Two random variables X and Y are distributed according tofx-yx-y-x-y-0-x-1- 0-y-10-otherwise-The probability PX-Y- 1 is -

P(x+y≤ 1)=∫Rf(x,y)dx dy =∫ _0^1∫ _0^1 x(x+y)dy dx =∫ _0^1 [ xy+y^22 ]_0^1 xdx =∫ _0^1 [ x(1 x)+(1 x)^22 ]dx = [ x^22 x^33 (1 x)^36 ]_0^1 =12 13+16 =13 ...

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Engineering Mathematics

Discrete and Continuous RV

Two random va...

Question

Two random variables X and Y are distributed according to

$f_{x, y} (x, y) = (\begin{matrix} (x, y), & 0 \leq x \leq 1, 0 \leq y \leq 1 0, & otherwise . \end{matrix}$

The probability $P (X + Y \leq 1) is$ .

0.33

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Solution

The correct option is A 0.33

$P (x + y \leq 1) = \int R f (x, y) d x d y$

$= \int_{0}^{1} \int_{0}^{1 - x} (x + y) d y d x$

$= \int_{0}^{1} {[x y + \frac{y^{2}}{2}]}_{0}^{1 - x} d x$

$= \int_{0}^{1} [x (1 - x) + \frac{(1 - x)^{2}}{2}] d x$

$= {[\frac{x^{2}}{2} - \frac{x^{3}}{3} - \frac{(1 - x)^{3}}{6}]}_{0}^{1}$

$= \frac{1}{2} - \frac{1}{3} + \frac{1}{6}$

$= \frac{1}{3}$

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Two random variables X and Y are distributed according to fx,y(x,y)=((x,y),0≤x≤1,0≤y≤10,otherwise. The probability P(X+Y≤1) is . 0.33

The correct option is A 0.33 P(x+y≤1)=∫Rf(x,y)dx dy =∫10∫1−x0(x+y)dy dx =∫10[xy+y22]1−x0dx =∫10[x(1−x)+(1−x)22]dx =[x22−x33−(1−x)36]10 =12−13+16 =13

Two random variables X and Y are distributed according to

$f_{x, y} (x, y) = (\begin{matrix} (x, y), & 0 \leq x \leq 1, 0 \leq y \leq 1 0, & otherwise . \end{matrix}$

The probability $P (X + Y \leq 1) is$ .

0.33