The independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max [X, Y] is less than 1/2 is

3/4

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9/16

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Solution

The correct option is B 9/16

Probability that max (x, y) is less than $\frac{1}{2}$

$= P [m a x (x, y) < \frac{1}{2}]$

$= P [- 1 \leq (x, y) < \frac{1}{2}]$

$= \frac{Area of Shaded Region}{Total Area}$

$= \frac{(\frac{3}{2}) (\frac{3}{2})}{2 \times 2} = \frac{9}{16}$

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The independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max [X, Y] is less than 1/2 is

The correct option is B 9/16 Probability that max (x, y) is less than 12 =P[max(x, y)<12] =P[−1≤(x, y)<12] =Area of Shaded RegionTotal Area =(32)(32)2×2=916

The correct option is B 9/16

Probability that max (x, y) is less than $\frac{1}{2}$

$= P [m a x (x, y) < \frac{1}{2}]$

$= P [- 1 \leq (x, y) < \frac{1}{2}]$

$= \frac{Area of Shaded Region}{Total Area}$

$= \frac{(\frac{3}{2}) (\frac{3}{2})}{2 \times 2} = \frac{9}{16}$