We have an interval $[-2,4]$ and we select two random numbers $x$ and $y$.
a) What is the probability of $$y-2x\geq-6$$ if $x > y$.
I change $$y-2x\geq-6$$ into $$y\geq 2x-6$$ and get a graph and 'paint' everything above the line. The painted area's surface is $32$ and then get the probability
$$\frac{32}{36}=\frac{8}{9}$$Correct answer is $$\frac{7}{9}$$What am I doing wrong here?
Probability of $y-2x\geq-6$, if $x>y$
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probability
inequality
graphing-functions
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0Do I understand this correctly: $x$ and $y$ are each uniformly distributed on $[-2,4]$ and independent, and you are asking about the probability of $y-2x\geq-6$ conditional on $x>y$? – 2017-01-16
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0@Casper yes, that's right – 2017-01-16
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0Then I think you forget to make use of the condition. The area where $x>y$ inside the square $[-2,4]\times[-2,4]$ is 18 not 36. – 2017-01-16
2 Answers
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This is a conditional probability, since we are given that $x < y$. The region defined by $x < y$ is precisely half of the $6 \times 6$ square. So we have: $$ \Pr[y \geq 2x - 6 \mid y < x] = \frac{\Pr[2x - 6 \leq y < x]}{\Pr[y < x]} = \frac{14/36}{18/36} = \frac{14}{18} = \frac{7}{9} $$
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You are using the wrong textbook. The author of the text must have carelessly decided that the area of the triangle is base times height, or that $-6+2\cdot 4 = 4$. You will have an easier time understanding where you go wrong in problems if the answer presented is the right answer.
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0Where did you get -6 + 2*4 = 4? – 2017-01-16
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0I didn't. But to explain the authors incorrect answer, I surmise that this is what he did. – 2017-01-17