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Difference between revisions of "2005 AMC 10B Problems/Problem 9"

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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>
 
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{\textbf{(D) }\ \dfrac{5}{9}}</math>
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Adding these two probabilities will give us our final answer. <math>\dfrac{4}{9}+\dfrac{1}{9}=\boxed{\textbf{(D) }\dfrac{5}{9}}</math>
  
 
If you run out of time, you can see it is obviously greater than <math>\frac{1}{2}</math> so it narrows guesses.
 
If you run out of time, you can see it is obviously greater than <math>\frac{1}{2}</math> so it narrows guesses.
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==Solution 2 (Complementary Counting)==
 
==Solution 2 (Complementary Counting)==
  
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We see can see that there are <math>2</math> ways to get an even number, and other ways get us an odd. Therefore, if we subtract the <math>P(\text{even})</math>, we will get <math>P(\text{odd})</math>. We can get an even either by getting <math>2</math> evens or <math>2</math> odds. Both cases give <math>\frac{2}{9}</math>, so <math>P(\text{odd})</math> is <math>1-2 \cdot \frac{2}{9}=\boxed{\text{(D)} \frac{5}{9}}</math>
+
We see can see that there are <math>2</math> ways to get an even number, and other ways get us an odd. Therefore, if we subtract the <math>P(\text{even})</math>, we will get <math>P(\text{odd})</math>. We can get an even either by getting <math>2</math> evens or <math>2</math> odds. Both cases give <math>\frac{2}{9}</math>, so <math>P(\text{odd})</math> is <math>1-2 \cdot \frac{2}{9}=\boxed{\textbf{(D) }\frac{5}{9}}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2005|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2005|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:33, 15 December 2021

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?

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

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. $\dfrac{4}{9}+\dfrac{1}{9}=\boxed{\textbf{(D) }\dfrac{5}{9}}$

If you run out of time, you can see it is obviously greater than $\frac{1}{2}$ so it narrows guesses.

Solution 2 (Complementary Counting)

We see can see that there are $2$ ways to get an even number, and other ways get us an odd. Therefore, if we subtract the $P(\text{even})$, we will get $P(\text{odd})$. We can get an even either by getting $2$ evens or $2$ odds. Both cases give $\frac{2}{9}$, so $P(\text{odd})$ is $1-2 \cdot \frac{2}{9}=\boxed{\textbf{(D) }\frac{5}{9}}$

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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