Difference between revisions of "2019 AMC 8 Problems/Problem 18"
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<math>\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}</math> | <math>\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}</math> | ||
− | ==Solution | + | ==Solution 1== |
− | We have | + | We have <math>2</math> dice with <math>2</math> evens and <math>4</math> odds on each die. For the sum to be even, the 2 rolls must be <math>2</math> odds or <math>2</math> evens. |
− | Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to | + | Ways to roll <math>2</math> odds (Case <math>1</math>): The total number of ways to obtain <math>2</math> odds on 2 rolls is <math>4*4=16</math>, as there are <math>4</math> possible odds on the first roll and <math>4</math> possible odds on the second roll. |
− | Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to | + | Ways to roll <math>2</math> evens (Case <math>2</math>): Similarly, we have <math>2*2=4</math> ways to obtain 2 evens. Probability is <math>\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>. |
− | + | ==Solution 2 (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 <math>\frac{1}{3}</math> , and the probability of an odd is <math>\frac{2}{3}</math>. We have to multiply by <math>2!</math> because the even and odd can be in any order. This gets us <math>\frac{4}{9}</math>, so the answer is <math>1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is <math>\frac{4}{6} \times \frac{4}{6}</math>. The probability of getting 2 evens is <math>\frac{2}{6} \times \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>. | ||
+ | |||
+ | ==Video Solution== | ||
− | + | ==Video Solution by Math-X (First understand the problem!!!)== | |
+ | https://youtu.be/IgpayYB48C4?si=UsI0Wu2OeYT813rn&t=5524 | ||
− | ~ | + | ~Math-X |
− | ==Solution | + | |
− | + | https://youtu.be/8fF55uF64mE | |
− | ==Solution | + | |
− | + | - Happytwin | |
+ | |||
+ | https://www.youtube.com/watch?v=_IK58KFUYpk ~David | ||
+ | |||
+ | https://www.youtube.com/watch?v=EoBZy_WYWEw | ||
+ | |||
+ | Associated video | ||
+ | |||
+ | https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | Solution detailing how to solve the problem: | ||
+ | |||
+ | https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/8gl4rCZMUFI | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/zrmHM_l1UkI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== | ||
+ | https://youtu.be/Xm4ZGND9WoY | ||
+ | |||
+ | ~Hayabusa1 | ||
==See Also== | ==See Also== |
Latest revision as of 18:29, 15 August 2024
Contents
- 1 Problem 18
- 2 Solution 1
- 3 Solution 2 (Complementary Counting)
- 4 Solution 3
- 5 Video Solution
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution
- 8 Video Solution
- 9 Video Solution (CREATIVE THINKING!!!)
- 10 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 11 See Also
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
We have dice with
evens and
odds on each die. For the sum to be even, the 2 rolls must be
odds or
evens.
Ways to roll odds (Case
): The total number of ways to obtain
odds on 2 rolls is
, as there are
possible odds on the first roll and
possible odds on the second roll.
Ways to roll evens (Case
): Similarly, we have
ways to obtain 2 evens. Probability is
, or
.
Solution 2 (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
.
Solution 3
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
=
.
Video Solution
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=UsI0Wu2OeYT813rn&t=5524
~Math-X
- Happytwin
https://www.youtube.com/watch?v=_IK58KFUYpk ~David
https://www.youtube.com/watch?v=EoBZy_WYWEw
Associated video
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s
Video Solution
Solution detailing how to solve the problem:
https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19
Video Solution
~savannahsolver
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
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.