Difference between revisions of "2019 AMC 8 Problems/Problem 18"
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==Solution 4== | ==Solution 4== | ||
− | To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>. | + | To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} * \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} * \frac{2}{6}</math>. If you add them together, you get <math>\frac{16}{36} + \frac{4}{36}</math> = <math>\boxed{(\textbf{C}) \frac{5}{9}}</math>. |
==Solution 5 (Casework)== | ==Solution 5 (Casework)== |
Revision as of 17:34, 26 October 2020
Contents
Problem 18
The faces of each of two fair dice are numbered , , , , , and . When the two dice are tossed, what is the probability that their sum will be an even number?
Solution 1
The approach to this problem: There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the numbers add up to an odd number and subtract the answer from .
How to solve the problem: The probability of getting an odd number first is . In order to make the sum odd, we must select an even number next. The probability of getting an even number is . Now we multiply the two fractions: . However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do . This is the probability of getting an odd number. In order to get the probability of getting an even number we do
- ViratKohli2018 (VK18)
Solution 2
We have a die with evens and odds on both dies. For the sum to be even, the rolls must consist of odds or evens.
Ways to roll odds (Case ): The total number of ways to roll odds is , as there are choices for the first odd on the first roll and choices for the second odd on the second roll.
Ways to roll evens (Case ): Similarly, we have ways to roll {36}=\frac{20}{36}=\frac{5}{9}\framebox{C}$.
Solution 3 (Complementary Counting)
We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is , and the probability of an odd is . We have to multiply by because the even and odd can be in any order. This gets us , so the answer is . - juliankuang
Solution 4
To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is . The probability of getting 2 evens is . If you add them together, you get = .
Solution 5 (Casework)
To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired .~binderclips1
Video Solution
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s ~ MathEx
Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.