This problem involves a Poisson distribution, where the mean ($\lambda$) represents the average number of events (crashes) in a fixed period. The probability mass function for a Poisson distribution is given by:

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

where $X$ is the random variable denoting the number of crashes, $\lambda$ is the mean, $k$ is the number of crashes, and $e$ is Euler's number ($\approx 2.718$).

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(a) At least two crashes in a semester

We are given $\lambda = 0.75$. To find the probability of at least two crashes in a semester ($P(X \geq 2)$):

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

1. Compute $P(X = 0)$: $P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-0.75}.$

2. Compute $P(X = 1)$: $P(X = 1) = \frac{\lambda^1 e^{-\lambda}}{1!} = 0.75 \cdot e^{-0.75}.$

Substitute these into $P(X \geq 2)$: $P(X \geq 2) = 1 - e^{-0.75} - 0.75 \cdot e^{-0.75}.$

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(b) At least two crashes in one year

In one year, there are 2 semesters. Assuming independence, the total crashes in a year are modeled by a Poisson distribution with mean: $\lambda_{\text{year}} = 2 \times 0.75 = 1.5.$

The probability of at least two crashes in a year is: $P(X \geq 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1).$

1. Compute $P(X = 0)$: $P(X = 0) = \frac{\lambda_{\text{year}}^0 e^{-\lambda_{\text{year}}}}{0!} = e^{-1.5}.$

2. Compute $P(X = 1)$: $P(X = 1) = \frac{\lambda_{\text{year}}^1 e^{-\lambda_{\text{year}}}}{1!} = 1.5 \cdot e^{-1.5}.$

Substitute these into $P(X \geq 2)$: $P(X \geq 2) = 1 - e^{-1.5} - 1.5 \cdot e^{-1.5}.$

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(c) Probability of zero crashes at least 90% in a semester

We want the probability of zero crashes ($P(X = 0)$) to be at least 90%: $P(X = 0) \geq 0.9.$

Using the formula for $P(X = 0)$: $P(X = 0) = e^{-\lambda}.$

Set this equal to 0.9 and solve for $\lambda$: $e^{-\lambda} \geq 0.9 \implies -\lambda \geq \ln(0.9) \implies \lambda \leq -\ln(0.9).$

Calculate: $\lambda \leq -\ln(0.9) \approx 0.105.$

Thus, the mean must be $\lambda \leq 0.105$.

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Summary of Results:

1. (a): At least two crashes in a semester:
   $P(X \geq 2) = 1 - e^{-0.75}(1 + 0.75) \approx 0.1738.$

2. (b): At least two crashes in one year: $P(X \geq 2) = 1 - e^{-1.5}(1 + 1.5) \approx 0.4422.$

3. (c): To ensure zero crashes with 90% probability in a semester, the mean must be:
   $\lambda \leq 0.105.$

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Would you like a step-by-step derivation or further clarifications? Here are 5 related questions for practice:

1. How would the probabilities change if the mean crashes per semester doubled?
2. How do you model such probabilities for overlapping time periods?
3. What is the expected value and variance for a Poisson process?
4. If crashes follow a different distribution (e.g., geometric), how does the calculation change?
5. How can you simulate this problem computationally?

Tip: The Poisson distribution is memoryless for independent intervals, meaning future events do not depend on past outcomes.