The random variable X is uniformly distributed over the interval [0,2]. This means that the probability density function (pdf) is 1/(b-a) = 1/2 for 0 ≤ X ≤ 2 and 0 otherwise.

The probability that X is less than or equal to 0.4 is the integral of the pdf from 0 to 0.4. Since the pdf is a constant 1/2, this integral is simply (0.4 - 0) * (1/2) = 0.2.

The probability that X is greater than 1.2 is the integral of the pdf from 1.2 to 2. Again, since the pdf is a constant 1/2, this integral is (2 - 1.2) * (1/2) = 0.4.

The events {X ≤ 0.4} and {X 1.2} are mutually exclusive, so the probability of their union is the sum of their probabilities. Therefore, P{X ≤ 0.4 or X 1.2} = 0.2 + 0.4 = 0.6.

Question

The random variable X is uniformly distributed over the interval [0,2]. This means that the probability density function (pdf) is 1/(b-a) = 1/2 for 0 ≤ X ≤ 2 and 0 otherwise.

The probability that X is less than or equal to 0.4 is the integral of the pdf from 0 to 0.4. Since the pdf is a constant 1/2, this integral is simply (0.4 - 0) * (1/2) = 0.2.

The probability that X is greater than 1.2 is the integral of the pdf from 1.2 to 2. Again, since the pdf is a constant 1/2, this integral is (2 - 1.2) * (1/2) = 0.4.

The events {X ≤ 0.4} and {X > 1.2} are mutually exclusive, so the probability of their union is the sum of their probabilities. Therefore, P{X ≤ 0.4 or X > 1.2} = 0.2 + 0.4 = 0.6.

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The random variable X is uniformly distributed over the interval [0,2]. This means that the probability density function (pdf) is 1/(b-a) = 1/2 for 0 ≤ X ≤ 2 and 0 otherwise.

The probability that X is less than or equal to 0.4 is the integral of the pdf from 0 to 0.4. Since the pdf is a constant 1/2, this integral is simply (0.4 - 0) * (1/2) = 0.2.

The probability that X is greater than 1.2 is the integral of the pdf from 1.2 to 2. Again, since the pdf is a constant 1/2, this integral is (2 - 1.2) * (1/2) = 0.4.

The events {X ≤ 0.4} and {X > 1.2} are mutually exclusive, so the probability of their union is the sum of their probabilities. Therefore, P{X ≤ 0.4 or X > 1.2} = 0.2 + 0.4 = 0.6.

Suppose that the random variable X is continuous and takes its values uniformly over the interval from 0 to 2. What is the value of the probability P{X ≤ 0.4 or X > 1.2}?

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