Difference between revisions of "1997 AHSME Problems/Problem 10"
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The overall probability is <math>\frac{1}{9} + \frac{4}{9} = \frac{5}{9}</math>, and the answer is <math>\boxed{D}</math>. | The overall probability is <math>\frac{1}{9} + \frac{4}{9} = \frac{5}{9}</math>, and the answer is <math>\boxed{D}</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1997|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:12, 5 July 2013
Problem
Two six-sided dice are fair in the sense that each face is equally likely to turn up. However, one of the dice has the 
replaced by 
and the other die has the replaced by . When these dice are rolled, what is the probability that the sum is an odd number?
Solution
On the first die, the chance of an odd number is 
, and the chance of an even number is 
.
On the second die, the chance of an odd number is , and the chance of an even number is 
.
To get an odd sum, we need exactly one even and one odd.
The odds of the first die being even and the second die being odd is 
The odds of the first die being odd and the second die being even is 
.
The overall probability is 
, and the answer is 
.
See also
| 1997 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
