Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

AoPS Wiki talk:Problem of the Day/June 27, 2011

Problem

AoPSWiki:Problem of the Day/June 27, 2011

Solutions

The product will be a multiple of $5$ only when the top is a multiple of $5$.

Solution 1

We will divide this into three cases.

CASE 1: THE 5 APPEARS ON THE LEFT DIE BUT NOT THE RIGHT

There is a $\dfrac{1}{6}$ chance of the left die getting the $5$, and a $\dfrac{5}{6}$ chance of the right die not getting a $5$, so the probability that both happen is $\dfrac{1}{6}\times\dfrac{5}{6}=\dfrac{5}{36}$.

CASE 2: THE 5 APPEARS ON THE RIGHT DIE BUT NOT THE LEFT

We can follow the same logic here. There is a $\dfrac{1}{6}$ chance of the right die getting the $5$, and a $\dfrac{5}{6}$ chance of the left die not rolling a $5$, so the probability that both happen is $\dfrac{1}{6}\times\dfrac{5}{6}=\dfrac{5}{36}$.

CASE 3: THE 5 APPEARS ON BOTH DIE

The probability this happens is $\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{36}$. (Do you see why?)

Adding these probabilities together, we find that the answer is $\dfrac{5}{36}+\dfrac{5}{36}+\dfrac{1}{36}=\boxed{\dfrac{11}{36}}$.

Solution 2

We will divide this into three cases once again, but this time the cases are a little different.

CASE 1: THE 5 APPEARS ON THE LEFT DIE

There is a $\dfrac{1}{6}$ chance of the left die getting the $5$, so this probability is $\dfrac{1}{6}$.

CASE 2: THE 5 APPEARS ON THE RIGHT DIE

As with the left die, there is a $\dfrac{1}{6}$ chance of rolling the $5$ on the top face.

CASE 3: THE 5 APPEARS ON BOTH DIE

The probability this happens is $\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{36}$.

Now, if we divide the cases this way, we have to be more careful about how we manipulate the probabilities. Both cases have the possibility of rolling two fives, so we have to subtract Case 3 once in order to not overcount based on PIE.

Following this logic, we have the total probability as $\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{36}=\dfrac{6}{36}+\dfrac{6}{36}-\dfrac{1}{36}=\boxed{\dfrac{11}{36}}$.

Solution 3

Make a sample space. Then, count in the sample space all multiples of five. There are eleven multiples, yielding a probability of $\boxed{\dfrac{11}{36}}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/June_27,_2011&oldid=39981"