Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number *X* has a Poisson distribution with parameter 𝜇 =0.5. (Round your answers to three decimal places.)

(a) What is the probability that a disk has exactly one missing pulse?


(b) What is the probability that a disk has at least two missing pulses?


(c) If two disks are independently selected, what is the probability that neither contains a missing pulse?

We are given a Poisson distribution with parameter \(\mu = 0.5\), where \(X\) represents the number of missing pulses on a disk. The probability mass function (PMF) for a Poisson distribution is given by:

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

where:
- \(\mu = 0.5\),
- \(k\) is the number of missing pulses (a non-negative integer),
- \(e\) is the base of the natural logarithm.

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

We need to calculate \(P(X = 1)\), which is the probability of exactly one missing pulse:

\[
P(X = 1) = \frac{0.5^1 e^{-0.5}}{1!}
\]

Let's compute this.

\[
P(X = 1) = \frac{0.5 e^{-0.5}}{1} = 0.5 \cdot e^{-0.5}
\]

\[
P(X = 1) \approx 0.5 \cdot 0.606531 = 0.303
\]

Thus, the probability that a disk has exactly one missing pulse is approximately:

\[
P(X = 1) = 0.303
\]

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

The probability of having at least two missing pulses is \(P(X \geq 2)\). To compute this, we can use the complement rule:

\[
P(X \geq 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1)
\]

We already know \(P(X = 1) \approx 0.303\). Now, let's calculate \(P(X = 0)\):

\[
P(X = 0) = \frac{0.5^0 e^{-0.5}}{0!} = e^{-0.5} = 0.606531
\]

Now we can find \(P(X \geq 2)\):

\[
P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) = 1 - (0.606531 + 0.303)
\]

\[
P(X \geq 2) = 1 - 0.909531 = 0.090
\]

Thus, the probability that a disk has at least two missing pulses is approximately:

\[
P(X \geq 2) = 0.090
\]

### (c) Probability that neither of two disks contains a missing pulse:

The probability that a disk contains no missing pulses is \(P(X = 0)\), which we have already computed:

\[
P(X = 0) = 0.606531
\]

Since the two disks are independently selected, the probability that neither contains a missing pulse is the product of the probabilities for each disk:

\[
P(\text{neither disk has a missing pulse}) = P(X = 0) \times P(X = 0)
\]

\[
P(\text{neither disk has a missing pulse}) = 0.606531 \times 0.606531 \approx 0.368
\]

Thus, the probability that neither of the two disks contains a missing pulse is approximately:

\[
P(\text{neither disk has a missing pulse}) = 0.368
\]

### Summary of results:

(a) The probability that a disk has exactly one missing pulse is \(P(X = 1) = 0.303\).

(b) The probability that a disk has at least two missing pulses is \(P(X \geq 2) = 0.090\).

(c) The probability that neither of two disks contains a missing pulse is \(P(\text{neither disk has a missing pulse}) = 0.368\).

---

Would you like more details on any of these steps or have any questions?

Here are 5 related questions to deepen your understanding:
1. What is the probability that a disk has zero missing pulses?
2. How does the probability change if the Poisson parameter \(\mu\) is increased to 1?
3. What is the expected number of missing pulses on a disk?
4. If five disks are selected, what is the probability that exactly three disks have no missing pulses?
5. How does the independence assumption between disks affect the calculations?

**Tip:** The Poisson distribution is often used to model rare events occurring over a fixed interval, such as missing pulses in a disk or defects in a manufactured item.

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number *X* has a Poisson distribution with parameter 𝜇 =0.5. (Round your answers to three decimal places.) (a) What is the probability that a disk has exactly one missing pulse? (b) What is the probability that a disk has at least two missing pulses? (c) If two disks are independently selected, what is the probability that neither contains a missing pulse?

Explore the Poisson distribution problem of counting missing pulses on a disk. Learn how to calculate the probability for exactly one missing pulse, at least two missing pulses, and the probability that neither of two independently selected disks contains missing pulses. This detailed solution covers probability concepts and Poisson distribution properties. Suitable for undergraduate students.

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