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

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A certain roller coaster has 3 cars, and a passenger is [#permalink]

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19 Nov 2007, 17:55
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-roller-coaster-has-3-cars-and-a-passenger-is-21226.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jul 2013, 09:37, edited 1 time in total.
Edited the question and added the OA.
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Re: A certain roller coaster has 3 cars, and a passenger is [#permalink]

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17 Jul 2013, 09:33
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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?

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.

$$q = 1 * \frac{1}{3} * \frac{1}{3}* 3 + 1 * \frac{1}{3} * \frac{1}{3} * 4 = \frac{1}{9} *3 + \frac{1}{9} * 4 = \frac{7}{9}$$

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]

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17 Jul 2013, 09:38
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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.

$$q = 1 * \frac{1}{3} * \frac{1}{3}* 3 + 1 * \frac{1}{3} * \frac{1}{3} * 4 = \frac{1}{9} *3 + \frac{1}{9} * 4 = \frac{7}{9}$$

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);

So $$P=\frac{3!}{3^3}=\frac{2}{9}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-roller-coaster-has-3-cars-and-a-passenger-is-21226.html
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27 Sep 2009, 22:27
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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

So the probability is
= 6/27
= 2/9

Ans is C
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19 Nov 2007, 18:37
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My ans is C:

1st ride - prob is 1/3
2nd ride - prob is 2/3
3rd ride - prob is 1 (as 1 car left ...)

P = 1/2 * 2/3 * 1 = 2/9
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26 Aug 2008, 21:00
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saviop 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

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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19 Nov 2007, 18:18
saviop 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

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

(3/3)*(2/3)*(1/3)=2/9
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26 Aug 2008, 21:15
$$(3*2*1)/3^3 = 2/9$$
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02 May 2011, 20:35
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Everytime the passenger rides a car, he has 3 choices each time. 3 x 3 x 3

To ride 3 times without repeating a car = 3 x 2 x 1

So (3 x 2 x 1)/(3 x 3 x 3) = 2/9
Re: Probability   [#permalink] 02 May 2011, 20:35
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