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A certain roller coaster has 3 cars, and a passenger is

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New post 30 Aug 2007, 03:39
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 the 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
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New post 30 Aug 2007, 03:58
Each time the passanger rides he has a 1/3 probability to ride each cart.
to ride in each cart after 3 rides he must ride in carts 1-2-3 or 1-3-2 or 2-1-3 or 2-3-1 or 3-1-2 or 3-2-1
favorable outcomes = 6 (3! he should ride each cart once so first time he has 3 choices, than 2, than 1)
possible outcomes 3*3*3=27 (meaning he can choose each of the 3 carts each time he rides)
so 6/27 = 3/9 = 1/3
Answer D
what is the OA?
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New post 30 Aug 2007, 04:13
agps wrote:
Answer D
what is the OA?


Answer is NOT D. Sorry. :lol:
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New post 30 Aug 2007, 04:21
solidcolor wrote:
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 the 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


I got B.

3 times 3 = 9 total possible outcomes. Among them only one is favourable. As long as the outcomes are 'equally likely' the result must be 1/9
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New post 30 Aug 2007, 04:49
solidcolor wrote:
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 the 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



Actually, considering order in which the roller coaster rides the cars is important, that is what agps did; the total # of possible outcomes is 24, and the # of favourbale outcomes is 3...but there is no among answer choices 1/8
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New post 30 Aug 2007, 06:13
I got B. But unfortunately it wasn't the right answer...

:roll:

Anyone?
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New post 30 Aug 2007, 06:58
It should be C.

Total # of outcomes = 3*3*3=27

# of favourable outcomes = 3*2*1=6
(In the first round, the passenger has 3 cars to choose from, in the second round he has just 2 and in the third round he has just 1 if the passenger has to sit in all 3 cars by the time he finishes 3 rounds)

Probability=6/27 = 2/9
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Subhen

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New post 30 Aug 2007, 15:19
subhen wrote:
It should be C.

Total # of outcomes = 3*3*3=27

# of favourable outcomes = 3*2*1=6
(In the first round, the passenger has 3 cars to choose from, in the second round he has just 2 and in the third round he has just 1 if the passenger has to sit in all 3 cars by the time he finishes 3 rounds)

Probability=6/27 = 2/9



I agree, C

That is a neat solution,
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New post 30 Aug 2007, 20:16
subhen wrote:
It should be C.

Total # of outcomes = 3*3*3=27

# of favourable outcomes = 3*2*1=6
(In the first round, the passenger has 3 cars to choose from, in the second round he has just 2 and in the third round he has just 1 if the passenger has to sit in all 3 cars by the time he finishes 3 rounds)

Probability=6/27 = 2/9


You've got it.

I have another explanation.

The first round, he can choose any car out of the three, 3 choices among 3 cars, P(1) = 3/3 = 1.
The second round, he has only 2 choices among the 3 cars, avoiding the car he took in the first round. P(2) = 2/3
The third round, he has only 1 choice left, avoiding the other 2 cars that he took in the first two rounds. P(3) = 1/3

Probability for "and" + "and + "and" should be
P = P(1) * P(2) * P(3)
= 1 * 2/3 * 1/3
= 2/9
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New post 30 Aug 2007, 20:20
Let ABC denotes the three cars. # of ways to arrange this word = 3! = 6
Each arrangement, P = 1/27. So total = 1/27 * 6 = 2/9
  [#permalink] 30 Aug 2007, 20:20
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