Difference between revisions of "2016 AMC 10A Problems/Problem 12"
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==Solution 3== | ==Solution 3== | ||
The probability that the product is odd, allowing duplication of the integers, is just <math>\left( \frac{1}{2} \right) ^3 = \frac{1}{8}</math>. Since forbidding duplication reduces the probability of all three integers being odd, we see <math>p<\dfrac{1}{8}</math> and our answer is <math>\boxed{\textbf{(A) }}</math>. | The probability that the product is odd, allowing duplication of the integers, is just <math>\left( \frac{1}{2} \right) ^3 = \frac{1}{8}</math>. Since forbidding duplication reduces the probability of all three integers being odd, we see <math>p<\dfrac{1}{8}</math> and our answer is <math>\boxed{\textbf{(A) }}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/dHY8gjoYFXU?t=300 | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=11|num-a=13}} | {{AMC10 box|year=2016|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 03:38, 19 May 2020
Problem
Three distinct integers are selected at random between 
and 
, inclusive. Which of the following is a correct statement about the probability 
that the product of the three integers is odd?
Solution 1
For the product to be odd, all three factors have to be odd. The probability of this is 
.
, but 
and 
are slightly less than 
. Thus, the whole product is slightly less than 
, so 
.
Solution 2
For the product to be odd, all three factors have to be odd. There are a total of 
ways to choose 3 numbers at random, and there are 
to choose 3 odd numbers. Therefore, the probability of choosing 3 odd numbers is 
. Simplifying this, we obtain 
, which is slightly less than 
, so our answer is .
Solution 3
The probability that the product is odd, allowing duplication of the integers, is just 
. Since forbidding duplication reduces the probability of all three integers being odd, we see 
and our answer is 
.
Video Solution
https://youtu.be/dHY8gjoYFXU?t=300
~IceMatrix
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
