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Difference between revisions of "2005 AMC 10B Problems/Problem 9"
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An odd sum requires either that the first die is even and the second is odd
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==Problem==
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or that the first die is odd and the second is even. The probability is
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One fair die has faces <math>1</math>, <math>1</math>, <math>2</math>, <math>2</math>, <math>3</math>, <math>3</math> and another has faces <math>4</math>, <math>4</math>, <math>5</math>, <math>5</math>, <math>6</math>, <math>6</math>. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?
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<math>(1/3*1/3)+(2/3*2/3)</math>=<math>1/9+4/9=5/9</math>='''D'''
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<math>\mathrm{(A)} \frac{1}{3} \qquad \mathrm{(B)} \frac{4}{9} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{5}{9} \qquad \mathrm{(E)} \frac{2}{3} </math>
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==Solution==
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In order to obtain an odd sum, exactly one out of the two dice must have an odd number. We can easily find the total probability using casework.
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'''Case 1''': The first die is odd and the second die is even.
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The probability of this happening is <math>\dfrac{4}{6}\times\dfrac{4}{6}=\dfrac{16}{36}=\dfrac{4}{9}</math>
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'''Case 2''': The first die is even and the second die is odd.
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The probability of this happening is <math>\dfrac{2}{6}\times\dfrac{2}{6}=\dfrac{4}{36}=\dfrac{1}{9}</math>
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Adding these two probabilities will give us our final answer. <math>\dfrac{4}{9}+\dfrac{1}{9}=\boxed{\mathrm{(D)}\ \dfrac{5}{9}}</math>
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==See Also==
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{{AMC10 box|year=2005|ab=B|num-b=8|num-a=10}}

Revision as of 18:04, 5 July 2011

Problem

One fair die has faces $1$, $1$, $2$, $2$, $3$, $3$ and another has faces $4$, $4$, $5$, $5$, $6$, $6$. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

$\mathrm{(A)} \frac{1}{3} \qquad \mathrm{(B)} \frac{4}{9} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{5}{9} \qquad \mathrm{(E)} \frac{2}{3}$

Solution

In order to obtain an odd sum, exactly one out of the two dice must have an odd number. We can easily find the total probability using casework.

Case 1: The first die is odd and the second die is even.

The probability of this happening is $\dfrac{4}{6}\times\dfrac{4}{6}=\dfrac{16}{36}=\dfrac{4}{9}$

Case 2: The first die is even and the second die is odd.

The probability of this happening is $\dfrac{2}{6}\times\dfrac{2}{6}=\dfrac{4}{36}=\dfrac{1}{9}$

Adding these two probabilities will give us our final answer. $\dfrac{4}{9}+\dfrac{1}{9}=\boxed{\mathrm{(D)}\ \dfrac{5}{9}}$


See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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
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