Final answer:
The probability of selecting a household with at most three cars, given only the number of households with three cars (as 1) and no data for 0, 1, or 2 cars, is 0.001 or 0.1%. More accurate calculations would be possible with complete data.
Step-by-step explanation:
To find the probability of randomly selecting a household that has at most three cars, we need to add up the probabilities of selecting a household with 0, 1, 2, or 3 cars. Let's denote the random variable X as the number of cars per household. Based on the data provided, we need to count the total number of households that fall into the categories of having 0, 1, 2, or 3 cars and then divide this by the total number of households.
From the data given, we have the following counts of households by the number of cars:
- 0 cars: Not provided, we'll assume it's 0
- 1 car: Not provided, we'll assume it's 0
- 2 cars: Not provided, we'll assume it's 0
- 3 cars: 1 household
- 4 or more cars: All other numbers provided
Since the question is about having at most three cars, we will only look at having 0 to 3 cars. The total count for these is 1 (since we have no data for 0, 1, or 2 cars, we assume they are 0). So the probability P(X ≤ 3) is:
P(X ≤ 3) = Number of households with ≤ 3 cars / Total number of households
P(X ≤ 3) = 1 / 1000
P(X ≤ 3) = 0.001 or 0.1%
This method is not entirely accurate as we don't have the actual numbers for households with 0, 1, and 2 cars, but typically these would be provided in the data set to accurately calculate the probability.