The number of times a laptop crashes in a month is a Poisson random
variable with parameter λ = 2. Suppose a new software reduces the Poisson parameter
to λ = 1.5 with 50% probability. If the laptop does not crash in a month, how likely
is it that the software was effective?

To solve this problem, we can use **Bayes' Theorem**. The goal is to find the probability that the software was effective (i.e., reduced the Poisson parameter to \( \lambda = 1.5 \)) given that the laptop did not crash in a month.

### Define the events:
- Let \( E \) be the event that the software is effective (with probability \( P(E) = 0.5 \)).
- Let \( N \) be the event that the laptop does not crash in a month (i.e., the number of crashes, \( X = 0 \)).

We are interested in finding \( P(E | N) \), which is the probability that the software was effective given that no crashes occurred. By **Bayes' Theorem**:

\[
P(E | N) = \frac{P(N | E) \cdot P(E)}{P(N)}
\]

### Step 1: Calculate \( P(N | E) \)
If the software is effective, the number of crashes follows a Poisson distribution with parameter \( \lambda = 1.5 \). The probability of zero crashes is given by the Poisson probability mass function:

\[
P(N | E) = P(X = 0 | \lambda = 1.5) = \frac{(1.5)^0 e^{-1.5}}{0!} = e^{-1.5}
\]

So,

\[
P(N | E) = e^{-1.5} \approx 0.2231
\]

### Step 2: Calculate \( P(N | E^c) \)
If the software is **not** effective, the number of crashes follows a Poisson distribution with parameter \( \lambda = 2 \). The probability of zero crashes is:

\[
P(N | E^c) = P(X = 0 | \lambda = 2) = \frac{(2)^0 e^{-2}}{0!} = e^{-2}
\]

So,

\[
P(N | E^c) = e^{-2} \approx 0.1353
\]

### Step 3: Calculate \( P(N) \)
Using the law of total probability, \( P(N) \) is the total probability of no crashes, considering both cases (software effective and not effective):

\[
P(N) = P(N | E) \cdot P(E) + P(N | E^c) \cdot P(E^c)
\]

Substitute the values:

\[
P(N) = (0.2231 \cdot 0.5) + (0.1353 \cdot 0.5) = 0.11155 + 0.06765 = 0.1792
\]

### Step 4: Apply Bayes' Theorem
Now, we can apply Bayes' Theorem to calculate \( P(E | N) \):

\[
P(E | N) = \frac{P(N | E) \cdot P(E)}{P(N)} = \frac{0.2231 \cdot 0.5}{0.1792} \approx \frac{0.11155}{0.1792} \approx 0.6226
\]

### Final Answer:
The probability that the software was effective given that the laptop did not crash in a month is approximately **0.6226**, or **62.26%**.

---

Would you like further details or explanations? Here are 5 related questions to expand your understanding:

1. How would the probability change if the new software had a higher probability of effectiveness, say 70%?
2. What would the probability \( P(E | N) \) be if \( \lambda \) for the ineffective software was 3 instead of 2?
3. How can we calculate the expected number of crashes in a month based on this model?
4. How is the Poisson distribution related to other discrete distributions like the binomial?
5. What assumptions are we making in using the Poisson distribution for modeling crashes?

**Tip**: In any Bayesian problem, clearly define your prior, likelihood, and evidence to avoid confusion in applying Bayes' Theorem.

The number of times a laptop crashes in a month is a Poisson random variable with parameter λ = 2. Suppose a new software reduces the Poisson parameter to λ = 1.5 with 50% probability. If the laptop does not crash in a month, how likely is it that the software was effective?

Learn how to calculate the probability that a software update was effective in reducing laptop crashes using Bayesian inference and the Poisson distribution. This problem involves Bayes' Theorem and key statistical concepts, suitable for students at the undergraduate level.

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