Difference between revisions of "2019 AMC 8 Problems/Problem 18"
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We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the 2 rolls be <math>2</math> odds or <math>2</math> evens. | We have a <math>2</math> die with <math>2</math> evens and <math>4</math> odds on both dies. For the sum to be even, the 2 rolls 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 obtain <math>2</math> odds 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> 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 obtain 2 evens. Probability is <math>\frac{20}{36}=\frac{5}{9}</math>, or <math>\framebox{C}</math>. | 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>. |
Revision as of 17:51, 21 July 2021
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
We have a die with
evens and
odds on both dies. For the sum to be even, the 2 rolls 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
https://youtu.be/8fF55uF64mE - Happytwin
Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw
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
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.