+0  
 
0
58
2
avatar+176 

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+3580 
+3

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

.
 Dec 13, 2018
edited by Rom  Dec 13, 2018
 #2
avatar+176 
+1

Thank you!!! :)

 Dec 13, 2018

9 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.