probability

Question

two independent random variables X and Y  are uniformly distributed in[-1,1] probability that max(X,Y) is less than 1/2

1)3/4

2) 9/16

3)1/4

4)2/3

Comments

1 3/4

Answer 1

X,Y is randomly distributed between [-1,1]

So, it is distributed over point (1,1),(1,-1),(-1,-1),(-1,1)

Total area covered 2*2=4

But we have to take probability of area less than 1/2 (i.e.ABCD) 

Area of ABCD=3/2*3/2

=9/4

P(max(X,Y)<1/2) =(9/4) /4 =9/16

Comments

think it this way :

since it is randomly distributed therefore the probability of x < 1/2 is 3/4 since there is an area of 3/4 on the left of 1/2 in the range [-1,1] similarly prabability of y < 1/2 is 3/4 (don't need to think it as x and y coordinates just imagine it to be the sample having value less than 1/2 in range (-1,1))

now, for max(x,y) < 1/2 will be possible only when both x and y < 1/2 therefore probability = 3/4)(3/4)

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random variables X and Y  are uniformly distributed in[-1,1]

This means X and Y can take values between -1 and 1. Consider all the possible values that the two variables can attain and plot them on the XY plane. You'll get the bigger square shown in the pic above. Point (-1, 1) shows the condition when X takes the value -1 and Y takes the value 1.

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Basics

Probability =favorable event/samples space

Probability OF (max(X Y))<0.5 --> This is possible only in shaded area so favorable event is shaded area 3/2*3/2

Sample space = Total area 2*2

Therefore, P(max(X Y))= 9/16

 

Answer 2

Two 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