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>.
 
Solution by Shaddoll
 
  
 
== 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}}

Latest revision as of 18:54, 18 October 2020

Problem

There is a $40\%$ chance of rain on Saturday and a $30\%$ 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 $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a+b$.

Solution

Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\dfrac{3}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}$ $\implies x=\dfrac{3}{14}$. Therefore, the probability that it doesn't rain on either day is $\left(1-\dfrac{3}{14}\right)\left(\dfrac{3}{5}\right)=\dfrac{33}{70}$. Therefore, the probability that rains on at least one of the days is $1-\dfrac{33}{70}=\dfrac{37}{70}$, so adding up the $2$ numbers, we have $37+70=\boxed{107}$.

See also

2016 AIME II (ProblemsAnswer KeyResources)
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

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