The random variables X and y are independent of each other and obey the uniform distribution in the interval [0,3]. Try to find P {min (x, y) ≤ 1} and P {max (x, y) > 1}

The random variables X and y are independent of each other and obey the uniform distribution in the interval [0,3]. Try to find P {min (x, y) ≤ 1} and P {max (x, y) > 1}

X and y are all subject to the uniform distribution on the interval [0, 3], so their probability density is f (x) = 13, 0 ≤ x ≤ 30, and the probability density f (x) (x) 13, 0 ≤ x ≤ 30, and other \8747; P {P {p {P {P {P {P, 0 ≤ x ≤ 30, other {P {P {(x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\= 1-13 · 13 = 89

In probability theory, P {max (x), what if we change Max to min?

Max (x, y) is a relatively large number
P {max (x, y) x > y or u > y > x
Min (x, y) is a relatively small number
P {min (x, y) U > y or Y > U > x

M = max (x, y), M = min (x, y) in probability and statistics,

M = max (x, y) means that M is a random variable
The greater of X and Y
M = min (x, y) means that M is a random variable
The smaller of X and Y

For random variables X, y, P (x ≥ 0, y ≥ 0) = 3 / 7, P (x ≥ 0) = P (Y ≥ 0) = 4 / 7, find P (max (x, y) ≥ 0), P (min (x, y)

P (max (x, y) ≥ 0) = P (x ≥ 0 or Y ≥ 0) = P (x ≥ 0) + P (Y ≥ 0) - P (x ≥ 0, y ≥ 0) = 4 / 7 + 4 / 7-3 / 7 = 5 / 7
P（min(X,Y)