Probability: Truck Parking

Hello everyone,

I am new to this forum and I'm here because I'm stuck on a probability question that I was working on in preparation of an upcoming test.

Problem:
A parking lot has 16 spaces in a row. Each of the twelve cars took one parking space, and their drivers chose spaces at random from among the available spaces. After that a truck arrived and it requires 2 adjacent spaces to park. What is the probability that the truck will be able to park?

I couldn't quite figure out how to mathematically take into account the adjacent parking spaces. I started with 16!/12! which is the number of combinations that the 12 cars can occupy the 16 spaces. Afterwards, I resorted to counting the number of configurations for which 2 adjacent spaces could exist but that got me nowhere. My approach now is perhaps to find the probability of the complement (the chance that the truck cannot park at all) but again that would mean the 4 empty spaces cannot be adjacent and I don't know how to account for that.

The answer is 17/28.

Thank you all in advance!

quex said:

Problem:
A parking lot has 16 spaces in a row. Each of the twelve cars took one parking space, and their drivers chose spaces at random from among the available spaces. After that a truck arrived and it requires 2 adjacent spaces to park. What is the probability that the truck will be able to park?
The answer is 17/28.

Click to expand...

I am giving you a model of the problem.
Suppose we have twelve \(\displaystyle C's\) for the cars and four \(\displaystyle E's\) for the empty spaces after the cars park.

Any arrangement of the string of letters \(\displaystyle CCECCCEECCCCCCCE\) is possible parking the cars.

If there are two adjacent \(\displaystyle E's\) then truck can park.

There are \(\displaystyle \dfrac{16!}{(4!)(12!)}\) ways to rearrange that string.
There are \(\displaystyle \dbinom{13}{4}\) ways to rearrange that string in which no two \(\displaystyle E's\) are together.

pka said:

I am giving you a model of the problem.
Suppose we have twelve \(\displaystyle C's\) for the cars and four \(\displaystyle E's\) for the empty spaces after the cars park.

Any arrangement of the string of letters \(\displaystyle CCECCCEECCCCCCCE\) is possible parking the cars.

If there are two adjacent \(\displaystyle E's\) then truck can park.

There are \(\displaystyle \dfrac{16!}{(4!)(12!)}\) ways to rearrange that string.
There are \(\displaystyle \dbinom{13}{4}\) ways to rearrange that string in which no two \(\displaystyle E's\) are together.

Click to expand...

Wow, thank you for your insight.

Since I posted, I got everything up until arranging the 4 empty spaces such that none of them touch. However, I do not understand how you found out that there were 13C4 ways of having the E's not touch. May I please ask how you came up with 13C4? What does the 13 and 4 mean?

Thank you for your answer!

quex said:

Hello everyone,
Problem:
A parking lot has 16 spaces in a row. Each of the twelve cars took one parking space, and their drivers chose spaces at random from among the available spaces. After that a truck arrived and it requires 2 adjacent spaces to park. What is the probability that the truck will be able to park?
...............................................
The answer is 17/28.
Thank you all in advance!

Click to expand...

Answer 17/28 is correct - [attachement deleted for security's sake]. Only this is the original aspect ratio 1105 / 1820.
Attached is generated table and there are formulas for control.

quex said:

I do not understand how you found out that there were 13C4 ways of having the E's not touch. May I please ask how you came up with 13C4? What does the 13 and 4 mean?

Click to expand...

Look at the string ____C____C____C____C____C____C____C____C____C____C____C____C___ .
The twelve C's create thirteen blanks. We can place the four E's into any four of those thirteen places and we will have a new string which represents a parking situation in which the truck cannot park.

The answer \(\displaystyle 1-\dfrac{\binom{13}{4}}{\binom{16}{4}}=\dfrac{17}{28}\)

BTW: I have no idea what reply #4 is about.