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 $P(\text{neither disk has a missing pulse}) = 0.368$.

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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.