Problem Solving

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

A) 0
B) 1/9
C) 2/9
D) 1/3
E) 1

OA - C

[spoiler]I get (1/3*2/3*2/3)*3 = 4/9[/spoiler]

The probability of picking one cart of the three possible is 1, since you have to pick one cart out of a number of three possible, but it doesn't matter which cart. Now, in order to pick another cart, you have a probability of 2/3, since there are two other carts to choose from (since you've already picked one of the three) out of a total of three. Now the third time the guy has only one cart left, since he's already ridden in the other two. The probability of picking this one is therefore 1/3. In the end you get total probability by multiplying the probabilities for the three cases: 1*2/3*1/3 = 2/9.
Answer C.

Hi Danaj,,,
Correct my approach...
for example i assign ABC for each cart...

I have only ABC is the only combination(i am not bothered abt the order) where i selected all the carts..

The following are the carts combinations where we haven't selected each cart....

AAA AAB AAC
BBB BBC BBA
CCC CCA CCB
total above 9 + 1(ABC)

But ABC is only our correction combination

so i guess the answer shld be 1/10....

Please correct my approach...

muralithe1 wrote:Hi Danaj,,,
Correct my approach...
for example i assign ABC for each cart...

I have only ABC is the only combination(i am not bothered abt the order) where i selected all the carts..

The following are the carts combinations where we haven't selected each cart....

AAA AAB AAC
BBB BBC BBA
CCC CCA CCB
total above 9 + 1(ABC)

But ABC is only our correction combination

so i guess the answer shld be 1/10....

Please correct my approach...

You've decided order doesn't matter. However, order does matter in this question.

If you wanted to list all of the possible "rides", you'd have a lot more.

AAA AAB ABA BAA AAC ACA CAA ABC ACB
BBB BBA BAB ABB BBC BCB CBB BAC BCA
CCC CCA CAC ACC CCB CBC BCC CAB CBA

27 possibilities, 6 match what we want.

Prob = (# of desired outcomes)/(Total # of possibilities) = 6/27 = 2/9

HI Stuart,
Thanks for replying...
Could you please explain me why order plays a role here..???

I thought we don;t have to bother of the oder of choosing the car...

Thanks in advance...

muralithe1 wrote:HI Stuart,
Thanks for replying...
Could you please explain me why order plays a role here..???

I thought we don;t have to bother of the oder of choosing the car...

Thanks in advance...

The passenger is riding the roller coaster 3 times. Since car 1/car 2/car 3 is a different arrangement than car 2/car 1/car 3, we have to count both of them to get an accurate picture of the possibilities.

Think of coin flip questions - if you want to know the probability of getting exactly 2 heads in 3 flips, you don't simply say:

well, that means I want HHT, so that's 1/2 * 1/2 * 1/2 = 1/8;

you also have to consider THH and HTH, since both of those also satisfy getting exactly 2 heads in 3 flips, and the answer is actually:

3 * 1/2 * 1/2 * 1/2 = 3/8.