# Two Problems on Probability

$1^{st}$ Question:

Consider two independent random variables $X$ and $Y$ having probability density functions uniform in the interval $[−1, 1]$. What is The probability that $X^2+Y^2\gt1$ ?

$2^{nd}$ Question:

What is the value of $\lambda$ such that $Pr(X\gt mean\{X\})=\frac{1}{e}$, where pdf of $X$ is $p_X(x)=\lambda e^{-\lambda x},\ x\geq0,\ λ\gt0$ ?

Here I'm not sure whether the $mean\{X\}$ represents expectation $E(X)$ or not.

(1) $$f\left( x,y \right)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4},~-1<x<1,~-1<y<1.$$ then $$P\left( {{X}^{2}}+{{Y}^{2}}>1 \right)=\frac{1}{4}\cdot \left( area \right)=1-\frac{\pi }{4}.$$

(2) $E(X)=1/\lambda$ $$P\left( X>\frac{1}{\lambda } \right)=\frac{1}{e}\Rightarrow \underset{\frac{1}{\lambda }}{\overset{\infty }{\mathop \int }}\,\lambda {{e}^{-\lambda x}}dx=\frac{1}{e}$$ for every $(0<\lambda<\infty)$.

• What do you mean by $f(x,y)$ here ? – dibyendu Oct 22 '13 at 19:46
• You got two variables, X and Y. – Rrjrjtlokrthjji Oct 22 '13 at 20:03
• He means the joint density function. The individual PDFs are independent so you multiply to get the joint. – Patrick Oct 22 '13 at 21:32

First question: Draw the square in which our joint density function lives. It is easier to find the probability that $X^2+Y^2\le 1$.

This is the part of the square that lies within the unit circle. The circle has area $\pi$, the square has area $4$, so the probability that $X^2+Y^2\le 1$ is $\frac{\pi}{4}$. Thus the desired probability is $1-\frac{\pi}{4}$.

Second question: Probably by mean they mean $E(X)$. This is $\frac{1}{\lambda}$. The probability that $X\gt t$ is $e^{-\lambda t}$. If $t=\frac{1}{\lambda}$, this is $e^{-1}$. Note that the result is independent of $\lambda$. So trick question, any $\lambda\gt 0$ has the required property.