Could someone please help? best 6 papers turned in by 28 classes are re-read by organizers to ensure marking is correct.

[babyblueyeon]

 

[khansaheb]

View attachment 37467

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[Agent Smith]

[imath]\frac{2}{28}[/imath]

[imath]\frac{2 \times 4}{28 \times 27}[/imath]

Right/wrong/both/neither?

 

[blamocur]

[imath]\frac{2}{28}[/imath]

[imath]\frac{2 \times 4}{28 \times 27}[/imath]

Right/wrong/both/neither?

The op says "six best papers", so I am assuming that only papers with the ratings of 83 and 76 are in the stack. I don't know why they are listing ratings, or even mentioning, the remaining 22 papers. Am I missing something in there?

 

[Agent Smith]

The op says "six best papers", so I am assuming that only papers with the ratings of 83 and 76 are in the stack. I don't know why they are listing ratings, or even mentioning, the remaining 22 papers. Am I missing something in there?

Dunno, combinatorics ain't my strong suit. I like the question format.

 

[Dr.Peterson]

The op says "six best papers", so I am assuming that only papers with the ratings of 83 and 76 are in the stack. I don't know why they are listing ratings, or even mentioning, the remaining 22 papers. Am I missing something in there?

Yes, that's how I read it, though one has to read through it a couple times to understand it. I imagine this is probably part of a longer problem, or series of problems, which gives more context, and the extraneous data are used in another question. It may be more easily understood in context.

So we have a stack containing 83 83 76 76 76 76 in some order, and want the probability that the first two, in order, are 83 76.

It's also worth noting that the problem calls for a specific method of solving, involving the number of favorable cases over all cases (not a product of probabilities, as AS has suggested); yet event that could be answered in different valid ways: considering all arrangements of the entire stack, or (as I would do it) only of the top two.

I don't like problems like that.

 

[The Highlander]

So why aren't the required answers just: \(\displaystyle 8, 720\,\,\&\, 1.1\dot1\%\)?

It's a pity the OP didn't provide the possible answers (it looks like there's a pull-down menu perhaps offering multiple choices?).

 

[khansaheb]

[imath]\frac{2}{28}[/imath]

[imath]\frac{2 \times 4}{28 \times 27}[/imath]

Right/wrong/both/neither?

Please tell us - how did you arrive at the expressions above.
Please show detailed work.

 

[Agent Smith]

Please tell us - how did you arrive at the expressions above.
Please show detailed work.

The number of ways that the event in question can happen = 2 × 4 [2 papers of score 83, 4 papers of score 76]

The sample space (for 2 first papers) = 28 × 27

? Non?

 

[Dr.Peterson]

The number of ways that the event in question can happen = 2 × 4 [2 papers of score 83, 4 papers of score 76]

The sample space (for 2 first papers) = 28 × 27

? Non?

No, there are only 6 papers in the stack, not all 28.

 

[The Highlander]

No, there are only 6 papers in the stack, not all 28.

So you would agree that the answers (sought by the question) are probably just: \(\displaystyle 8, 720\,\,\&\, 1.1\dot1\%\)? (720 being ₆P₆.)

Or am I missing something?

 

[blamocur]

Or am I missing something?

I believe you are: 8 is the number of ways to pick the top 2 papers, but it does not take into account all the ways the remaining 4 papers can be arranged. But 720 = 6! counts those arrangements as different cases.

 

[The Highlander]

I believe you are: 8 is the number of ways to pick the top 2 papers, but it does not take into account all the ways the remaining 4 papers can be arranged. But 720 = 6! counts those arrangements as different cases.

That seemed reasonable to me. (There are ₆P₆ (ie: 720) different ways to arrange the stack of six papers of which 8 would meet the stipulated criterion.)

Are you suggesting that the 8 ways to arrange the 'successful outcomes' should be compared(?) to the 24 ways to arrange the remaining 4 papers to give the number of possible outcomes as 192 (8 × 24) and a resulting probability of \(\displaystyle 4.1\dot6\%\)?

 

[Dr.Peterson]

View attachment 37467

Let's look at a couple ways to solve this, since it's been a month.

First, if we consider only the top two papers in the stack, which are the top six papers (all either 83 or 76), there are [imath]2\times4=8[/imath] favorable cases, and [imath]_6P_2=6\times5=30[/imath] total cases, giving probability [imath]\frac{8}{30}=\frac{4}{15}=0.266...[/imath].

Second, if we consider all 6 papers in the stack, there are [imath]2\times4\times4!=192[/imath] favorable cases, out of [imath]6!=720[/imath] total cases, giving probability [imath]\frac{192}{720}=\frac{4}{15}=0.266...[/imath].

Third, using conditional probability, the probability that the first is 83 is 2/6, and the probability that the second is 76, given that the first was 83, is 4/5, giving a total probability of [imath]\frac{2}{6}\times\frac{4}{5}=\frac{4}{15}=0.266...[/imath].

It's very easy to confuse the various methods (by not focusing on a specific model), or to make other mistakes; I didn't actually get it right until one of these corrected my work on another. In combinatorics and probability, I don't trust my answer on a complicated problem (even as simple as this one) until I get the same answer two ways! And I find it important to write out each complete paragraph as I did here, stating exactly what I am thinking about.

 

[blamocur]

Are you suggesting that the 8 ways to arrange the 'successful outcomes' should be compared(?) to the 24 ways to arrange the remaining 4 papers to give the number of possible outcomes as 192 (8 × 24) and a resulting probability of 4.16˙%\displaystyle 4.1\dot6\%4.16˙%?

Almost: got each of the 8 cases of picking the first 2 there are 24 cases of arranging the rest. Also, I get a different number for 192/720

 

[The Highlander]

Let's look at a couple ways to solve this, since it's been a month.

First, if we consider only the top two papers in the stack, which are the top six papers (all either 83 or 76), there are [imath]2\times4=8[/imath] favorable cases, and [imath]_6P_2=6\times5=30[/imath] total cases, giving probability [imath]\frac{8}{30}=\frac{4}{15}=0.266...[/imath].

Second, if we consider all 6 papers in the stack, there are [imath]2\times4\times4!=192[/imath] favorable cases, out of [imath]6!=720[/imath] total cases, giving probability [imath]\frac{192}{720}=\frac{4}{15}=0.266...[/imath].

Third, using conditional probability, the probability that the first is 83 is 2/6, and the probability that the second is 76, given that the first was 83, is 4/5, giving a total probability of [imath]\frac{2}{6}\times\frac{4}{5}=\frac{4}{15}=0.266...[/imath].

It's very easy to confuse the various methods (by not focusing on a specific model), or to make other mistakes; I didn't actually get it right until one of these corrected my work on another. In combinatorics and probability, I don't trust my answer on a complicated problem (even as simple as this one) until I get the same answer two ways! And I find it important to write out each complete paragraph as I did here, stating exactly what I am thinking about.

Almost: got each of the 8 cases of picking the first 2 there are 24 cases of arranging the rest. Also, I get a different number for 192/720

So the correct answers (to go into the provided boxes) are: \(\displaystyle 192, 720\,\,\&\, 26.\dot6\%\).

Thank you @Dr.Peterson & @blamocur.

 

[Dr.Peterson]

So the correct answers (to go into the provided boxes) are: \(\displaystyle 192, 720\,\,\&\, 26.\dot6\%\).

Or it could be 8, 30, and 0.266...,though they more likely expect your set. This was the point of post #6:

It's also worth noting that the problem calls for a specific method of solving, involving the number of favorable cases over all cases (not a product of probabilities, as AS has suggested); yet [even] that could be answered in different valid ways: considering all arrangements of the entire stack, or (as I would do it) only of the top two.

 

[Agent Smith]

No, there are only 6 papers in the stack, not all 28.

There are a total of 28 papers.
My computation is only for the first 2 papers (the topmost 2).

There are [imath]28 \times 27[/imath] different permutations of 2 papers.

There are [imath]2 \times 4[/imath] different permutations with an 83 on top and 76 right below it.

---

Were you expecting the following answer?

How many ways can 6 papers be stacked out of 28?
[imath]^{28}P_6[/imath]

How many permutations of 6 papers can there be with the topmost 83 and the one below it 76?
[imath]2 \times 4 \times ^{26}P_4[/imath]

 

[Dr.Peterson]

There are a total of 28 papers.
My computation is only for the first 2 papers (the topmost 2).

There are [imath]28 \times 27[/imath] different permutations of 2 papers.

There are [imath]2 \times 4[/imath] different permutations with an 83 on top and 76 right below it.

Read the problem carefully:



The other 22 papers are somewhere else; they don't come into the problem at all. They can't be in the stack.

 

[Agent Smith]

Read the problem carefully:

View attachment 37694

The other 22 papers are somewhere else; they don't come into the problem at all. They can't be in the stack.

Read the problem carefully:

View attachment 37694

The other 22 papers are somewhere else; they don't come into the problem at all. They can't be in the stack.

Gracias, muchas, your guidance is much appreciated.

[imath]\frac{2 \times 4 \times ^4P_4}{^6P_6} = \frac{192}{720} = 0.2\overline 6 [/imath]