We are asked to find the probability that a real number picked uniformly at random from the interval $[-12, 12]$ lies in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ along with the entire interval $[-12, 12]$ (i.e., $(-1/2, 1/2) \cup [-12, 12]$).

Step 1: Understanding the Problem

The interval $[-12, 12]$ already contains $\left(-\frac{1}{2}, \frac{1}{2}\right)$ as a subset because $-\frac{1}{2}$ and $\frac{1}{2}$ lie within $[-12, 12]$. This means that picking a number from the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is equivalent to focusing on that portion of the larger interval.

Step 2: Calculate the Length of the Intervals

The length of the total interval is: $\text{Length of } [-12, 12] = 12 - (-12) = 24.$

The length of the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is: $\text{Length of } \left(-\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1.$

Step 3: Probability Calculation

The probability of picking a number from $\left(-\frac{1}{2}, \frac{1}{2}\right)$ out of $[-12, 12]$ is simply the ratio of the lengths of these intervals: $\text{Probability} = \frac{\text{Length of } \left(-\frac{1}{2}, \frac{1}{2}\right)}{\text{Length of } [-12, 12]} = \frac{1}{24}.$

Step 4: Rounding

Now, we round the result to four decimal places: $\frac{1}{24} \approx 0.0417.$

Final Answer:

The probability that a randomly selected number from the interval $[-12, 12]$ lies in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is approximately $\boxed{0.0417}$.

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Would you like further details or clarification? Here are 5 related questions to expand on this concept:

1. How would the probability change if the interval was $[-6, 6]$ instead of $[-12, 12]$?
2. What is the probability of selecting a number from the interval $[2, 5]$ within $[-12, 12]$?
3. If the interval were $[-12, 24]$, how would the probability calculation change?
4. How does this problem relate to the concept of probability density in continuous distributions?
5. Can you explain how to compute the probability if the selection were from a different distribution, such as a normal distribution?

Tip: When calculating probabilities for continuous distributions, always relate the desired interval to the total range of possible outcomes to get a proper ratio.