123. Starting next to your house, there is a strip of 30 feet of street parking

Question

123. Starting next to your house, there is a strip of 30 feet of street parking where you’d like to park. By the time you arrive there will only be a scooter parked there. It will be parked perpendicular to the curb, so you disregard its width, treating it as an infinitely thin point. But the scooter could be anywhere, and you regard its possible locations on the strip as uniformly distributed. You will need at least 18 feet of free space to park. What is the probability that you will be able to park right next to your house? 124. What is the probability that you will be able to park somewhere on the strip? 125. Your neighbors have moved their vehicle, so that there is 45 feet of street parking, and the scooter is gone, too. But a 20 foot truck could be parked anywhere (uniformly distributed) on this strip. What is the probability that you will be able to park right next to your house? 126. What is the probability that you will be able to park somewhere on the strip?

Explanation / Answer

Q.123 THe parking is of 30 feet of street parking in front of my house. Lets make these points as(0,0) and ( 30,0) and as it is uniform probability of parking any vehicle on that strip. So, i need 18 feet street parking for my vehicle. So, the scooter will devide the front street parking in 2 parts. Let's say these two parts lengths are x and y feet. so, the max value of x and y should be greater than 18 feet.

PDF of sccoter location fZ(z) = 1/30 ; 0 <= z <= 30

FZ(z) = z/30

Here x + y = 30 and we have to find P(Max(x,y) >=18) or say P[ Max.(x, 30-x) >= 18]

so for values x < 15, Max(x, 30-x) = 30-x [ so 30- x>18 => x =<12]

so P[X=<12] = 12/30 = 0.4

similarly for values X>15, Max(x,30-x) = x [ so x >18]

so P(X>=18) = (30-18)/30 = 0.4

so P[ Max.(x, 30-x) >= 18] = 0.4 + 0.4 = 0.8

Q.124 as the scooter is an infinity thin point so, probability of that i will be able to park somewhere on the strip

= (30 -0)/30 = 1

Q.125 A 20 feet truck could be parked anywhere uniformly distributed. So, out of the available 45 feet of street parking, only 25 feet is available for parking. So, the truck will devide the front street parking in 3 parts. Let's say these three parts will have lengths are 20, x and y feet. So, what we have to calculate P[ Max(x,y)] >=18]

PDF of truck location fZ(Z) = 1/45 ; 0 <= z <= 45

FZ(z) = z/45

Here x + y = 45- 20 = 25 and we have to find P(Max(x,y) >=18) or say P[ Max.(x, 25-x) >= 18]

so for values x < 12.5, Max(x, 25-x) = 25-x [ so 25- x>18 => x =<7]

so P[X=<12.5] = 7/45

similarly,

so for values x >= 12.5, Max(x, 25-x) = x [ so x>18 ]

so P[ X>=12.5] = 7/45

so,  P[ Max.(x, y) >= 18] = 7/45 + 7/45 =14/45 = 0.311

Q.126

As the truck has 20 feet in length, probability of that i will be able to park somewhere on the strip

= (45-20)/45 = 0.555

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