probability

[lolita20]

Hi im new here and I really need help with this question

1) 6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A and B occupy the first two positions.


2) 6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A is placed somewhere before B in the queue.

 

[Dr.Peterson]

Hi im new here and I really need help with this question

1) 6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A and B occupy the first two positions.


2) 6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A is placed somewhere before B in the queue.

In order to give appropriate help, we need to have some idea of what you are able to do, and where you are stuck. Did you read this?

READ BEFORE POSTING

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If you can't even get started, tell us what you have learned been taught, that this is intended to exercise.

 

[lolita20]

In order to give appropriate help, we need to have some idea of what you are able to do, and where you are stuck. Did you read this?

READ BEFORE POSTING

Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. A summary is available here, but please read the complete guidelines below at your earliest...

www.freemathhelp.com

If you can't even get started, tell us what you have learned been taught, that this is intended to exercise.

thanks , I solved this problem but im not sure if it is correct,

6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A is placed somewhere before B in the queue.
solution in picture:

 

[lolita20]

In order to give appropriate help, we need to have some idea of what you are able to do, and where you are stuck. Did you read this?

READ BEFORE POSTING

Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. A summary is available here, but please read the complete guidelines below at your earliest...

www.freemathhelp.com

If you can't even get started, tell us what you have learned been taught, that this is intended to exercise.

here is the answer for this
6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A and B occupy the first two positions.
is this correct?

 

[pka]

2) 6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A is placed somewhere before B in the queue.

In any line of these six people \(A\) will either be before \(B\) or after \(B\). Think about it. What is the probability \(A\) before \(B\)?

 

[Dr.Peterson]

here is the answer for this
6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A and B occupy the first two positions.
is this correct?

Could you make that a little easier to read -- and a little less verbose? Maybe paste it in as text, rather than as an image?

What if B, rather than A, is in the first position?

thanks , I solved this problem but im not sure if it is correct,

6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A is placed somewhere before B in the queue.
solution in picture:

Did you do the thinking pka suggested (after you had written)? This doesn't take nearly so much work.

 

[Steven G]

In addition to what pka suggested, shouldn't the probability that A is before B be the same probability as B is before A. What should these two probabilities add up to?

 

[JeffM]

How many ways can you order k distinct things without replacement!

This is one of the combinatoric formulas, you should memorize.

[MATH]k![/MATH]
If you have one item, you can order it in only one way. or 1!.

With two items, you can order it CD or DC, two ways or 2!.

With three items, you can order them CDE, CED, DCE, DEC, ECD, or EDC, six ways or 3!.

With four items, you can order them CDEF. CDFE, CEDF, CEFD, CFDE, CFED, DCEF, DCFE, DECF, DEFC, DFCE, DFEC, ECDE, ECED, EDCF, EDFC, EFCD, EFDC, FCDE, FCED, FDCE, FDEC, FECD, or FEDC, 24 ways or 4!.

How many distinct ways can A and B occupy the first two spots: 2!.

So the number of distinct ways that A and B can occupy the first two spots is 2! = 2. The number of ways that that C, D, E, and F can occupy the last four spots is 4! The number of distinct ways that the six can occupy all six spots? 6!

So the probability that A and B are in the first two spots is

[MATH]\dfrac{2! * 4!}{6!} = \dfrac{2 * 4!}{6 * 5 * 4!} = \dfrac{1}{15}.[/MATH]
We can get that answer a different way.

What is the probability of getting A in the first spot and B in the second spot?

[MATH]\dfrac{1}{6} * \dfrac{1}{5} = \dfrac{1}{30}.[/MATH]
Where did that come from?

What is the probability of getting B in the first spot and A in the second spot.

[MATH]\dfrac{1}{6} * \dfrac{1}{5} = \dfrac{1}{30}.[/MATH]
But A first and B second and B first and A second are mutually exclusive so the probability of one or the other is simply the sum of
the two probabilities.

[MATH]\dfrac{1}{15} + \dfrac{1}{15} = \dfrac{2}{30} = \dfrac{1}{15}.[/MATH]
There are usually several ways to attack such problems. Try to think of two simple ones and see if the answers agree. If they do agree, you are very likely correct.

 

[lolita20]

How many ways can you order k distinct things without replacement!

This is one of the combinatoric formulas, you should memorize.

[MATH]k![/MATH]
If you have one item, you can order it in only one way. or 1!.

With two items, you can order it CD or DC, two ways or 2!.

With three items, you can order them CDE, CED, DCE, DEC, ECD, or EDC, six ways or 3!.

With four items, you can order them CDEF. CDFE, CEDF, CEFD, CFDE, CFED, DCEF, DCFE, DECF, DEFC, DFCE, DFEC, ECDE, ECED, EDCF, EDFC, EFCD, EFDC, FCDE, FCED, FDCE, FDEC, FECD, or FEDC, 24 ways or 4!.

How many distinct ways can A and B occupy the first two spots: 2!.

So the number of distinct ways that A and B can occupy the first two spots is 2! = 2. The number of ways that that C, D, E, and F can occupy the last four spots is 4! The number of distinct ways that the six can occupy all six spots? 6!

So the probability that A and B are in the first two spots is

[MATH]\dfrac{2! * 4!}{6!} = \dfrac{2 * 4!}{6 * 5 * 4!} = \dfrac{1}{15}.[/MATH]
We can get that answer a different way.

What is the probability of getting A in the first spot and B in the second spot?

[MATH]\dfrac{1}{6} * \dfrac{1}{5} = \dfrac{1}{30}.[/MATH]
Where did that come from?

What is the probability of getting B in the first spot and A in the second spot.

[MATH]\dfrac{1}{6} * \dfrac{1}{5} = \dfrac{1}{30}.[/MATH]
But A first and B second and B first and A second are mutually exclusive so the probability of one or the other is simply the sum of
the two probabilities.

[MATH]\dfrac{1}{15} + \dfrac{1}{15} = \dfrac{2}{30} = \dfrac{1}{15}.[/MATH]
There are usually several ways to attack such problems. Try to think of two simple ones and see if the answers agree. If they do agree, you are very likely correct.

thanks for clarifying

But it really hard for me to understand your writing
i just used this answer but not sure if it correct
to remind you im a begginer in math and the english is not our first language here is the answer

 

[Dr.Peterson]

thanks for clarifying

But it really hard for me to understand your writing
i just used this answer but not sure if it correct
to remind you im a begginer in math and the english is not our first language here is the answer

Your work for the first part is correct.

For the second, you appear to be assuming that A is in the first position, which is not required. Or perhaps you are just ignoring the need to place all 6 people.

One good method, which has been suggested, is to simply observe that for each arrangement in which A is before B, there is another in which B is before A (formed by interchanging the two). This shows that there are equal numbers in each case, so exactly half of all outcomes have A before B.

If you want a more direct approach, you could modify what you did. If A is in the first position, then there are 5 places to put B following it: A _ _ _ _ _. Then there are 4! ways to fill in the rest. If A is in the second position, ... . Keep going like that, and add them up.

Or, you could count the number of ways to choose two places in which A and B will go, and then put A and B in those two places and in that order (which can be done in only one way). Finish that.

I like finding multiple ways to get the same answer; these three are very different!

 

[pka]

2) 6 people, A, B, C, D, E, F are placed randomly in a queue. What is the probability that A is placed somewhere before B in the queue.

The answer for #2 ia \(0.5\) For more help, Explain why!

 

[lolita20]

The answer for #2 ia \(0.5\) For more help, Explain why!

because A B C D E F can be placed in 6! ways.
there are 10 * 4! ways to place A somewhere before B
there are 5! ways to place A exactly before B
so (10 + 5) * 4! = 360 ways to place A before B

the probabbility A placed before B is = 360/6! = 1/2= 0.5
???????? is this the reason?

 

[JeffM]

thanks for clarifying

But it really hard for me to understand your writing
i just used this answer but not sure if it correct
to remind you im a begginer in math and the english is not our first language here is the answer

It is not always easy for us to give an answer that is right for a particular student because students do not always tell us what they know.

First, I provided two answers to part 1 of your question.

Second, looking at the hand written sheet that you provided, you seem to have used the first solution I provided. Is there some specific aspect about it you do not understand?

In many cases, there is an intuitive way to determine an answer in probability and an efficient way. I showed you two ways, each reaching the same result. If one way seems obscure, does the other one make sense?

 

[ISTER_REG]

A _ _ _ _ _ --> 5 choices for B
X A _ _ _ _ --> 4 choices for B
X X A _ _ _ --> 3 choices for B
X X X A _ _ --> 2 choices for B
X X X X A _ --> 1 choice for B
5+4+3+2+1 = 15 choices

There are in every choice for A and B, 4! permutations of the remaining letters, so the number of arrangements is 15*4! = 360.
The number of total arrangements are 6!, so 15*4!/6! = 1/2 = 0.5

Short question: Where did you found this interesting questions? From a textbook? Please name it

 

[pka]

A _ _ _ _ _ --> 5 choices for B
X A _ _ _ _ --> 4 choices for B
X X A _ _ _ --> 3 choices for B
X X X A _ _ --> 2 choices for B
X X X X A _ --> 1 choice for B
5+4+3+2+1 = 15 choices
There are in every choice for A and B, 4! permutations of the remaining letters, so the number of arrangements is 15*4! = 360.
The number of total arrangements are 6!, so 15*4!/6! = 1/2 = 0.5

@ISTER_REG, You are missing the whole point of this discussion
Suppose that we have a collections of \(10^2\) distinct symbols among which are \(\alpha~\&~\beta\) the Greek letters.
If we randomly form a queue of those hundred symbols, what is the probability that in the queue \(\alpha\) appears before \(\beta~?\)
Well, the answer is \(0.5\). Any calculations are a waste of time. In any such queue \(\alpha\) is either to the right of or to the left of \(\beta~.\)
Save calculations for a more interesting question: in such a queue what is the probability that \(\alpha~\& ~\beta\) are separated by exactly ten other symbols?

 

[JeffM]

@pka and @ISTER_REG

Pka’s theorem is valid and pretty. And it may seem intuitive to us, but intuition is subjective. It may not be intuitive to a beginner. In any case, math students are not usually encouraged to rely on intuition. Don’t we need a proof of pka’s theorem?

Let n be the number of distinct symbols, with n > 1. Alpha and beta are among those symbols. What want to know is

[MATH]p = \text {P} ( \beta \text { follows } \alpha ).[/MATH] But

[MATH]p = \sum_{j=1}^n \text {P}(\alpha \text { is in position } j) * \text {P} ( \beta \text { in a later position given } \alpha \text { in position } j).[/MATH]
The probability that beta is in a later position given that alpha is in the final position must be zero. Furthermore, the probability that alpha is in a specific location is 1/n because assignment is random.

[MATH]\therefore p = \dfrac{1}{n} * \sum_{j=1}^{n-1} \text {P} ( \beta \text { in a later position given } \alpha \text { in position } j).[/MATH]
If beta follows alpha, which is in the j position, it must be the case that, of the n - 2 symbols different from alpha and beta, j - 1 of them precede alpha so the conditional probability that beta follows alpha is

[MATH]\dbinom{n - 2}{j - 1} \div \dbinom{n - 1}{j - 1} = \dfrac{(n - 2)!}{(j - 1)! * \{(n - 2) - (j - 1)\}} \div \dfrac{(n - 1)!}{(j - 1)! * \{(n - 1) - (j - 1)\}!} = [/MATH]
[MATH]\dfrac{(n - 2)!}{(j - 1)! * (n - 1 - j)!} * \dfrac{(j - 1)! * (n - j)!}{(n - 1)!} = \dfrac{n - j}{n - 1}. [/MATH]
[MATH]\therefore p = \dfrac{1}{n} * \sum_{j=1}^{n-1} \dfrac{n - j}{n - 1} = \dfrac{1}{n(n - 1)} * \left \{ \left ( \sum_{j=1}^{n-1} n \right ) - \left ( \sum_{j=1}^{n-1} j \right ) \right \} = \dfrac{1}{n(n - 1)} * \left ( n(n - 1) - \dfrac{(n - 1)\{(n - 1) + 1\}}{2} \right ) \implies [/MATH]
[MATH]p = \dfrac{n(n - 1)}{n(n - 1)} * \left (1 - \dfrac{1}{2} \right ) = \dfrac{1}{2}.[/MATH]
Lovely general solution.

But I doubt a beginner can prove it; indeed, a beginner may not even be able to follow the proof.

I think ISTER_REG was right on the money with the answer in #14.

 

[Dr.Peterson]

Don’t we need a proof of pka’s theorem?

I gave the outline of a simple proof in #10:

One good method, which has been suggested, is to simply observe that for each arrangement in which A is before B, there is another in which B is before A (formed by interchanging the two). This shows that there are equal numbers in each case, so exactly half of all outcomes have A before B.

In an actual proof, I would mention a one-to-one correspondence.

On the other hand, I entirely agree that a beginner is probably expected to use more direct methods, which I also suggested in #10; and ISTER_REG's approach is one of them. Although pka's approach is an excellent one, it is not necessarily right to say,

You are missing the whole point of this discussion

We don't know the context of the problem, so we can't say what was intended. And to me, the whole point of the discussion is that there are multiple ways to solve such a problem. I think all the methods that have been mentioned are good; in an advanced class, I would expect the more sophisticated method to be seen immediately, but this is probably not that class. (That's why I asked about context in #2.)

 

[JeffM]

I gave the outline of a simple proof in #10:

In an actual proof, I would mention a one-to-one correspondence.

Sorry. I missed that.

Yours is a far more elegant proof of pka's theorem than mine. A beginning student would likely follow that proof though I would not expect a beginner to find it.

I repeat that I had no doubts about the validity of pka's answer. I just doubted that a self-declared "beginner" could comprehend it.