Exam question (1 Viewer)

[Drongoski]

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.

Questions on probability of this type can be very tricky; even experienced ones can get it wrong. I think answer is:

​

This is a case of binomial probability, n=5 trials; p = prob of success = 2/3, q = prob of failure = 1/3. So problem becomes one of number of successes (i.e. out of 5 prize awards, how many times the 6 women as a group, succeeds in getting the prize). Here we are looking for the prob. that out of the 5 prizes given out (each called a trial or an experiment), the women as a block gets exactly 3 - i.e. 3 successes. Hope this explains it. I could be wrong!

 

[hermand]

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

hmm yeah i suppose. i don't really remember doing perms and combs but that was just what i remembered [i think - i could have made it all up haha] and the way i did it made logical sense in my head haha.

hmmmmmm maybe i do need to multiply. i was confussed as to which one i was supposed to do when i was writing the solution. i'll try that and see what answer it gives me.

life is too hard without a calculator. haha.

 

[Timothy.Siu]

Questions on probability of this type can be very tricky; even experienced ones can get it wrong. I think answer is:

​

This is a case of binomial probability, n=5 trials; p = prob of success = 2/3, q = prob of failure = 1/3. So problem becomes one of number of successes (i.e. out of 5 prize awards, how many times the 6 women as a group, succeeds in getting the prize). Here we are looking for the prob. that out of the 5 prizes given out (each called a trial or an experiment), the women as a block gets exactly 3 - i.e. 3 successes. Hope this explains it. I could be wrong!

i think u shud have read the posts above =)

 

[x3.eddayyeeee]

would that combination and permutation stuff work here?
because i dont think they allow for repititions. ie. doesnt allow person to get more than one award?

 

[lolokay]

the trick is to include the prizes in with the permutation/combination

 

[Trebla]

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.

This EXACT question is from the Cambridge Study Guide LOL!!! It was set out in the book as follows:

Five prizes are awarded at random to members of a group consisting of 3 men and 6 women. Find the probability that exactly three of the prizes are awarded to women:
(i) If there is a restriction of at most one prize per person
(ii) If there is no restriction of the number of prizes per person

Answers:
(i) 10/21
(ii) 80/243

The way the answer to (i) was found according to the study guide was:
(⁶C₃ x ³C₂) / ⁹C₅ = 10/21

The way the answer to (ii) was found according to the study guide was:
⁵C₃ (2/3)³ (1/3)² = 80/243

The 10/21 answer arises when there is a restriction of at most one prize per person (which was part (i) of that question as it appears on the book) Perhaps the person who set the exam, misread the answer as part (i) from that book rather than part (ii) LOL

 

[lyounamu]

This EXACT question is from the Cambridge Study Guide LOL!!! It was set out in the book as follows:

Five prizes are awarded at random to members of a group consisting of 3 men and 6 women. Find the probability that exactly three of the prizes are awarded to women:
(i) If there is a restriction of at most one prize per person
(ii) If there is no restriction of the number of prizes per person

Answers:
(i) 10/21
(ii) 80/243

The way the answer to (i) was found according to the study guide was:
(⁶C₃ x ³C₂) / ⁹C₅ = 10/21

The way the answer to (ii) was found according to the study guide was:
⁵C₃ (2/3)³ (1/3)² = 80/243

The 10/21 answer arises when there is a restriction of at most one prize per person (which was part (i) of that question as it appears on the book) Perhaps the person who set the exam, misread the answer as part (i) from that book rather than part (ii) LOL

I see it now.

That's where I got 10/21!!! (lol, that was actually the first part of the question)

Yeah the answer to this question was 80/243...

Thanks Trebla.

Actually that question was 2-part question.

The earlier answer that I posted up was a mix-up that I just had.

Yeah, I got those 2 answers. and others in my class were disputing that this answers were wrong...hahah

 

[lolokay]

wait so 80/243 is actually correct?

 

[Trebla]

Yes, according to the original source of that question...

 

[lolokay]

yes it is correct. there's no reason it shouldn't be as it is just a simple binomial probability (after all of that...)

1/3 chance boy, 2/3 change girl, 5 prizes

i see my error now. i assumed all combinations were equally likely (but they're not. need to keep this in mind in future and not overcomplicate things) which i could sort of see myself doing at the start and so changed my mind...

too tired for this -.-

at least we have the numbers of combinations now lol.