Therefore, the event $ z \geq 100 $ has probability zero. - Richter Guitar

Understanding the Context

Understanding Continuous Random Variables

First, it’s essential to distinguish between discrete and continuous random variables. For discrete variables (like the roll of a die), probabilities are assigned to distinct outcomes, such as $ P(X = 6) = \frac{1}{6} $. In contrast, continuous random variables (such as time, temperature, or measurement errors) take values over an interval, like all real numbers within $ [a, b] $. Because of this continuous nature, outcomes at single points — and even over intervals — often have zero probability.

The Probability of Single Points Is Zero

Key Insights

In continuous probability, the probability that $ Z $ takes any exact value $ z $ is zero:

$$
P(Z = z) = 0 \quad \ ext{for any real number } z.
$$

This occurs because the “width” of a single point is zero. The probability over an interval is defined as the integral of the probability density function (PDF), $ f_Z(z) $, over that interval:

$$
P(a \leq Z \leq b) = \int_a^b f_Z(z) \, dz.
$$

Since the integral sums up infinitesimal probabilities over small intervals, the total probability of hitting exactly one value — say, $ z = 100 $ — amounts to zero.

Final Thoughts

Why $ z \geq 100 $ Has Probability Zero

Now, consider the event $ z \geq 100 $. Geometrically, this corresponds to the probability of $ Z $ being in the unbounded tail extending from 100 to infinity:

$$
P(Z \geq 100) = \int_{100}^{\infty} f_Z(z) \, dz.
$$

Even if $ f_Z(z) $ is small beyond 100, integrating over an infinite range causes the total probability to approach zero — provided the integral converges. For well-behaved distributions (such as normal, exponential, or uniform on finite intervals extended safely), this integral is finite, confirming:

$$
P(Z \geq 100) = 0.
$$