Difference between revisions of "2005 AMC 10B Problems/Problem 9"
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==Solution 2 (Complementary Counting)== | ==Solution 2 (Complementary Counting)== | ||
− | 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{\text{(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}} |
Revision as of 12:52, 22 December 2019
Problem
One fair die has faces ,
,
,
,
,
and another has faces
,
,
,
,
,
. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?
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
Case 2: The first die is even and the second die is odd.
The probability of this happening is
Adding these two probabilities will give us our final answer.
If you run out of time, you can see it is obviously greater than 1 so it narrows guesses.
Solution 2 (Complementary Counting)
We see can see that there are ways to get an even number, and other ways get us an odd. Therefore, if we subtract the
, we will get
. We can get an even either by getting
evens or
odds. Both cases give
, so
is
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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