PROBLEM 4: PARKING LOT PROBLEM (3 points possible)

Mary and Tom park their cars in an empty parking lot with n≥2 consecutive parking spaces (i.e, n spaces in a row, where only one car fits in each space). Mary and Tom pick parking spaces at random. (All pairs of parking spaces are equally likely.) What is the probability that there is at most one empty parking space between them? (Express your answer using standard notation.)

(4*n-6)/(n*(n-1))

To solve this problem, we need to determine the total number of possible outcomes and the number of favorable outcomes.

First, let's consider the position of Mary's car. She can choose any of the n parking spaces, so there are n possible positions for her car.

Now, let's consider the position of Tom's car. Since there must be at most one empty parking space between them, Tom's car can be placed in three different ways:
1. In the parking space immediately before Mary's car.
2. In the parking space immediately after Mary's car.
3. In any other parking space between Mary's car and the end of the parking lot.

For case 1, if Mary's car is in the first parking space (position 1), then there is no parking space before her car, so Tom's car cannot be placed in the space before her. Therefore, we discard this case.

For case 2, if Mary's car is in the last parking space (position n), then there is no parking space after her car, so Tom's car cannot be placed in the space after her. Therefore, we discard this case as well.

For case 3, if Mary's car is in any other position (2 to n-1), then there are two parking spaces available before her car and n-2 parking spaces available after her car. So, in this case, Tom's car can be placed in (2 + (n-2)) = (n) positions.

Therefore, the total number of favorable outcomes is (n-2). And since Mary can choose any of the n parking spaces, there are a total of n * (n-2) possible outcomes.

To calculate the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
P = (n-2) / (n * (n-2))
Simplifying the expression, we get:
P = 1 / n

So, the probability that there is at most one empty parking space between Mary and Tom is 1/n, where n is the number of consecutive parking spaces.

To solve this problem, we can consider the different scenarios where Mary and Tom can park their cars with at most one empty space between them.

Case 1: If Mary and Tom park their cars in adjacent spaces.
In this case, there is no empty space between them. The probability of this happening is 1/(n-1), as there are (n-1) choices for the space that Mary can choose.

Case 2: If Mary and Tom park their cars with one empty space between them.
In this case, there are two possible sub-cases:
a) Mary parks her car on the left of the empty space.
b) Mary parks her car on the right of the empty space.

For sub-case a), there are (n-2) spaces that Mary can choose from on the left side of the empty space. The probability of this happening is (n-2)/(n-1).

For sub-case b), similarly, there are (n-2) spaces that Mary can choose from on the right side of the empty space. The probability of this happening is also (n-2)/(n-1).

Since the sub-cases are mutually exclusive, we can simply add the probabilities together: (n-2)/(n-1) + (n-2)/(n-1) = 2(n-2)/(n-1).

Therefore, the probability of having at most one empty space between Mary and Tom's cars is:
P = 1/(n-1) + 2(n-2)/(n-1) = (1 + 2(n-2))/(n-1) = (2n-3)/(n-1).

Hence, the probability of having at most one empty space between them is (2n-3)/(n-1).