# The probability of sum $x+y$ to be greater than $20$

The variable $$x$$ takes a value between $$0$$ and $$10$$ with uniform probability distribution.The variable $$y$$ takes a value between $$0$$ and $$20$$ with uniform probability distribution. The probability of the sum of variables $$(x+y)$$ being greater than $$20$$ is

a) $$0.5\ \quad$$ b) $$0\ \quad$$ c) $$0.25\ \quad$$ d) $$0.33\ \quad$$

My try:

for uniform probability distribution, $$f(x)=\frac{1}{10-0}=\frac{1}{10}$$

$$f(y)=\frac{1}{20-0}=\frac{1}{20}$$

$$f(x+y) =\frac{1}{\infty-20}=0$$

$$\therefore P(x+y>20)=\int_0^{\infty}f(t)dt=0$$

I am not sure if i am right. Please correct me if i am wrong. please help me I am weak in probability problems.

• Unfortunately, this is quite wrong. In order to answer this question, you need to know whether or not the variables are independent. Assuming they are, $P(X+Y>20)$ is found by doing a double integral of the joint pdf of $X$ and $Y$ (which since they are independent, is of the form $f_X(x)f_Y(y)$) over the region in the plane defined by $x+y>20$. – Mike Earnest Feb 7 at 22:33

First, we have $$P[x\gt\alpha]=\frac{10-\alpha}{10}[0\le\alpha\le10]$$ Therefore, \begin{align} \frac1{20}\int_0^{20}P[x\gt20-y]\,\mathrm{d}y &=\frac1{20}\int_{10}^{20}\frac{y-10}{10}\,\mathrm{d}y\\ &=\frac1{200}\int_0^{10}y\,\mathrm{d}y\\[3pt] &=\frac14 \end{align}
We can also notice that $$(x,y)$$ is uniformly distributed over a $$10\times20$$ rectangle and the triangle over which $$x+y\gt20$$ is one quarter of that rectangle.