Question 1199110: A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 50
kilometers of highway on a blizzard day. Assume that the occurrence of accidents along the highway is modeled by a Poisson process.
For a stretch of highway that is 20 kilometers long, consider the following:
(a) What is the probability that at least one accident will occur on a given blizzard day?
(b) Suppose there are five blizzard days this winter. What is the probability that two out of these five blizzard days are accident free?
Assume that accident occurrence between blizzard days are statistically independent
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **a) Probability of at least one accident on a given blizzard day**
* **Calculate the average number of accidents for the 20 km stretch:**
* If 1 accident occurs per 50 km, then the average number of accidents for 20 km is:
* (20 km / 50 km) * 1 accident = 0.4 accidents
* **Define the random variable:** Let X be the number of accidents on the 20 km stretch. X follows a Poisson distribution with mean λ = 0.4.
* **Probability of no accidents:**
* P(X = 0) = (e^(-λ) * λ^x) / x!
* P(X = 0) = (e^(-0.4) * 0.4^0) / 0!
* P(X = 0) = e^(-0.4) ≈ 0.6703
* **Probability of at least one accident:**
* P(X ≥ 1) = 1 - P(X = 0)
* P(X ≥ 1) = 1 - 0.6703
* P(X ≥ 1) ≈ 0.3297
**Therefore, the probability that at least one accident will occur on a given blizzard day on this 20 km stretch of highway is approximately 0.3297.**
**b) Probability of two out of five blizzard days being accident-free**
* **Probability of no accidents on a single day:**
* P(no accident) = 0.6703 (calculated in part a)
* **Probability of two accident-free days out of five:**
* This follows a binomial distribution with:
* n = 5 (number of trials - blizzard days)
* p = 0.6703 (probability of success - no accident)
* k = 2 (number of successes - accident-free days)
* **Binomial Probability Formula:**
* P(X = k) = (nCk) * p^k * (1-p)^(n-k)
* where nCk = n! / (k! * (n-k)!)
* **Calculate the probability:**
* P(X = 2) = (5C2) * (0.6703)^2 * (1 - 0.6703)^(5-2)
* P(X = 2) = 10 * (0.6703)^2 * (0.3297)^3
* P(X = 2) ≈ 0.2601
**Therefore, the probability that two out of the five blizzard days are accident-free is approximately 0.2601.**
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