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Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 27, 2011"

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{{:AoPSWiki:Problem of the Day/June 27, 2011}}
 
{{:AoPSWiki:Problem of the Day/June 27, 2011}}
 
==Solutions==
 
==Solutions==
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{{potd_solution}}
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The product will be a multiple of <math>5</math> only when the top is a multiple of <math>5</math>.
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We will divide this into three cases.
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'''CASE 1: THE 5 APPEARS ON THE LEFT DIE BUT NOT THE RIGHT'''
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There is a <math>\dfrac{1}{6}</math> chance of the left die getting the <math>5</math>, and a <math>\dfrac{5}{6}</math> chance of the right die not getting a <math>5</math>, so the probability that both happen is <math>\dfrac{1}{6}\times\dfrac{5}{6}=\dfrac{5}{36}</math>.
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'''CASE 2: THE 5 APPEARS ON THE RIGHT DIE BUT NOT THE LEFT'''
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We can follow the same logic here.  There is a <math>\dfrac{1}{6}</math> chance of the right die getting the <math>5</math>, and a <math>\dfrac{5}{6}</math> chance of the left die not rolling a <math>5</math>, so the probability that both happen is <math>\dfrac{1}{6}\times\dfrac{5}{6}=\dfrac{5}{36}</math>.
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'''CASE 2: THE 5 APPEARS ON BOTH DIE'''
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The probability this happens is <math>\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{36}</math>.  (Do you see why?)
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Adding these probabilities together, we find that the answer is <math>\dfrac{5}{6}+\dfrac{5}{6}+\dfrac{1}{6}=\boxed{\dfrac{11}{36}}</math>.

Revision as of 20:32, 26 June 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 $5$.

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 2: 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 $\dfrac{5}{6}+\dfrac{5}{6}+\dfrac{1}{6}=\boxed{\dfrac{11}{36}}$.

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