Question 1199110
**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.**