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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
Posts: 1550
Location: Dhaka
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
2
11
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Difficulty:

55% (hard)

Question Stats:

59% (01:33) correct 41% (01:24) wrong based on 361 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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Joined: 02 Sep 2009
Posts: 55228
Re: A certain roller coaster has 3 cars, and a passenger is equally likely  [#permalink]

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21 Feb 2012, 01:30
4
9
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 equally likely  [#permalink]

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28 Feb 2012, 11:24
7
+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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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)

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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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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 equally likely  [#permalink]

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07 Apr 2019, 12:26
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Re: A certain roller coaster has 3 cars, and a passenger is equally likely   [#permalink] 07 Apr 2019, 12:26
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