Difference between revisions of "2006 AMC 12B Problems/Problem 17"

(Solution)
 
Line 35: Line 35:
 
\boxed{\frac{8}{63}}
 
\boxed{\frac{8}{63}}
 
</cmath>
 
</cmath>
 +
 +
See also [[2016 AIME I Problems/Problem 2]]
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2006|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:10, 9 June 2016

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?

$\mathrm{(A)}\ \frac 4{63} \qquad \mathrm{(B)}\ \frac 18  \qquad \mathrm{(C)}\ \frac 8{63} \qquad \mathrm{(D)}\ \frac 16 \qquad \mathrm{(E)}\ \frac 27$

Solution

The probability of getting an $x$ on one of these dice is $\frac{x}{21}$.

The probability of getting $1$ on the first and $6$ on the second die is $\frac 1{21}\cdot\frac 6{21}$. Similarly we can express the probabilities for the other five ways how we can get a total $7$. (Note that we only need the first three, the other three are symmetric.)

Summing these, the probability of getting a total $7$ is: \[2\cdot\left( \frac 1{21}\cdot\frac 6{21} + \frac 2{21}\cdot\frac 5{21} + \frac 3{21}\cdot\frac 4{21} \right) =  \frac{56}{441} = \boxed{\frac{8}{63}}\]

See also 2016 AIME I Problems/Problem 2

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS