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Probability Question - Please help

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Probability Question - Please help [#permalink] New post 30 Aug 2009, 00:09
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A roller coaster has 3 cars and a passenger is equally likely to ride in any 1 of the 3 cars each time that a passenger rides the roller coaster. If he rides 3 times, what is the probability the he will ride in each of the 3 cars?

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Re: Probability Question - Please help [#permalink] New post 31 Aug 2009, 05:31
konst75 wrote:
A roller coaster has 3 cars and a passenger is equally likely to ride in any 1 of the 3 cars each time that a passenger rides the roller coaster. If he rides 3 times, what is the probability the he will ride in each of the 3 cars?

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I am only guessing this approach. This problem is very much like saying you are flipping three coins, 3 times, what is the probability of getting a heads on each coin only in each flip? First flip only first coin gives heads, second flip only second coin gives heads and third flip only third coin gives heads.

The probability of taking one specific seat among the 3 seats is 1/3, and that of not taking it is 2/3. So you have a bernoulli's trials when 3 times, he might take that seat or not. This is probability with just one seat, with 3 seats, multiply with 3 in the end. The answer must be 3C3(1/3)^3(2/3)^0(3)??
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Re: Probability Question - Please help [#permalink] New post 31 Aug 2009, 05:53
is it 1/6 let me know i will explain my mind here...but not sure
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Re: Probability Question - Please help [#permalink] New post 31 Aug 2009, 09:00
I believe the answer for this problem was 2/9...
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Re: Probability Question - Please help [#permalink] New post 31 Aug 2009, 09:52
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Probability of getting in cars 1,2,or 3 in first try = 3/3
Probability of getting in car(other than car occupied in first try) in second try = 2/3
Probability of getting in car left after first and second try = 1/3

Total probability = (3/3)(2/3)(1/3) = 2/9
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Re: Probability Question - Please help [#permalink] New post 07 Sep 2009, 12:09
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Number the cars as 1,2&3. So possible outcomes are: 123 132 231 213 312 321 - total 6 ways
Total number of possible outcomes : 3*3*3 = 27

probability : 6/27 = 2/9
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Re: Probability Question - Please help [#permalink] New post 08 Sep 2009, 13:16
getmba wrote:
Probability of getting in cars 1,2,or 3 in first try = 3/3
Probability of getting in car(other than car occupied in first try) in second try = 2/3
Probability of getting in car left after first and second try = 1/3

Total probability = (3/3)(2/3)(1/3) = 2/9


Nice explanation!
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Re: Probability Question - Please help [#permalink] New post 16 Sep 2009, 10:14
Soln: Total number of possible ways in which he could have travelled is, 3 * 3 * 3 = 27.

Now to find out the number of possible ways in which he would have travelled in each car once,
the first time he could have travelled in any of the 3 cars = 3 ways
the second time he could have travelled in any of the other 2 cars = 2 ways
the third time he could have travelled in the left over 1 car = 1 ways
Hence total number of ways = 3 * 2 * 1 = 6

Probabality = (Total no. of favourable outcomes)/(Total no. of possible outcomes)
= 6/27
= 2/9
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Re: Probability Question - Please help [#permalink] New post 02 May 2011, 01:00
Interesting numerical.

I ended up with a 4/9 instead. :)
[(a b! c!) + (a! b c! ) + (a! b! c)] *3
a=1/3 a! = 2/3.
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Re: Probability Question - Please help   [#permalink] 02 May 2011, 01:00
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