The question in the image asks about probabilities based on the uniform distribution of a random variable $X$ over the interval [0, 2].

The uniform distribution over [0, 2] means that the probability density function (PDF) is constant across this interval. Specifically, the PDF is $f(x) = \frac{1}{2}$ for $0 \leq x \leq 2$.

Now let's solve each probability:

1. $P(X > 1.46)$

This is the probability that $X$ takes a value greater than 1.46. For a uniform distribution, this probability is simply the fraction of the remaining interval after 1.46:

$P(X > 1.46) = \frac{2 - 1.46}{2} = \frac{0.54}{2} = 0.27$

2. $P(X < 0.08)$

This is the probability that $X$ takes a value less than 0.08. Again, for a uniform distribution, this probability is proportional to the length of the interval from 0 to 0.08:

$P(X < 0.08) = \frac{0.08 - 0}{2} = \frac{0.08}{2} = 0.04$

Therefore, the answers are:

• $P(X > 1.46) = 0.27$
• $P(X < 0.08) = 0.04$

Let me know if you'd like further details or explanations!

Related questions:

1. How is the probability density function (PDF) defined for a uniform distribution?
2. How do you calculate probabilities for other types of continuous distributions (e.g., normal distribution)?
3. What is the difference between a cumulative distribution function (CDF) and a probability density function (PDF)?
4. Can uniform distributions be defined on infinite intervals, or are they always finite?
5. How do you find the mean and variance of a uniform distribution?

Tip: For uniform distributions, calculating probabilities always involves simple proportions of the interval length.