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

Answer:

Therefore P(max(X, Y) > 2) = 0.667

Step-by-step explanation:

X is a uniformly distributed random variable over the interval [0, 4]

height of X  = 1 / 4

X is a uniformly distributed random variable over the interval [0, 6]

height of Y = 1 /6

P(max(X, Y) > 2)  = max{$\frac{1}{4} \times (4 - 2), \frac{1}{6} \times (6 - 2)$]  = max[ 0.5 , 0.667]  = 0.667