Exam question (1 Viewer)

[lyounamu]

In my 3 Unit exam, one of the question was like this:

A group consisting of 3 men and 6 women attends a prize giving ceremony.

If 5 prizes are awarded at random to members of the group, find the probability that exactly 3 of the prizes are awarded to women if there is no restriction on the number of prizes per person.

This question was quite controversial as some people in my year have noted that this is not a permutation question, rather, Binomial Probability question. I personally don't agree with this. I got this question right but there has been a backlash from people that the solution is incorrect.

Please post up your solution. I want to see if yours corresponds to mine and if you can, please provide your own explanation. Thanks.

 

[x3.eddayyeeee]

i think i agree with you.
is the answer 8/243 but?

 

[hermand]

okay, it's been over a year and a half since i did perms and combs, so i'm waaaaaayyy rusty & feel free to tell me i'm totally wrong, this is all from memory, but i'd think it was a probability question that uses combinations [assuming the awards are generic, and therefore not being a permutation, as it does not matter which specific award you get].

essentially there are fifteen men and thirty women, as each actual man/woman can receive up to five awards, therefore i multiplied the original amounts by five.

now i don't have a calculator to use at the moment, i'll have to give you my solution in unsimplified form [yes i do four unit maths and my calculator battery is being used for the electronic monopoly game therefore i haven't had a calculator to use all year lol.].

think i'm doing this the long way but ohwell.

to give 0 prizes to 30 women, there is one way. & to give 5 prizes to 15 men is 15C5.
to give 1 prize to 30 women, there is 30 ways. & to give 4 prizes to 15 men is 15C4.
to give 2 prizes to 30 women, there is 30C2 ways. & to give 3 prizes to 15 men is 15C3.
to give 3 prizes to 30 women, there is 30C3 ways. & to give 2 prizes to 15 men is 15C2.
to give 4 prizes to 30 women, there is 30C4 ways. & to give 1 prize to 15 men, there is 15 ways.
to give 5 prizes to 30 women, there is 30C5 ways. & to give 0 prizes to 15 men, there is 1 way.

therefore, with the bold line on top [as this is the case we want] and the addition of all possible combinations on the bottom, we get......

and whatever that adds up to on a working calculator haha.

hope this remotely helped?

 

[lolokay]

I just tried it and got 112/429. no idea if that's right though

 

[lyounamu]

Please provide your solution as in working out as well.

@x3...: Solution please?
@hermand: you interpreted my question incorrectly. There aren't 30 women to start with.

 

[lyounamu]

no wait i looked at it again and i think it's 80/243 ?

5C3 (1/3)²(2/3)³

just from binomial probability

Yep, yep.

That's one of the solution that was disputed.

Thanks.

Other solutions are welcome. ^.^

 

[lolokay]

woops, deleted it
\\

no wait i looked at it again and i think it's 80/243 ?

5C3 (1/3)²(2/3)³

just from binomial probability

 

[hermand]

Please provide your solution as in working out as well.

@x3...: Solution please?
@hermand: you interpreted my question incorrectly. There aren't 30 women to start with.

yeah i said that in my explanation, there aren't thirty women, nor are there fifteen men, but essentially there are thirty chances for women to get an award and fifteen for the men, as each person can receive up to five awards.

 

[Timothy.Siu]

(2/3)^3 x(1/3)^2 x 5C2

is what i got.

 

[lolokay]

ahh the binomial one is not correct, because you can't multiply by 5C3 if one person got multiple awards edit:lol yes you can..

@hermand, i think you should be multiplying those probablilities (15c and 30 c pairs) not adding. and 15c2*30c3 should be added on the bottom as well

apart from that i can't see what would be wrong with it, apart from it being a long calculation.
*actually, it wouldn't be correct.. because of the way you broke one person into 5 people: it doesn't matter which of these "parts" gets the award in reality, but your method does take into account which "part" gets is


I think I should show the working for my original answer:

to find the total number of prize combinations we may first arrange the 9 members in a row, and if a person gets a prize we put this prize right after them.
now, the first spot must be taken by a person, but all other 13 spots can be either a person or prize. so there are 13C5 combinations (pick 5 spots to be prizes)

we now break them up into boys and girls, and using similar reasoning get 4C2 for the 3 boys to get 2 prizes and 8C3 for the 6 girls to get 3 prizes. multiplying these and dividing by total number of combinations gives (I think) 112/429

however this way seems quite complex for a 3u exam question. i see nothing wrong with it though

 

[lyounamu]

ahh the binomial one is not correct, because you can't multiply by 5C3 if one person got multiple awards

@hermand, i think you should be multiplying those probablilities (15c and 30 c pairs) not adding. and 15c2*30c3 should be added on the bottom as well

apart from that i can't see what would be wrong with it, apart from it being a long calculation.
*actually, it wouldn't be correct.. because of the way you broke one person into 5 people: it doesn't matter which of these "parts" gets the award in reality, but your method does take into account which "part" gets is


I think I should show the working for my original answer:

Yeah, this question is little bit messed up.

There are so many disputes over this question.

Apparently the "correct answer" is 10/21.

 

[lolokay]

10/21?! surely not. that sounds much too high (the binomial one -overcounts- some combinations)

 

[lolokay]

I'm convinced that my 112/429 method is the correct one

 

[lyounamu]

10/21?! surely not. that sounds much too high (the binomial one -overcounts- some combinations)

Hm, maybe the wording is little bit confusing I guess...

 

[Timothy.Siu]

ah i see, that answer seems a bit strange though, looks simple.

 

[lolokay]

i think they should just remove this question from the exam

 

[lyounamu]

ah i see, that answer seems a bit strange though, looks simple.

Yeah, it does.

I don't know.

The question of the original wording is confusing. I think in the exam, it was presented in that way but interpreted differently by the exam marker.

 

[Timothy.Siu]

lolokays solution seems good by me.

 

[hermand]

@hermand, i think you should be multiplying those probablilities (15c and 30 c pairs) not adding. and 15c2*30c3 should be added on the bottom as well

apart from that i can't see what would be wrong with it, apart from it being a long calculation.
*actually, it wouldn't be correct.. because of the way you broke one person into 5 people: it doesn't matter which of these "parts" gets the award in reality, but your method does take into account which "part" gets it.

they're not probabilities, they're combinations, ie, that's the amount of combinations possible to get those awards, therefore adding makes sense.

okay, so it's not "people" as such, but the chance they have to get an award. how do you allow for the one person to get all five awards otherwise?

 

[lolokay]

sorry, i meant combinations ^ but you should still be multiplying them (unless i misinterpreted your method?)

you allow for a person to get more than one award using the method i posted