Exam probability

Route1

Have an exam tomorrow, been provided with 8 potential questions of which 4 will come up. I'm thinking of learning only 5 is that sufficient what is the mathematical probability that I get caught out?

Capt'n Midnight

Depends, what's the pass mark ?

You won't get 100% for each question, what % do you think you get for each question ?

Route1

I have them pre written, just the case of rote learning them.

The Backwards Man

100%

You've already been caught out.

Alf Stewart.

Route1 wrote: »

Have an exam tomorrow, been provided with 8 potential questions of which 4 will come up. I'm thinking of learning only 5 is that sufficient what is the mathematical probability that I get caught out?

**adopts Irish accent**

Ahem....um.... one in free bloody fiddy, mate.

Badly Drunk Boy

Instead of learning just 5 of the 8, get off the internet immediately so you might learn 6 or 7 of them! :pac:

Route1

No, I like my short cuts.

failinis

Learn off 5 by heart and maybe do short bullet points on the remaining ones to glance at.
There were a possible 20 essay questions for a history exam - (As in do 6 out of the above) so I done a similar thing.

Worst case you get 3 unknown and 1 known
Giving a max of 25% if you know it inside out
Learn em all or at least 5 in depth and bits of the other 3

Happened to me in an exam before

Antar Bolaeisk

Don't bother learning any, just sit here and enjoy watching the thread collapse as AH wrestles with probability, combinations, permutations and odds. It'll be way more fun. You know it will.

What's the exam on?

stoplooklisten

If a bus leaves main street at 5.40, how many apples does the driver need for his break?



find x







x

Antar Bolaeisk

Antar Bolaeisk wrote: »

Don't bother learning any, just sit here and enjoy watching the thread collapse as AH wrestles with probability, combinations, permutations and odds. It'll be way more fun. You know it will.

What's the exam on?

how funny if it were on probability and combinations ...

Route1

come on people.

stoplooklisten

Route1 wrote: »

come on people.

That kind of thing would get you thrown out :eek:

The Backwards Man

Route1 wrote: »

come on people.

We're not the ones that didn't study for our exam:p

the groutch

Route1 wrote: »

Have an exam tomorrow, been provided with 8 potential questions of which 4 will come up. I'm thinking of learning only 5 is that sufficient what is the mathematical probability that I get caught out?

hope it's not a maths exam, or else you're screwed.

Route1

the groutch wrote: »

hope it's not a maths exam, or else you're screwed.

It's not m8.

maudgonner

The exam is tomorrow. It's too late to learn anything now. Phone in a bomb threat in the morning. It worked for that lad from Intel who felt like pulling a sickie.

Invest the time between now and 9am in perfecting your Nordie/Arab accent. Or if you're in Dublin and want to take advantage of the gang wars, watch Love/Hate instead and copy them. You'll be grand.

thee glitz

Doing 5 out of 8, you're guaranteed 1 will come up.

There's 3% chance it will just be 1,
34% you'll know 2,
51% chance of 3 and
over 11% that 4 of your 5 come up.

best of luck.

Capt'n Midnight

https://www.youtube.com/watch?v=tTv5ckMe_2M

Here's one to learn off.

XR3i

have u considered burning down the exam hall?

zl1whqvjs75cdy

How many questions do you need to answer?

Antar Bolaeisk

Route1 wrote: »

come on people.

What I can tell you is that there's an inverse correlation between the length of time you spend on here versus the probability of you passing the exam.

Donald Trump

thee glitz wrote: »

Doing 5 out of 8, you're guaranteed 1 will come up.

There's 3% chance it will just be 1,
34% you'll know 2,
51% chance of 3 and
over 11% that 4 of your 5 come up.

best of luck.

It is late, I am sleepy and stand open to correction but I think the correct answers are respectively
7.14% One of your 5 appears
42.86% Two of your 5 appear
42.86% Three of your 5 appear
7.14% four of your five appear

Think of the final case. you don't care about the order of specific questions. We'll call the questions you don't know "red" and the ones you do know "green"

There are 8 in total - three red and five green.

the probability that the first question will be a green is 5/8
the probability that the second will be a green is then 4/7
the probability that the third will be a green in then 3/6
and the probability that the last will be a green is 2/5


My guess is what you did (at least for the first one) is to draw with replacement. That is to assume that each question could appear more than once. If you assume that you get 2.9% .

Eugene Norman

md23040 wrote: »

Your above assertions are correct but your follow through wrong.

There is 100% chance of one question coming ùp
There is an 78.57% chance of two questions coming up
There is an 69% chance that three questions will come up
There is a 61.75% chance that four questions will come up.

In relation to 78.57% in second scenario -he has 100% chance of getting a question coming up, then he has a 4 in 7 chance of another question arising or 57% chance

(100+57) ÷ 2= 78.5%

So much for the wisdom of the crowd.

Your working out is not very clear.

Are you saying there's a 61% chance 4 of his 5 questions will come up? That's clearly wrong.

Edit:

Unlike Donald who was right, it is your original assumption that is incorrect. It's not 100% that 1 question will turn up. It's 7.14%. It's 100% that one or more will turn up.

You can see that in Donald's probabilities, the sum of which are 100%. He didn't even try to get that answer but it is there.

everlast75

Route1 wrote: »

No, I like my short cuts.

apt username then so...

cowboyBuilder

I remember when I was in college , some subject, 5 questions on the exam - each one topic each worth 20%.

Pass mark 40%.

I studied 2 topics - but really well ..

BOOM I got 40%.

I winged my way through college I tell ya!!!


Dem were De days!!!

The_Valeyard

Use fire.

Princess Consuela Bananahammock

Route1 wrote: »

Have an exam tomorrow, been provided with 8 potential questions of which 4 will come up. I'm thinking of learning only 5 is that sufficient what is the mathematical probability that I get caught out?

If this is a maths or probablilty exam, I think you should just give up!

HensVassal

Route1 wrote: »

Have an exam tomorrow, been provided with 8 potential questions of which 4 will come up. I'm thinking of learning only 5 is that sufficient what is the mathematical probability that I get caught out?

If 4 are definitely coming up and you learn 5 then best case scenario you'll have bagged the 4 and will get 4 correct. Do you only need to answer 4? If so you get 100%

Worst case scenario is that the 3 you ignored will come up which means you'll only have learned 1 of the 4 (if you learn the 5 you're talking about) and you get 25%, so you fail.

Only way to be sure it to learn all 8.

artanevilla

If you work the minimum amount now, and just about pass your exams, it will work out fine for the future.

You'll just about be able to pay your mortgage.

Well at the end of the day, its 50/50; you either pass or you fail

The Randy Riverbeast

If you learn 6 you'll be fine. At most 2 questions you don't know will come up and that's 50% letting you pass with some room left.

Capt'n Midnight

So what questions came up ?

Also we still don't know either
the pass mark
or how well the OP could do on the questions , I'm guessing this isn't multiple choice

5rtytry56

What is the pass rate OP? ie, the number of candidates to pass, as opposed to the number of failures.

How do you think you performed in the exam?

thee glitz

Donald Trump wrote: »

It is late, I am sleepy and stand open to correction but I think the correct answers are respectively
7.14% One of your 5 appears
42.86% Two of your 5 appear
42.86% Three of your 5 appear
7.14% four of your five appear

Think of the final case. you don't care about the order of specific questions. We'll call the questions you don't know "red" and the ones you do know "green"

There are 8 in total - three red and five green.

the probability that the first question will be a green is 5/8
the probability that the second will be a green is then 4/7
the probability that the third will be a green in then 3/6
and the probability that the last will be a green is 2/5


My guess is what you did (at least for the first one) is to draw with replacement. That is to assume that each question could appear more than once. If you assume that you get 2.9% .

No, I fecked it up by saying 1 was guaranteed to come up, so worked off the remaining 7.
At risk of further embarassment, I think my calculations are correct once the first question
has come up as studied for. Looks like you got it right.