SOLUTION: Records indicate that , on average , 3.2 breakdowns per day occur on an urban highway during the morning rush hour. Assume that the distribution is poisson. Find the probability th

Question 1193322: Records indicate that , on average , 3.2 breakdowns per day occur on an urban highway during the morning rush hour. Assume that the distribution is poisson. Find the probability that on any given day there will be fewer than two breakdowns on this highway during the morning rush.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
**1. Define**
* Let X be the random variable representing the number of breakdowns per day.
* X follows a Poisson distribution with an average rate (λ) of 3.2 breakdowns per day.
**2. Calculate the Probability**
* We need to find the probability of having fewer than two breakdowns, which means 0 or 1 breakdown.
* This can be calculated using the Poisson probability mass function (PMF):
* P(X = k) = (λ^k * e^(-λ)) / k!
where:
* k is the number of occurrences (0 or 1 in this case)
* λ is the average rate of occurrences (3.2)
* e is the base of the natural logarithm (approximately 2.71828)
* k! is the factorial of k (0! = 1, 1! = 1)
* P(X < 2) = P(X = 0) + P(X = 1)
* P(X = 0) = (3.2^0 * e^(-3.2)) / 0! = e^(-3.2) ≈ 0.0408
* P(X = 1) = (3.2^1 * e^(-3.2)) / 1! = 3.2 * e^(-3.2) ≈ 0.1306
* P(X < 2) = 0.0408 + 0.1306 = 0.1714
**3. Conclusion**
The probability that on any given day there will be fewer than two breakdowns on this highway during the morning rush hour is approximately **0.1714 (or 17.14%)**.

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