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
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:
8 answers
P(n)=(n-4)*4 + 10 = 4*(n-3/2)
(4*n-6)/(n*(n-1))
This is the correct answer (4*n-6)/(n*(n-1))
(2*n - 3)*2 / ((n-1)*n)
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.
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.
Hmm what grade is this im in 7th grade so i can check my notes to see
where does n(n-1) comes from?
8 answers