Difference between revisions of "2021 Fall AMC 10B Problems/Problem 14"
(→See Also) |
(→Solution) |
||
Line 22: | Line 22: | ||
Our answer is <math>1-\frac{5}{64}=\boxed{\textbf{(E)}\frac{59}{64}}</math>. | Our answer is <math>1-\frac{5}{64}=\boxed{\textbf{(E)}\frac{59}{64}}</math>. | ||
+ | |||
+ | ~kingofpineapplz | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=15|num-b=13}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=15|num-b=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:26, 23 November 2021
Problem 14
Una rolls standard
-sided dice simultaneously and calculates the product of the
numbers obtained. What is the probability that the product is divisible by
Solution
We will first find the probability that the product is not divisible by 4. We have 2 cases.
Case 1: The product is not divisible by .
We need every number to be odd, and since the chance we roll an odd number is our probability is
Case 2: The product is divisible by , but not by
.
We need numbers to be odd, and one to be divisible by
, but not by
. There is a
chance that an odd number is rolled, a
chance that we roll a number satisfying the second condition (only
and
work), and 6 ways to choose the order in which the even number appears.
Our probability is
Therefore, the probability the product is not divisible by is
.
Our answer is .
~kingofpineapplz
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.