Difference between revisions of "2019 AMC 8 Problems/Problem 18"
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==Video Solution== | ==Video Solution== | ||
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+ | https://www.youtube.com/watch?v=EoBZy_WYWEw | ||
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+ | Associated video | ||
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s | https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s | ||
https://www.youtube.com/watch?v=_IK58KFUYpk | https://www.youtube.com/watch?v=_IK58KFUYpk | ||
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== Video Solution == | == Video Solution == |
Revision as of 02:51, 20 December 2022
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
- Happytwin
https://www.youtube.com/watch?v=EoBZy_WYWEw
Associated video
https://www.youtube.com/watch?v=H52AqAl4nt4&t=2s
https://www.youtube.com/watch?v=_IK58KFUYpk
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=94D1UnH7seo&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=19
Video Solution
~savannahsolver
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.