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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.

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  • $\begingroup$ 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 $x+y>20$. $\endgroup$ – Mike Earnest Feb 7 at 22:33
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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.

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