$	$10$	$6$

Add them. $6 + 10 + 12 + 12 + 10 + 6 = 56$ , so the chance is $\frac{56}{441}$ .
Simplify. $56 = 7 \cdot 8$ and $441 = 7 \cdot 63$ , so $\frac{56}{441} = \frac{8}{63}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{8}{63}}$ .

Why this works

A ratio of chances becomes real chances once you divide each part by the total of the ratio, because all the chances together must make $1$ . After that it is an ordinary "two dice add to $7$ " question, except each outcome has its own weight. The table is symmetric, so you can work out three products and double.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The six probabilities are proportional to $1, 2, \ldots, 6$ , which add to $21$ , so face $k$ comes up with probability $\frac{k}{21}$ on each die.

A total of $7$ arises from the ordered outcomes $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ . The dice are independent, so the outcome $(j, 7-j)$ has probability $\frac{j(7-j)}{21^2}$ . Adding the six:
$\frac{1\cdot6 + 2\cdot5 + 3\cdot4 + 4\cdot3 + 5\cdot2 + 6\cdot1}{441} = \frac{6 + 10 + 12 + 12 + 10 + 6}{441} = \frac{56}{441}.$
Since $56 = 7 \cdot 8$ and $441 = 7 \cdot 63$ , this reduces to $\frac{8}{63}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{8}{63}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques