There are 3 cars going on a autobahn. At the end of the autobahn there are 3 turnpikes. What is the probability that one of them will pass through a different turnpike than the others and the other 2 will pass through the same turnpike.

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- Jan 28th 2013, 09:02 AM #1

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- Jan 28th 2013, 03:48 PM #2
## Re: need help

The first car has 3 turnpike he can take, the second car as 2 way or choosing a turnpike and the third one must take the same as the second one, so he only has 1 choice.

You end up with 3*2*1=3!=6.

Now since we've probability, we'll guess each car as the same %chance of going in each turnpike, so we must count how many possible way there is.

Every car as 3 choices so you end up with 3*3*3=27

So the probabily is 6/27=2/9

That look good to me. But I might be wrong, so be sure to double check

- Jan 28th 2013, 06:32 PM #3

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## Re: need help

Hello, kastamonu!

There are 3 cars going on a autobahn. At the end of the autobahn there are 3 turnpikes.

What is the probability that one of them will pass through a different turnpike than the others

and the other two will pass through the same turnpike?

Each car has a choice of 3 turnpikes.

. . There are possible outcomes.

Suppose all three use theturnpike.*same*

. . There are 3 ways:

Suppose all three useturnpikes.*different*

. . There are ways:

Hence, there are: . ways.

- Jan 28th 2013, 11:29 PM #4

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- Jan 28th 2013, 11:29 PM #5

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- Jan 29th 2013, 10:28 AM #6

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## Re: need help

Sorry, but Soroban's answer is the correct one. Barioth's answer is the probability the a

**specified**one of the cars, specifically, the one he calls "the first car", takes one of the turnpike and the other two both take the same one of the other two. Soroban's answer is the probability that "one car takes one turn pike and the other two take the same one of the other two" which was the question.

Because there are 3 cars, and any one can be the car that goes "alone", so you can get the correct answer by multiplying Barioth's answer by 3: 3(2/9)= 2/3.

- Jan 29th 2013, 10:42 AM #7

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- Jan 29th 2013, 05:11 PM #8