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# There are 3 boogies in a fun-train and a passenger is equall

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There are 3 boogies in a fun-train and a passenger is equall [#permalink]

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16 Jun 2004, 21:57
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There are 3 boogies in a fun-train and a passenger is equally to ride in any one of the 3 boogies each time he rides the fun-train. If a certain passenger is to ride the fun-train 3 times, what is the probability that the client would ride in each of the three boogies.

OA:
[Reveal] Spoiler:
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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Last edited by Bunuel on 10 Apr 2013, 01:01, edited 2 times in total.
Renamed the topic and edited the question.
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17 Jun 2004, 07:26
Mark his attempts as 1,2,3
For each attempts he can choose among cars a,b,c
So there are 3 vriations for each attempt possible
Total attempts = 9
out of this only one combination makes him ride different cars
p = 1/9
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17 Jun 2004, 08:08
Hi Anand

Sorry, thats not the answer I have here.....could you give another try.
I have changed the question a little bit, instead of cars, i have made it boogies in a train.

Try again. Thanks Anand.
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17 Jun 2004, 08:18
Hi Halle

Thats not the answer either .....
I shall post the answer here, and let you guys give me an explaination ...that would help me better and u all too.

The correct answer is 2/9 (no idea how this comes).
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17 Jun 2004, 10:13
I did it with brute force
aaa
aab
aac
aba
abb
abc
aca
acb
acc

You have 9 combinations and two of them have a,b,c
Since 3 different cars are there
P = 6/27 = 2/9
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17 Jun 2004, 11:09
Actually 3!/(3^3) is correct. I think halle did a silly mistake in his computation.
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17 Jun 2004, 18:28
Hi Anand

What is a better way to do ... use the way you have done or use the short-cut (formulae) as Halle has done. Well, i do not know the formulae... whats the approcah for such problems., If someone sees a problem like this, where do we start with. A tip in this, would clear my fear on this pal.

thanks
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17 Jun 2004, 19:24
Hi,

I told you about brute force. What would you do if there were 10 cars.
You cannot use brute force. Just go through these problems at the club and make notes. You should be able to do good. More problems you solve more you will learn. You can try to remember the formulae but problems will generally be twisted in which case u need to apply more than one formula.

- Anand.

Let me know if you still need explanation of this problem.
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17 Jun 2004, 19:56

One question, how did Halle out the formulae 3!/3^3 (it should not matter if there are 3 cars, or 6 cars, right?. It value in the formulae changes.

The question, how do u translate those words in to value and formulaes. (if time permits, let me know those tricks)

Regards
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18 Jun 2004, 13:41
Hi carsen,

Yeah it should not matter how many cars are there. The only way I can explain is you try to pick a problem and solve it. I will try to find where u are going wrong and advise you.

For the above problem assume u have 10 cars (A,B,C....J ) or you could number them from 0 to 9
Each time you try to rent a car I will give you whatever is available.
It is possible that in three times you get car number 5 or 4 or anything from 0 to 9
So no of combinations is simply 10 * 10 * 10
If you had rented 5 times then no of combinations is simply
10*10*10*10*10 or 10^5

So the formula is (no of variations) ^ ( no of repeatations )
Something like how many different numbers can you make using 2 digits.

Now you have to choose 3 different cars from 10. You do that in 10C3 ways.
Say you chose 3,4,5
But in first attempt you can get 3 second 4 and third 5 or
first 5, second 3 and third 4
Therefore there are actually 3! variations.
So Desired combinations = (Choosing 3 cars ) * (arranging 3 cars )
= 10C3 * 3!
Total combinations = 10^3

so P = (10C3 * 3! ) / 10^3

In the above example you will have

3C3 * 3! / 3^3

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18 Jun 2004, 14:22
number of ways for all three = 3*2*1
all possible ways = 3*3*3 (note, repetition is allowed).
P=3*2*1/3*3*3=2/9
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21 Jun 2004, 22:59
Thanks all of you ..

Anand, Thanks for the detail explaination on the 10 cars.

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22 Jun 2004, 03:09
carsen wrote:
Hi

There are 3 boogies in a fun-train and a passenger is equally to ride in any one of the 3 boogies each time he rides the fun-train. If a certain passenger is to ride the fun-train 3 times, what is the probability that the client would ride in each of the three boogies.

Again, please explain the steps to obtain the solution.

Regards

P(first time he rides in 1st boogie, second time he rides in the second, third time he rides in the third boogie) = 1/3*1/3*1/3 = 1/27.

P(each time he rides in different boogies) = 3!*P(first = 1, second = 2, third = 3) = 6/27 = 2/9.
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09 Apr 2013, 15:59
I think it should be 1/3^3 the way i did is P(1st time)= 1/3;P(2)= 1/3;P(3)=1/3 now P(AandBandC)= 1/3*1/3*1/3 = 1/3^3 What is wrong in this?
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10 Apr 2013, 01:02
arkle wrote:
I think it should be 1/3^3 the way i did is P(1st time)= 1/3;P(2)= 1/3;P(3)=1/3 now P(AandBandC)= 1/3*1/3*1/3 = 1/3^3 What is wrong in this?

Check here: a-certain-roller-coaster-has-3-cars-and-a-passenger-is-21226.html

Hope it helps.
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Re: probability   [#permalink] 10 Apr 2013, 01:02
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