#1

This is purely a hypothetical question, but I can't think of any formula that deals with it (granted, I don't actually know a lot of maths)

Say I had a test, where 20 items need to be revised. You are informed that you will be tested on 3 of these items.

What is the minimum amount of objects that need to be revised to make it statistically still in your favour (i.e. over 50%) that the objects you revised WILL come up?

Say I had a test, where 20 items need to be revised. You are informed that you will be tested on 3 of these items.

What is the minimum amount of objects that need to be revised to make it statistically still in your favour (i.e. over 50%) that the objects you revised WILL come up?

*Last edited by fender_696 at Oct 20, 2009,*

#2

all of them is the only way you'll get 100%

#3

erm sounds like u want me to calculate the amount of items u need to study to pass?

meh, sorry i'm doing lambda calculus atm, not in the mood to calculate chances

meh, sorry i'm doing lambda calculus atm, not in the mood to calculate chances

*Last edited by numinis at Oct 19, 2009,*

#4

Statisticians*

#5

11 minimum. Simple really.

#6

Ok, I've done the calculations...

If you revise twenty you're 100% guaranteed that you will have revised the right three.

If you revise twenty you're 100% guaranteed that you will have revised the right three.

#7

well, better than a pass, assuming 40% is a passerm sounds like u want me to calculate the amount of items u need to study to pass?

meh, sorry i'm doing lambda calculus atm, not in the mood to calculate chances

#8

Statisticians*

My dad is a statistician. He'd be so pissed if he read that that guy misspelled it.

#9

^ you have a better chance than me!

please, i have been up half the night trying to work this out

all i realised is that there are 4 variables:

1. how many you have revised

2. how many you haven't

3. how many are picked

4. how many they can be picked out of

please, i have been up half the night trying to work this out

all i realised is that there are 4 variables:

1. how many you have revised

2. how many you haven't

3. how many are picked

4. how many they can be picked out of

#10

11 minimum. Simple really.

This.

You are looking for the least amount to study, while keeping it in statistical favour that 3 you have studied are more likely to come up, than not.

So if you study 11. Then there is a greater chance that the 3 that come up will come from the 11 you have studied, as opposed to the 9 that you havent.

#11

not exactlyThis.

You are looking for the least amount to study, while keeping it in statistical favour that 3 you have studied are more likely to come up, than not.

So if you study 11. Then there is a greater chance that the 3 that come up will come from the 11 you have studied, as opposed to the 9 that you havent.

it's the chance that all 3 WILL come up. obviously there is no way of knowing this, so what is the minimum that have to be revised so that it is still in your favour that the 3 will come up

i reckon it's around 16-17

again, i'm not very good at maths, but there must be a formula, or graph

#12

not exactly

it's the chance that all 3 WILL come up. obviously there is no way of knowing this, so what is the minimum that have to be revised so that it is still in your favour that the 3 will come up

i reckon it's around 16-17

again, i'm not very good at maths, but there must be a formula, or graph

The only way you can have 3 that you have studied, definitely come up, is if you study all 20.

There is no other way around it.

#13

^ i have had the test, so it is not that anymore

but what is it so that it is over 50% that the 3 will come i.e. it's logically sound to revise this many

i know there is a formula, and it's winding me up

but what is it so that it is over 50% that the 3 will come i.e. it's logically sound to revise this many

i know there is a formula, and it's winding me up

#14

19/20

That way at least two of the topics will be in your favor, making it more than 66% in your favor?

Or im completely off the mark

That way at least two of the topics will be in your favor, making it more than 66% in your favor?

Or im completely off the mark

#15

You can solve the hypergeometric distribution's probability mass function with the given parameters, when the probability is 0,5.

And yes, when you revise 16 items, the probability is allmost 50%, 49,1% to be exact. Revising 17 items pushes it up to 59,6%.

And yes, when you revise 16 items, the probability is allmost 50%, 49,1% to be exact. Revising 17 items pushes it up to 59,6%.

#16

The only way you can have 3 that you have studied, definitely come up, is if you study all 20.

There is no other way around it.

Aye.

However, if you were in an exam that you were told that, from 12 questions, there will be 8 on the paper and you must answer 4, then you can have a good look at a minimum number of questions less than the 12 questions. The minimum there will be 8 questions you must revise.

number of questions to revise = questions to be answered + (number of possible questions - number of questions on the paper)

so for 20 questions answering 3:

x = 3 + (20-3)

x= 3+17

x=20

for my 12 question example ( 12 possible, 8 on paper, 4 to answer):

x= 4 + (12-8)

x= 4 + 4

x= 8

That's for questions you have revised to be guaranteed to be on the paper.

Additionally, if you only want to get a maximum mark of 66.6%, then you substitute the questions to answer value with the number of questions that will get you above a certain percentage, assuming that questions are weighted equally. So instead of 3, you'd put 2, making it 19. It's much easier to revise smaller amounts for more questions IMO.

#17

that's what my friend says

20 -100%

19 - 85%

18 -71.5%

17 - 59.6%

16 - 49.1%

he is going to tell me the formula later on today

he also says:

However, if you only revised 15, there is a 85.9% chance that at least 2 of the questions will come up!!

20 -100%

19 - 85%

18 -71.5%

17 - 59.6%

16 - 49.1%

he is going to tell me the formula later on today

he also says:

However, if you only revised 15, there is a 85.9% chance that at least 2 of the questions will come up!!

#18

that's what my friend says

20 -100%

19 - 85%

18 -71.5%

17 - 59.6%

16 - 49.1%

he is going to tell me the formula later on today

he also says:

However, if you only revised 15, there is a 85.9% chance that at least 2 of the questions will come up!!

what he did thar be probability.

study 19 subjects

19/20 x 18/19 x 17/18 = 85%

Cant think of a formula though

#19

On a side, it's a bad idea to work to the probabilities. Sod's law being the day it's extremely important, is the day none of the questions you revised comes up. A lad on my uni course played the percentages. Now he has to redo his final year. Work the formula I used. In the original 20 questions example, revising all 20 means you can revise less across the whole board- to get the higher marks, you need to go above and beyond. To get above 50% overall marks from two questions, you need to do really bloody good and revise a shedload more across all 19 you've revised. Revise all 20, you can pretty much revise basics for each question, and get above 50% for less work.

#20

It has been a while since I did stats but maybe try Chi Square test to determine the probability of an act occuring by assessing an observed amount of actions, against the expected amount of actions. Not sure if it will be what you are looking for.

Delirium makes a good point about sample size though. 20 is not much to work with so you results might not be valid. Sample sized has to be assessed based on the population size and if the population size is too small the results are unreliable...I think.

I took a quick lookie and the formula for CHI SQUARE might be:

=(expected results - observed results) x the power of 2 / expected results.

Upon further review, I am likely wrong about using the chi square but you can look into it anyway.

Delirium makes a good point about sample size though. 20 is not much to work with so you results might not be valid. Sample sized has to be assessed based on the population size and if the population size is too small the results are unreliable...I think.

I took a quick lookie and the formula for CHI SQUARE might be:

=(expected results - observed results) x the power of 2 / expected results.

Upon further review, I am likely wrong about using the chi square but you can look into it anyway.

*Last edited by TwistedLogic at Oct 20, 2009,*