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Senior Manager
Joined: 25 Dec 2003
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There are 3 boogies in a funtrain and a passenger is equall [#permalink]
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16 Jun 2004, 21:57
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88% (02:02) correct
13% (02:13) wrong based on 13 sessions
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There are 3 boogies in a funtrain and a passenger is equally to ride in any one of the 3 boogies each time he rides the funtrain. If a certain passenger is to ride the funtrain 3 times, what is the probability that the client would ride in each of the three boogies. OA: OPEN DISCUSSION OF THIS QUESTION IS HERE: acertainrollercoasterhas3carsandapassengeris21226.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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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



Senior Manager
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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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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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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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Actually 3!/(3^3) is correct. I think halle did a silly mistake in his computation.



Senior Manager
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Hi Anand
What is a better way to do ... use the way you have done or use the shortcut (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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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.



Senior Manager
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Thanks for the reply
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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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
I hope this is helpful.



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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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Senior Manager
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Thanks all of you ..
Anand, Thanks for the detail explaination on the 10 cars.
The answer is correct.
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Re: probability [#permalink]
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22 Jun 2004, 03:09
carsen wrote: Hi
Please explain your answer stepbystep... thanks guys
There are 3 boogies in a funtrain and a passenger is equally to ride in any one of the 3 boogies each time he rides the funtrain. If a certain passenger is to ride the funtrain 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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Re: probability [#permalink]
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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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Re: probability [#permalink]
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10 Apr 2013, 01:02











