outcomes problem (QR)

[ds2012tg]

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Hi guys, can somebody help me with this problem? I cant determine if I should use the combinations or the permutations formula. Thanks!!!!!

Four students are racing in a track and field competition. If you are told that the student from North High School comes in second, how many possible outcomes to the race are there?

 

[aqz]

Neither.

_ _ _ _ you know that one student is going to be in second so _ x _ _

That gives you 3 choices for first, 2 choices for third, 1 choice for last = 3*2*1 = 6

 

[Smurf12]

Permutation order does matter: i.e. Out of ten people only three to be chosen for three position: president, vp, secretary(president, vp, secretary are not he same).

another example: consider the order of letters A, B where order does matter: AB, BA are not the same

Combination order does not matter: I.e. Out of ten people only three to be chosen for three position in the senate(either position in the senate are the same).

another example: consider the order of letter A,B where order does not matter: AB, BA are the same
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For the sake of the problem use permutations instead since it's a race and there are 1st, 2nd, etc positions, but this problem has other stipulations you have to consider.

 

[ds2012tg]

thanks a lot!!!! i was driving myself crazy with the formulas.

 

[ds2012tg]

thanks a lot!!!!!!!!

 

[Moka11]

Reviving old thread..So if in 1st place its 3*2*1 and same goes for 3rd and 4th wouldnt this be 6*6*6?

 

[KyoPhan]

Reviving old thread..So if in 1st place its 3*2*1 and same goes for 3rd and 4th wouldnt this be 6*6*6?

In selecting for 1st place out of 3 people, there are 3 possible outcomes for 1st place only. 1st place can be person A, B or C.
AXBC
AXCB

BXAC
BXCA

CXAB
CXBA

This problem is basically saying how many ways can you arrange 3 people, since the 2nd place position is not adjustable.


You might be confusing it with a this sort of problem.
"Your keyboard only has the letters A, B, and C. How many ways can you type a 4 letter word."

In this case, we can recycle A, B, and C. It is not restricted to one use like in our race problem. We can have AAAA, AABB, ACBA, etc...To solve this problem, we have to consider each slot separately.
_ _ _ _

The first slot can have 3 possible outcomes; A, B, or C.
The same thing is true for the 2nd slot, 3rd slot, and the 4th.

So the number of ways is just 3x3x3x3.
Which turns out to be 81.

 

[oms47]

Hey,

Can I get some help with this one. "If 1/3 of cars driven by residents in a specific town are japanese made, what is the probability of seeing at least one japanese car out of every three cars on the road?"

Here order doesn't matter so we use the combinations formula? I'm lost now.

Here is my thinking... probability * total combinations for seeing atleast one car

therefore: 2*2*2 = 8 total combinations (2 for the first, 2 for the 2nd, 2 for the third)

How do you figure out probability?
.333*.6666*.6666=?
or is it
.3333*.33333*.6666=?
or is it
.3333*.6666*.6666=?
or all the above and why?

Thanks!

 

[KyoPhan]

Hey,

Can I get some help with this one. "If 1/3 of cars driven by residents in a specific town are japanese made, what is the probability of seeing at least one japanese car out of every three cars on the road?"

Here order doesn't matter so we use the combinations formula? I'm lost now.

Here is my thinking... probability * total combinations for seeing atleast one car

therefore: 2*2*2 = 8 total combinations (2 for the first, 2 for the 2nd, 2 for the third)

How do you figure out probability?
.333*.6666*.6666=?
or is it
.3333*.33333*.6666=?
or is it
.3333*.6666*.6666=?
or all the above and why?

Thanks!

Edit: Fixed stupid math mistake.
I'm not sure how you would do it with combinations. But for me the easiest way to think about it is to do the following:
The only combination you DON'T include is when all of the cars are NOT japanese made.

The probability of 1 car not being japanese made is 2/3! You get this buy doing (1-1/3). So for the probability that ALL 3 cars are non-japanese has to be ( 2/3 x 2/3 x 2/3 ). This comes out to be a 8/27, meaning that there is a 8/27 chance that all are non-japanese made.

Thus 1-(8/27), will have atleast 1 japanese made car. This comes out to be 19/27.

 

[oms47]

the answer is 19/27

 

[KyoPhan]

An example of when order DOESN'T matter:
We have 8 people available. How many ways can we select 5 people to make a basket ball team. Here, the order in which we pick out people doesn't matter. I can pick in these orders:
A,B,C,D,E
or
A,C,E,D,B

These 2 combination of people are the same! So they only count as 1 combination.
If we use the formula, it'll be n!/(n-r)!r!
Thus the answer would be 8!/(8-5)!5!, or 8!/(3!)5!. Simplifying it will give (8x7x6)/(3x2)

Now if I adjust the problem a little bit to make it so that order DOES matter.

We have 8 people available. How many ways can I line up 5 people.
Here order matters.

A,B,C,D,E
or
A,C,E,D,B

These 2 ways count as 2 combinations since they are arranged differently.
Now our formula will be n!/(n-r)!
8!/(8-5)! or 8!/3!. Simplifying will give us (8x7x6x5x4)/(3x2).

 

[KyoPhan]

Edit:
Fixed it. Did a stupid math mistake.
I said 2/3 x 2/3 x 2/3 = 6/27 earlier.
should have been 8/27.

 

[Moka11]

Edit: Fixed stupid math mistake.
I'm not sure how you would do it with combinations. But for me the easiest way to think about it is to do the following:
The only combination you DON'T include is when all of the cars are NOT japanese made.

The probability of 1 car not being japanese made is 2/3! You get this buy doing (1-1/3). So for the probability that ALL 3 cars are non-japanese has to be ( 2/3 x 2/3 x 2/3 ). This comes out to be a 8/27, meaning that there is a 8/27 chance that all our non-japanese made.

Thus 1-(8/27), will have atleast 1 japanese made car. This comes out to be 19/27.

Thanks for answering both our Q's. 👍