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A certain roller coaster has 3 cars, and a passenger is [#permalink]
19 Nov 2007, 17:55

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Question Stats:

61% (01:32) correct
39% (00:35) wrong based on 115 sessions

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 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

prob that any car 3/3
prob that any of left 2 cars 2/3
prob of last one 1/3

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

GOOD QUESTION.

SAY 123 are car numbers.. possible combinations 123,132,231,213,321,312

p= 3! * (1/3^3) = 6/27 = 2/9

C

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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

Soln: If he is to ride 3 times and since he can choose any of the 3 cars each time, total number of ways is = 3 * 3 * 3 = 27

Now the number of ways if he is to choose a different car each time is = 3 * 2 * 1 = 6

Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]
17 Jul 2013, 09:33

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This post received KUDOS

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?

I couldn´t solve it using the reverse combination approach. Anyone knows how?. But here are the other 3 ways:

Probability approach:

First ride: he can choose between 3 cars: 1/3 Second ride: he can choose only (3 less 1) cars: 2/3 Third ride: he can choose the only one in which he hadn´t ride.

Then: \frac{1}{3}*\frac{2}{3} *1=\frac{2}{9}

Reverse probability approach:

P = 1-q. q = probability that he rides only in one or only in two cars, but not in the three.

First term I multiply by 3 because he could use A, B or C. Second term multuply by four because he could use A&B, A&C, C&B or B&C.

P=1-\frac{7}{9}= \frac{2}{9}

Combinatory approach:

C^3_1 * C^2_1 * C^1_1 = 3*2*1= 6

Total combinations: (C^3_1)^3 = 27

\frac{6}{27}=\frac{2}{9}

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Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]
17 Jul 2013, 09:38

1

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Expert's post

Maxirosario2012 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?

I couldn´t solve it using the reverse combination approach. Anyone knows how?. But here are the other 3 ways:

Probability approach:

First ride: he can choose between 3 cars: 1/3 Second ride: he can choose only (3 less 1) cars: 2/3 Third ride: he can choose the only one in which he hadn´t ride.

Then: \frac{1}{3}*\frac{2}{3} *1=\frac{2}{9}

Reverse probability approach:

P = 1-q. q = probability that he rides only in one or only in two cars, but not in the three.

First term I multiply by 3 because he could use A, B or C. Second term multuply by four because he could use A&B, A&C, C&B or B&C.

P=1-\frac{7}{9}= \frac{2}{9}

Combinatory approach:

C^3_1 * C^2_1 * C^1_1 = 3*2*1= 6

Total combinations: (C^3_1)^3 = 27

\frac{6}{27}=\frac{2}{9}

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

First time a passenger can ride any car: p=1; Second time the passenger should ride another car: p=2/3; Third time the passenger should ride not the 2 cars he had already driven: p=1/3.

So P=1*\frac{2}{3}*\frac{1}{3}=\frac{2}{9}.

OR: Total # of ways the passenger can drive 3 cars (for 3 rides) is 3^3 (for each ride passenger has 3 options); # of ways the passenger can drive 3 different cars is 3! (ABC, ACB, BAC, BCA, CAB, CBA);