Question 1193322
**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%)**.