Difference between revisions of "2016 AIME II Problems/Problem 2"

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==Problem==
 
There is a <math>40\%</math> chance of rain on Saturday and a <math>30\%</math> 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 <math>\frac{a}{b}</math>, where <math>a</math> and <math>b</math> are relatively prime positive integers. Find <math>a+b</math>.
 
There is a <math>40\%</math> chance of rain on Saturday and a <math>30\%</math> 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 <math>\frac{a}{b}</math>, where <math>a</math> and <math>b</math> are relatively prime positive integers. Find <math>a+b</math>.
 
==Solution==
 
==Solution==

Revision as of 17:14, 22 March 2018

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 $37+70=\boxed{107}$.

Solution by Shaddoll

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