Probability question

[TallBill]

Say that there is a class of 50 students, and the top 10 receive honors. There are four instructors.

One particular instructor has 15 of the 50 students, and five of those 15 makes the top 10 overall. The other five honors students come from the pool of 35 students.

How likely would this be?

 

[DaTT]

Is 100% the correct answer?

 

[z1ggy]

Yah... Feeling like this is a trick question or something because it seems too obvious....

Stop that.

 

[kranky]

You'd have to make a lot of assumptions.

Is the quality of students equally distributed across all four instructors?
Are all four instructors equally skilled?
Is the type of student (non-traditional/traditional, night class vs. day class vs. Saturday class) equal across all four instructors?

It might be presumptuous to assume one of the four instructors is an easy grader.

 

[TallBill]

You'd have to make a lot of assumptions.

Is the quality of students equally distributed across all four instructors?
Are all four instructors equally skilled?
Is the type of student (non-traditional/traditional, night class vs. day class vs. Saturday class) equal across all four instructors?

It might be presumptuous to assume one of the four instructors is an easy grader.

This is exactly what happened.

Basically I'm number 11, and one instructor is known for being ridiculously easy.

Only 1 person from my group of 16 made the top 10.

 

[Possessed Freak]

So... 50 balls, 10 are red, 40 blue. You pick 15 balls at random and EXACTLY 5 are red?

 

[TallBill]

So... 50 balls, 10 are red, 40 blue. You pick 15 balls at random and EXACTLY 5 are red?

Yes. Or alternatively, 5 or more are red.

I'm drawing a blank on how to calculate this.

 

[glenn1]

Yes. Or alternatively, 5 or more are red.

I'm drawing a blank on how to calculate this.

Here you go.

http://www.wikihow.com/Calculate-Probability

 

[magomago]

how likely would WHAT be? If its that the other 5 top students come from the remaining 35 students, then DUH there is 100% certainity that the other 5 students must have been selected from the remaining 35 students.

Looks like this is a great example of teaching conditional probability, and making students remember to actually read the question to see what is being asked.

 

[TallBill]

how likely would WHAT be? If its that the other 5 top students come from the remaining 35 students, then DUH there is 100% certainity that the other 5 students must have been selected from the remaining 35 students.

Looks like this is a great example of teaching conditional probability, and making students remember to actually read the question to see what is being asked.

That five came from 15, and five came from 35.

 

[ElFenix]

you switch doors




do you mean class like a high school class?

 

[magomago]

Yes. Or alternatively, 5 or more are red.

I'm drawing a blank on how to calculate this.

Hypergeometric distribution. This is what you use for probability without replacement when your sample sizes are pretty small.

http://en.wikipedia.org/wiki/Hypergeometric_distribution

Its the first equation on that page. It looks complex, but just remember that the numbers in the parenthesis is an easily calculated binomial coefficient; calculate it first equation on "Factorial formula"
http://en.wikipedia.org/wiki/Binomial_coefficient

Edit:

For the example provides, there is a 9.49% (lets just say about 10%) chance that exactly 5 balls are red when 15 balls are drawn, in a population of 50 balls where 40 are blue and 10 are red.

Here is the matlab code (nchooseK simply does the math for a binomial coefficient):
N= 50
K=10
n=15
k=5

nchoosek(K,k) * nchoosek(N-K,n-k) / nchoosek(N,n)

 

[TallBill]

you switch doors




do you mean class like a high school class?

Work related class.

 

[Newell Steamer]

Ppppfffftt!!

This is totes easy - you just ask; which road would your brother pick? and then choose the opposite!!

BAM.

 

[dullard]

Hypergeometric distribution. This is what you use for probability without replacement when your sample sizes are pretty small.

http://en.wikipedia.org/wiki/Hypergeometric_distribution

Its the first equation on that page. It looks complex, but just remember that the numbers in the parenthesis is an easily calculated binomial coefficient; calculate it first equation on "Factorial formula"
http://en.wikipedia.org/wiki/Binomial_coefficient

Or just link to a calculator:
http://stattrek.com/online-calculator/hypergeometric.aspx

Population size: 50
Number of successes in that 50 person population: 10
Sample size: 15
Number of successes in that 15 person sample: 5

Hypergeometric Probability: P(X = 5) is 9.49%. There is a 9.49% chance of exactly 5 honor students in that easy instructor's class of 15, if randomly assigned.

Cumulative Probability: P(X > 5) is 12.49%. There is a 12.49% chance of at least 5 honor students in that easy instructor's class, if randomly assigned.

 

[ElFenix]

Work related class.

i didn't convey the question very well, but i suppose i have my answer. i was envisioning something like one of my history survey classes in college where there were different graders and the prof calibrated the graders by having them all grade the same 10 papers each.

some sort of TPS class i'd assume the testing would be the same for everyone, even with different instructors.

 

[magomago]

Or just link to a calculator:
http://stattrek.com/online-calculator/hypergeometric.aspx

Population size: 50
Number of successes in that 50 person population: 10
Sample size: 15
Number of successes in that 15 person sample: 5

Cumulative Probability: P(X > 5) is 12.49%

Sweetheart I'm teaching the man how to fish!

 

[Mark R]

There are C(50,10) ways of choosing 10 people out of 50.
There are C(15,5) ways of choosing exactly 5 out of 15, and C(35, 5) for exactly 5 out of 35.

The probability of there being exactly 5 in each group is therefore (C(15,5) * C(35,5)) / C(50,10) = 0.95%

 

[magomago]

There are C(50,10) ways of choosing 10 people out of 50.
There are C(15,5) ways of choosing exactly 5 out of 15, and C(35, 5) for exactly 5 out of 35.

The probability of there being exactly 5 in each group is therefore (C(15,5) * C(35,5)) / C(50,10) = 0.95%

Me thinks you have a decimal point issue.

 

[TallBill]

Or just link to a calculator:
http://stattrek.com/online-calculator/hypergeometric.aspx

Population size: 50
Number of successes in that 50 person population: 10
Sample size: 15
Number of successes in that 15 person sample: 5

Hypergeometric Probability: P(X = 5) is 9.49%. There is a 9.49% chance of exactly 5 honor students in that easy instructor's class of 15, if randomly assigned.

Cumulative Probability: P(X > 5) is 12.49%. There is a 12.49% chance of at least 5 honor students in that easy instructor's class, if randomly assigned.

Thanks man. Would have done the leg work myself, but using a phone on shitty WiFi is less than desirable.

Most of the grades are subjective, the instructors grade their own groups on a wide variety of tasks. On the computerized testing, scores were far more proportional.

 

[TallBill]

i didn't convey the question very well, but i suppose i have my answer. i was envisioning something like one of my history survey classes in college where there were different graders and the prof calibrated the graders by having them all grade the same 10 papers each.

some sort of TPS class i'd assume the testing would be the same for everyone, even with different instructors.

Heh I wish, I would have done better this way. There were far too many things that had to be evaluated though. Interesting concept though.

 

[Schmide]

40c10 * 10c5

847,660,528 * 252 = 1 in 213,610,453,056

Or so

 

[rudeguy]

TB I'm sure I don't have to tell you that shit like this is never fair. Its much more about who swallowed the load than who did better.

For future reference: http://www.wolframalpha.com/

There is an Android app for you too. Figure any problem you can think of.

 

[TallBill]

TB I'm sure I don't have to tell you that shit like this is never fair. Its much more about who swallowed the load than who did better.

For future reference: http://www.wolframalpha.com/

There is an Android app for you too. Figure any problem you can think of.

Haha I know, which is why I don't care. My real supervisors know my abilities. It does hurt when going for promotion, but I'll never kiss ass.

 

[PlanetJosh]

There's politics in ranking students in college classes?