Respuesta :
Answer:
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.
Probability of an event is the measure of its chance of occurrence. The probability that there is at most one empty parking space between the parked cars is [tex]P(E) = P(A) + P(B) = \dfrac{2(2n-3)}{n(n-1)}[/tex]
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
What is chain rule in probability?
For two events A and B, by chain rule, we have:
[tex]P(A \cap B) = P(B)P(A|B) = P(A)P(A|B)[/tex]
What is law of total probability?
Suppose that the sample space is divided in n mutual exclusive and exhaustive events tagged as
[tex]B_i \: ; i \in \{1,2,3.., n\}[/tex]
Then, suppose there is event A in sample space.
Then probability of A's occurrence can be given as
[tex]P(A) = \sum_{i=1}^n P(A \cap B_i)[/tex]
Using the chain rule, we get
[tex]P(A) = \sum_{i=1}^n P(A \cap B_i) = \sum_{i=1}^n P(A)P(B_i|A) = \sum_{i=1}^nP(B_i)P(A|B_i)[/tex]
For the given case, we can assume two events as:
- A = both cars parked together
- B = both cars parked with 1 empty parking space between them
- C = both cars parked with > 1 empty parking space between them
- E= both cars are at most having 1 empty parking space between them.
Then [tex]P(C \cap E) = 0[/tex]
And thus, [tex]P(E) = P(E \cap A) + P(E \cap B) + 0 = P(A) + P(B)[/tex]
(since A is subset of event E and B is subset of event E).
Now calculating the probabilities of A and B event, we get:
n(A) = number of ways A can occur = n-1(if we start from first parking space, then both cars are adjacent to each other, so 1 way(assuming no distinction between them(we can assume so, since space between cars doesn't care about order in which they're arranged)), and thus, total n-1 ways(as we can't go ahead of n-1 as there is 2nd are on nth space).
and n(S) = number of ways of choosing 2 parking spaces = [tex]^nC_2 = \dfrac{n(n-1)}{2}[/tex] ways.
Also, n(B) = n-2 ways similar to above logic,
Thus, we get:
[tex]P(A) = \dfrac{n-1}{n(n-1)/2}\\P(B) = \dfrac{n-2}{n(n-1)/2}\\\\P(E) = P(A) + P(B) = \dfrac{2(2n-3)}{n(n-1)}[/tex]
Thus, the probability that there is at most one empty parking space between the parked cars is [tex]P(E) = P(A) + P(B) = \dfrac{2(2n-3)}{n(n-1)}[/tex]
Learn more about probability here:
brainly.com/question/1210781
