Mary and Tom park their cars in an empty parking lot with n >_ 2 consecutive parking spaces (i.e, 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.)
unanswered
Answer:
(4*n-6)/(n*(n-1))
Since there are n parking lots and each pair are equally likely. We should note the order of pair matters here: we should choose permutation over combination here or let say Tom has n parking choices and then Mary will only have n-1 parking choices: total possible parking pairs = n*(n-1)
Next question is in how many ways they can park their car in a row or no spaces between them: for first and last parking space we will have another car parked only on one side and all other parking space they can parked on both sides. Mathematically, 2*1+ (n-2)*2.
Next question is in how many ways the cars can be parked with one space between them: for the first 2 and last 2 parking space, we can have space only on one side of parking space and for all other parking spaces, we can leave a space on both sides. Mathematically, 2*1+2*1+(n-4)*2
Possible ways that two cars can be parked with at most one space between them: 2+(n-2)*2+4+(n-4)*2 = 4n-6
Prob = (4n-6)/n(n-1). This is the correct answer.
This is the correct answer (4*n-6)/(n*(n-1))
(2*n - 3)*2 / ((n-1)*n)
Hmm what grade is this im in 7th grade so i can check my notes to see
where does n(n-1) comes from?
To find the probability that there is at most one empty parking space between Mary and Tom's cars, we need to consider the different scenarios that can occur.
Let's assume that Mary and Tom choose consecutive parking spaces, without any empty spaces between them. In this case, there is only one possibility for their selection.
Now, let's consider the scenario where there is exactly one empty parking space between Mary and Tom's cars. To determine the probability of this happening, we need to calculate the number of possible selections for Mary and Tom that satisfy this condition.
To do this, we can fix the position of one of the cars and consider the different placements for the other car. Let's assume that Mary's car is placed first. In this case, the empty parking space can either be to the left of Mary's car or to the right.
If the empty space is to the left of Mary's car, Tom's car can be placed in any of the (n - 3) remaining spaces. This is because there are (n - 1) spaces left after placing Mary's car and the added constraint of one empty space leaves (n - 3) spaces available for Tom's car.
Similarly, if the empty space is to the right of Mary's car, Tom's car can also be placed in any of the (n - 3) remaining spaces.
Therefore, the total number of selections where there is exactly one empty space between Mary and Tom's cars is 2 * (n - 3).
Lastly, we need to consider the scenario where there are two or more empty spaces between Mary and Tom's cars. In this case, there are no possible selections that satisfy the condition.
To find the probability, we need to divide the number of favorable outcomes (scenarios with at most one empty space) by the total number of possible outcomes (scenarios with any selection).
The total number of possible outcomes is given by the number of ways to select two parking spaces out of the available n parking spaces, which is represented by n choose 2, written as C(n, 2) = n! / ((2!)(n - 2)!).
Therefore, the probability is:
P(there is at most one empty space) = (1 + 2 * (n - 3)) / (n! / ((2!)(n - 2)!))
Simplifying this expression will give us the final answer.
P(n)=(n-4)*4 + 10 = 4*(n-3/2)
For each parking space, there are four possible parking place to be ocupied, except for the first and the last, which has just 2 places, and the second and the "before the last", which have 3 possibilities. So, we have a total P(n) of
P(n)=(n-4)*4 + 12 = 4*(n-1)
possible combinations. This, over the total possibilities, which is
T(n) = (n-1)*n
You just have to divide P(n)/T(n) and you get the answer