Difference between revisions of "2019 AMC 12B Problems/Problem 16"
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==Solution== | ==Solution== | ||
| + | We solve for the probability by doing <math>\frac{1-(Probability of Equality)}{2}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
Revision as of 12:27, 14 February 2019
Problem
Solution
We solve for the probability by doing 
.
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 12 Problems and Solutions | |
