Difference between revisions of "2006 AMC 10B Problems/Problem 21"

Revision as of 08:28, 27 October 2021

Problem

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?

$\mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7}$

Solution

Let $x$ be the probability of rolling a $1$ . The probabilities of rolling a $2$ , $3$ , $4$ , $5$ , and $6$ are $2x$ , $3x$ , $4x$ , $5x$ , and $6x$ , respectively.

The sum of the probabilities of rolling each number must equal 1, so

$x+2x+3x+4x+5x+6x=1$

$21x=1$

$x=\frac{1}{21}$

So the probabilities of rolling a $1$ , $2$ , $3$ , $4$ , $5$ , and $6$ are respectively $\frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}$ , and $\frac{6}{21}$ .

The possible combinations of two rolls that total $7$ are: $(1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ; (6,1)$

The probability P of rolling a total of $7$ on the two dice is equal to the sum of the probabilities of rolling each combination.

$P = \frac{1}{21}\cdot\frac{6}{21}+\frac{2}{21}\cdot\frac{5}{21}+\frac{3}{21}\cdot\frac{4}{21}+\frac{4}{21}\cdot\frac{3}{21}+\frac{5}{21}\cdot\frac{2}{21}+\frac{6}{21}\cdot\frac{1}{21}=\frac{8}{63} \Rightarrow C$

Solution 2 (Not as bashy)

(Alcumus solution) On each die the probability of rolling $k$ , for $1\leq k\leq 6$ , is $\frac{k}{1+2+3+4+5+6}=\frac{k}{21}.$ There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs $(1,6)$ , $(2,5)$ , $(3,4)$ , $(4,3)$ , $(5,2)$ , and $(6,1)$ . Thus the probability of rolling a total of 7 is $\frac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}=\frac{56}{21^2}=\boxed{\frac{8}{63}}.$

Soilution 3 (intuitive)

There are $6$ ways to get the sum of $7$ of the dice. Let's do case by case.

Case $1$ : $\frac {1}{21} \cdot \frac {6}{21} = \frac {6}{441}$ .

Case $2$ : $\frac {2}{21} \cdot \frac {5}{21} = \frac {10}{441}$ .

Case $3$ : $\frac {3}{21} \cdot \frac {4}{21} = \frac {12}{441}$ .

The rest of the cases are symmetric to these cases above. We have $2 \cdot \frac {28}{441}$ . We have $\frac {56}{441} = \frac {8}{63}$ . Therefore, our answer is $\boxed {\frac {8}{63}}$

~Arcticturn

Video Solution

https://youtu.be/IRyWOZQMTV8?t=3057

~ pi_is_3.14

@@ Line 34: / Line 34: @@
 ==Soilution 3 (intuitive)==
+There are <math>6</math> ways to get the sum of <math>7</math> of the dice. Let's do case by case.
+Case <math>1</math>: <math>\frac {1}{21} \cdot \frac {6}{21} = \frac {6}{441}</math>.
+Case <math>2</math>: <math>\frac {2}{21} \cdot \frac {5}{21} = \frac {10}{441}</math>.
+Case <math>3</math>: <math>\frac {3}{21} \cdot \frac {4}{21} = \frac {12}{441}</math>.
+The rest of the cases are symmetric to these cases above. We have <math>2 \cdot \frac {28}{441}</math>. We have <math>\frac {56}{441} = \frac {8}{63}</math>. Therefore, our answer is <math>\boxed {\frac {8}{63}}</math>
+~Arcticturn
 == Video Solution ==

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search