Difference between revisions of "2016 AIME II Problems/Problem 2"
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==Solution== | ==Solution== | ||
Let <math>x</math> be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have <math>\dfrac{3}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}</math> <math> \implies x=\dfrac{3}{14}</math>. Therefore, the probability that it doesn't rain on either day is <math>\left(1-\dfrac{3}{14}\right)\left(\dfrac{3}{5}\right)=\dfrac{33}{70}</math>. Therefore, the probability that rains on at least one of the days is <math>1-\dfrac{33}{70}=\dfrac{37}{70}</math>, so adding up the <math>2</math> numbers, we have <math>37+70=\boxed{107}</math>. | Let <math>x</math> be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have <math>\dfrac{3}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}</math> <math> \implies x=\dfrac{3}{14}</math>. Therefore, the probability that it doesn't rain on either day is <math>\left(1-\dfrac{3}{14}\right)\left(\dfrac{3}{5}\right)=\dfrac{33}{70}</math>. Therefore, the probability that rains on at least one of the days is <math>1-\dfrac{33}{70}=\dfrac{37}{70}</math>, so adding up the <math>2</math> numbers, we have <math>37+70=\boxed{107}</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=1|num-a=3}} | {{AIME box|year=2016|n=II|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:54, 18 October 2020
Problem
There is a 
chance of rain on Saturday and a 
chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
Let 
be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have 
. Therefore, the probability that it doesn't rain on either day is 
. Therefore, the probability that rains on at least one of the days is 
, so adding up the 
numbers, we have 
.
See also
| 2016 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
