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A point (x,y) is randomly selected such that 0 \le x \le 3  and . 0 \le y \le 6. What is the probability that x + y \le 4? Express your answer as a common fraction.

 

 

I tried 2/9 and apparently that wasn't correct.

 Dec 13, 2018
edited by ANotSmartPerson  Dec 13, 2018
 #1
avatar+5095 
+3

\(\text{The point }p \text{ is described by a 2D joint uniform distribution}\\ p(x,y) = \dfrac{1}{18},~0\leq x \leq 3,~0\leq y\leq 6\\ P[x+y \leq 4] = P[y\leq 4-x] = \\ \displaystyle \int_0^3 \int_0^{4-x}~\dfrac{1}{18}~dy~dx = \\ \displaystyle \int_0^3 \dfrac{4-x}{18}~dx = \\ \left. \dfrac{4x-\frac{x^2}{2}}{18}\right |_0^3 = \dfrac{12-\frac 9 2}{18} = \dfrac{5}{12}\)

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 Dec 13, 2018
edited by Rom  Dec 13, 2018
 #2
avatar+238 
+1

Thank you!!! :)

 Dec 13, 2018

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