Question

Let X and Y be a uniformly distributed random variable over the interval [0, 4] and [0, 6] respectively. If X and Y are independent
events, then compute the probability, PP(max(X, Y ) > 2)

Answer 1

5/6 is the  probability, PP(max(X, Y ) > 2)

Step-by-step explanation:

Probability of X > 2  =   2/4  = 1/2

Probability of Y > 2  =  4/6  = 2/3

PP(max(X, Y ) > 2)  = P(X > 2)  + P(Y > 2)  - P ( X  ∩ Y > 2)

= (1/2)  + (2/3)  - (1/2)(2/3)

= 1/2  + 2/3  - 1/3

= (3 + 4 - 2)/6

= 5/6

5/6 is the  probability, PP(max(X, Y ) > 2)

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Answer 2

Answer:

Let X and Y be a uniformly distributed random variable over the interval [0, 4] and [0, 6] respectively. If X and Y are independent events, then compute the probability, P(max(X,Y)>3)

a.1/6

b.5/6

c.2/3

d.1/2

e.2/6

f.5/8