Roller coaster

[toqueerahmed]

a roller coaster has 3 cars and a passenger is equally likely to ride in any 1 of the 3 cars each time that the passenger rides the roller coaster. if a certain passenger is to ride the roller coaster 3 times, what is the probability that a passenger will ride in each of the 3 cars.

0
1/9
2/9
1/3
1

source: Power PREP

[atul.prasad]

Hi,

Total possible ways in which a passenger can ride 3 times is :
3*3*3, since all choices are equally likely.

Now, if we number the chairs 1,2,3 , then the passenger would have ridden on each of them if she/he chose it in the order 1,2,3 or 2,1,3 .. (essentially 3!) ways.

So the probability should be 6/27 = 2/9

[jnelson0612]

atul, once again this is an excellent explanation. I have nothing to add. Nice work!

[shaji]

Hi,

Total possible ways in which a passenger can ride 3 times is :er
3*3*3, since all choices are equally likely.

Now, if we number the chairs 1,2,3 , then the passenger would have ridden on each of them if she/he chose it in the order 1,2,3 or 2,1,3 .. (essentially 3!) ways.

So the probability should be 6/27 = 2/9

Hi Atul
The probability is indeed 2/9 to occupy three cars atleast once. Consider the probability if passenger rides the coaster 4 times i.e. just once more than he did? An interesting problem.

[atul.prasad]

Hi Shaji,

So if I interpret your question correctly,
we need to find out the probability that the passenger rides in all cars at least once.

total possible ways in which the passenger can take the ride is:
3*3*3*3 (since all cars are equally likely and the passenger shall ride 4 times)

To count total number of ways in which passenger rides in all cars at least once, we should note that the passenger shall choose any of the cars twice.
Hence the total number of times is :
when 1st car was repeated = (4!/2!)[permutations of 1,2,3,1]
when 2nd car was repeated = (4!/2!)[permutations of 1,2,3,2]
when 3rd car was repeated = (4!/2!)[permutations of 1,2,3,3]

total = 3*12

hence the probability should be 36/81 = 4/9

[shaji]

Hi Atul

Yes! indeed you are correct. You will notice that the passenger has doubled his chances by merely trying one more time i.e. Three attempts, prob=2/9 and 4 attempts, Prob=4/9. This is an interesting observation.

Therefore, you may find it stimulating to find the minimum number of times the passenger should attempt the ride if he/she needs to be atleast 90% sure of riding all the cars atleast once.

This is a problem with practical applications

Hi Shaji,

So if I interpret your question correctly,
we need to find out the probability that the passenger rides in all cars at least once.

total possible ways in which the passenger can take the ride is:
3*3*3*3 (since all cars are equally likely and the passenger shall ride 4 times)

To count total number of ways in which passenger rides in all cars at least once, we should note that the passenger shall choose any of the cars twice.
Hence the total number of times is :
when 1st car was repeated = (4!/2!)[permutations of 1,2,3,1]
when 2nd car was repeated = (4!/2!)[permutations of 1,2,3,2]
when 3rd car was repeated = (4!/2!)[permutations of 1,2,3,3]

total = 3*12

hence the probability should be 36/81 = 4/9

[atul.prasad]

wow! thats an interesting observation.

[danielpatinkin]

a roller coaster has 3 cars and a passenger is equally likely to ride in any 1 of the 3 cars each time that the passenger rides the roller coaster. if a certain passenger is to ride the roller coaster 3 times, what is the probability that a passenger will ride in each of the 3 cars.

0
1/9
2/9
1/3
1

source: Power PREP

Brilliant stuff guys!
An alternative way to answer the original question is to multiply probabilities. The probability that the passenger will sit in the 1st car, then the 2nd car, then the 3rd car is 1/3 * 1/3 * 1/3, which equals 1/27. However, because he can choose any order in which to accomplish this, you multiply that probability by 3!, giving you 6/27 or 2/9!