$T w o i n d e p e n d e n t r a n d o m v a r i b l e s X a n d Y a r e i d e n t i c a l a n d u n i f o r m l y d i s t r i b u t e d i n t h e r a n g e [0, 2] . T h e p r o b a b i l i t y t h a t m i n [X, Y] i s l e s s t h a n 1 i s e q u a l t o$ ___________

0.75

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Solution

The correct option is A 0.75
$f_{x} (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{1}{2}; f o r 0 < x < 2 0; o t h e r w i s e \end{matrix} a n d f_{y} (y) = ⎧ ⎨ ⎩ \begin{matrix} \frac{1}{2}; 0 < y < 2 0 o t h e r w i s e \end{matrix} P {m i n [x, y] < 1} = {(X < 1) \cap (Y > X)} + P {(Y < 1) \cap (X > Y)} = \int_{0}^{x = 1} \int_{y = x}^{y = 2} (\frac{1}{4}) d y d x ∴ X a n d Y a r e i d e n t i c a l = 2 \int_{x = 0}^{x = 1} {[\frac{1}{4} y]}_{y = x}^{y = 2} d x = \frac{2}{4} \int_{0}^{x = 1} (2 - x) d x = \frac{1}{2} {[2 x - \frac{x^{2}}{2}]}_{x = 0}^{x = 1} = \frac{1}{2} \times \frac{3}{2} = 0.75$

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