Given that the number of missing pulses $X$ follows a Poisson distribution with parameter $\mu = 0.1$, the probability mass function (PMF) of a Poisson-distributed random variable is given by:

$P(X = k) = \frac{\mu^k e^{-\mu}}{k!}$

where $k$ is the number of occurrences (in this case, missing pulses), $\mu = 0.1$, and $e$ is the base of the natural logarithm.

(a) Probability that the disk has exactly one missing pulse:

Here, $k = 1$. Substituting into the Poisson PMF formula:

$P(X = 1) = \frac{(0.1)^1 e^{-0.1}}{1!}$

Let me calculate this.

(b) Probability that the disk has at least two missing pulses:

The probability of having at least two missing pulses is:

$P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

We already know how to compute $P(X = 1)$. Now, let's calculate $P(X = 0)$:

$P(X = 0) = \frac{(0.1)^0 e^{-0.1}}{0!} = e^{-0.1}$

Let me compute the probabilities for both parts (a) and (b).### (a) The probability that the disk has exactly one missing pulse is approximately:

$P(X = 1) = 0.090$

(b) The probability that the disk has at least two missing pulses is approximately:

$P(X \geq 2) = 0.005$

Would you like any further clarification or additional details on these calculations?

Here are some related questions:

1. What is the probability that a disk has no missing pulses?
2. How would the results change if $\mu = 0.2$?
3. What is the expected number of missing pulses?
4. How does the Poisson distribution differ from the binomial distribution?
5. What is the probability of having exactly two missing pulses?

Tip: The sum of probabilities for all possible outcomes of a Poisson distribution (0, 1, 2, ...) equals 1, which is a characteristic of all probability distributions.