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 distinct parking spaces are equally likely.) what is the probability that there is at most one empty parking space between them?

  p(at most one space between) = (4n-6)/(n(n-1))

Step-by-step explanation:

There are n-1 ways the cars can be parked next to each other, and n-2 ways they can be parked with one empty space between. So, the total number of ways the cars can be parked with at most one empty space is ...

  (n -1) +(n -2) = 2n-3

The number of ways that 2 cars can be parked in n spaces is ...

  (n)(n -1)/2

So, the probability is ...

  (2n-3)/((n(n-1)/2) = (4n -6)/(n(n -1))

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If the cars are considered distinguishable and order matters, then the number of ways they can be parked will double. The factor of 2 cancels in the final probability ratio, so the answer remains the same.

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Check

For n=2 or 3, p=1 as you expect.

For n=4, p=5/6, since there is only one of the 6 ways the cars can be parked that has 2 spaces between.

okay so 15.15 *()5: +5/6

step-by-step explanation:

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.