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

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Joined: 28 May 2005
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A certain roller coaster has 3 cars, and a passenger is equally likely  [#permalink]

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Updated on: 07 Apr 2019, 12:31
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55% (hard)

Question Stats:

58% (01:33) correct 42% (01:28) wrong based on 461 sessions

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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

Originally posted by nakib77 on 13 Oct 2005, 13:49.
Last edited by Bunuel on 07 Apr 2019, 12:31, edited 3 times in total.
Edited the OA
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Posts: 59586
Re: A certain roller coaster has 3 cars, and a passenger is equally likely  [#permalink]

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21 Feb 2012, 01:30
6
12
nakib77 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

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

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

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17 Jul 2013, 10:33
7
4
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 equally likely  [#permalink]

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28 Feb 2012, 11:24
8
1
+1 C

Let's suppose that the cars are A, B, and C.
Now, let's find the probability of picking the cars in this order: A-B-C

$$\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}$$

However, there could be more way to organize A-B-C. The numbers of ways is 3! = 6

Then, $$\frac{1}{27} * 6 = \frac{6}{27} = \frac{2}{9}$$

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Joined: 27 Oct 2008
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27 Sep 2009, 23:27
3
5
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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26 Aug 2008, 22:00
2
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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Re: A certain roller coaster has 3 cars, and a passenger is equally likely  [#permalink]

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04 Jun 2012, 05:30
2
Hi,

Probability = (favorable cases)/(total number of cases)

Total number of ways in which a person can ride car = 3*3*3 = 27
(In first ride he has 3 options to sit, in second right again he has 3 seats available to sit and so on)

Number of favorable cases, i.e., when he rides on different cars;
He can choose seat car in 3 ways in his 1st ride.
He can choose seat car in 2 ways in his 2nd ride.
He can choose seat car in 1 ways in his 3rd ride.
So, 3*2*1 = 6 ways

Thus, probability of choosing different seats = 6/27 = 2/9
(C)

Regards,
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Posts: 8281
Re: A certain roller coaster has 3 cars, and a passenger is equally likely  [#permalink]

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08 Feb 2016, 02:59
2
1
ParmarKP wrote:

I did -

For first ride the passenger has to choose one from three, so the probability would be - 1/3
For the next ride he has to choose one out of the two left, so the probability would be - 1/2
And last because he is left with only one ride to choose from to have ridden each once, probability would be - 1/1

Total probability = 1/3*1/2*1 = 1/6

I understood the solution but need to understand why this is incorrect. Thanks.

HI,
why you are going wrong is because you are not applying the info of Q correctly..
Quote:
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

so each time he plans to take the ride he can take any of the three again...
BUT in your solution, you assume that a ride once taken cannot be taken again..

so the solution on your lines would be:-
For first ride the passenger has to choose one from three, so the probability would be - 1/3
For the next ride too, he has to choose one out of the three, so the probability would be - 1/3
And last because he is left with only one ride to choose, but he has to choose again from all three, probability would be - 1/3

Total probability = 1/3*1/3*1/3 = 1/27..
but within these three the order can differ and there will be 3! ways..

hope it helps
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19 Nov 2007, 19:18
1
1
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
Manager
Joined: 03 Sep 2006
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19 Nov 2007, 19:37
1
1
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
Manager
Joined: 09 Aug 2010
Posts: 74

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02 May 2011, 21:35
1
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
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Posts: 11
Re: A certain roller coaster has 3 cars, and a passenger is equally likely  [#permalink]

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08 Feb 2016, 02:48
Bunuel wrote:
nakib77 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

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

I did -

For first ride the passenger has to choose one from three, so the probability would be - 1/3
For the next ride he has to choose one out of the two left, so the probability would be - 1/2
And last because he is left with only one ride to choose from to have ridden each once, probability would be - 1/1

Total probability = 1/3*1/2*1 = 1/6

I understood the solution but need to understand why this is incorrect. Thanks.
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Re: A certain roller coaster has 3 cars, and a passenger is equally likely  [#permalink]

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02 Apr 2016, 18:36
nakib77 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

pretty decent one...
a good question to remind everyone to read everything thoroughly.
we need a passenger to ride all 3 cars.
first try, he can choose any of the cars, so probability is 1.
second try, he can choose any of the two he did not ride, so probability is 2/3
third try, there is only one left out of three, so probability is 1/3
the overall probability:
1 * 2/3 * 1/3 = 2/9

C

if to think logically, we can eliminate right away A and E
among B, C, and D
B 1/9 is the probability if it takes either only 1 car or 1 car first try, then another car two times. so not good
D is way too much, so as well can be eliminated.
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Re: A certain roller coaster has 3 cars, and a passenger is  [#permalink]

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14 Oct 2019, 06:53
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Re: A certain roller coaster has 3 cars, and a passenger is   [#permalink] 14 Oct 2019, 06:53
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