91Ó°ÊÓ

When dealing with probability calculations in Poisson distributions, it is crucial to become familiar with the formula that helps compute the likelihood of different outcomes. The Poisson distribution is utilized when you are studying the occurrence of events over a fixed interval or region, assuming these events happen independently.

For this distribution, the probability of observing exactly \(k\) occurrences (or missing pulses in our context) is given by:
\[ P(X=k) = \frac{e^{-\mu} \cdot \mu^k}{k!} \]
where:

• \(\mu\) is the average number of occurrences over the interval.
• \(e\) is the base of the natural logarithm, approximately equal to 2.718.
• \(k!\) represents the factorial of \(k\), which means \(k \times (k-1) \times (k-2) \cdots \times 1\).

For example, if you want to find the probability of having exactly one missing pulse on the disk with \(\mu = 2\), you calculate as follows:
\[ P(X=1) = \frac{e^{-2} \cdot 2^1}{1!} = e^{-2} \cdot 2 \]
This value approximates to 0.2707, indicating a 27.07% chance of finding exactly one missing pulse.

In probability, understanding the concept of independent events is essential, especially when analyzing situations like selecting multiple disks independently. Two events, A and B, are said to be independent if the occurrence of one does not influence the other. Mathematically, this translates to:
\[ P(A \text{ and } B) = P(A) \times P(B) \]
In the given exercise, when selecting two disks independently, we want the probability that neither disk has a missing pulse. If the probability for one disk to have zero missing pulses is \(P(X=0) = e^{-2}\), then for two independent disks, you simply square this probability because each event (the status of each disk) is independent.

Thus, the probability that neither disk has a missing pulse becomes:
\[ P(\text{neither has a missing pulse}) = (e^{-2})^2 = e^{-4} \]
This results in approximately 0.0183, or a 1.83% chance that neither disk shows a missing pulse if they are selected independently.

When analyzing events like missing pulses in disks, the Poisson distribution acts as a very useful model, especially when these events are rare or infrequent. Here, missing pulses represent the events of interest occurring over a disk.

In the context of the exercise, knowing the average rate of missing pulses \(\mu\) helps to predict the number of missing pulses you might observe on any single disk.

If concerned about having at least two missing pulses, you calculate this probability by considering all possibilities, such as zero or one missing pulse, and subtracting their combined probability from 1. This yields the probability that a disk has at least two missing pulses:
\[ P(X \geq 2) = 1 - P(X=0) - P(X=1) \]

Calculating these individually using the Poisson formula:

• \( P(X=0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2} \)
• \( P(X=1) = \frac{e^{-2} \cdot 2^1}{1!} = e^{-2} \cdot 2 \)

Subtracting the sum of these probabilities from 1 gives you around 0.594, indicating a 59.4% chance of having at least two missing pulses on the disk.