The question is about a uniform distribution of a random variable $X$, and you are asked to find the probability $P(X > 1.6)$ or $P(X < 0.12)$ from the given uniform distribution over the interval [0, 2].

For a uniform distribution over an interval $[a, b]$, the probability density function (PDF) is given by:

$f(x) = \frac{1}{b - a}$

Here, $a = 0$ and $b = 2$, so the PDF is:

$f(x) = \frac{1}{2 - 0} = \frac{1}{2}$

The probability of $X$ being in any subinterval is simply the length of that subinterval multiplied by the PDF. Thus, to compute the probabilities:

1. $P(X > 1.6)$:

The length of the interval $[1.6, 2]$ is:

$2 - 1.6 = 0.4$

So, the probability is:

$P(X > 1.6) = 0.4 \times \frac{1}{2} = 0.2$

2. $P(X < 0.12)$:

The length of the interval $[0, 0.12]$ is:

$0.12 - 0 = 0.12$

So, the probability is:

$P(X < 0.12) = 0.12 \times \frac{1}{2} = 0.06$

3. For $P(X > 1.6)$ or $P(X < 0.12)$:

Since these two events are mutually exclusive (they can't happen at the same time), you can add the probabilities:

$P(X > 1.6 \text{ or } X < 0.12) = P(X > 1.6) + P(X < 0.12) = 0.2 + 0.06 = 0.26$

Thus, the final answer is 0.26.

Would you like more details or have any questions? Here are some related questions to consider:

1. How do we calculate the probability for a uniform distribution over different intervals?
2. What are the properties of a uniform distribution?
3. Can uniform distributions be discrete or only continuous?
4. How would the probability change if the interval for $X$ were different, like $[0, 3]$?
5. What are some real-world applications of uniform distributions?

Tip: Always visualize the interval on the number line for uniform distributions to better understand the probabilities associated with different regions.