Probabilities & Statistics 1 - OCR (not MEI) - jan 25th 2012 discussion

Well it may seem that way but I'm just a concerned parent.

If you don't trust me then fine - all you need to answer is the first bit - what marks were the last couple of questions worth so I can work out the potential impact on her score.

I've no need to cheat to get a score on these exams as I'm fairly confident with a few days prep I'd get 100% in them all if I needed to take them but with a First Class honours degree in microelectronic systems design - I'll give that a pass and not waste my time.

If you are stuck on a questions just ask me.

P(X=x) =

Sorry, I'll try to explain better.
Last part of question 8 you have to find probability that there are more red than blue (or the other way around I can't remember). For the part before that you had to work out the probability that there are 11 of each. so you do P(R=11) 22C11 x 0.5^11 x 0.5^11. Which comes to about 0.168.
The only possibilities are that Red and Blue have the same number, there are more red, or there are more blue. Because p = 0.5 the probability that there are more red or more blue are the same. So you do 1 - 0.168, then divide by two because the other two outcomes are equally likely.

And for the last part of question 9, the only possible combinations of making 28 and 9,8,7,4 and 9,8,6,5. so to work out out the probability that it's 9,8,7,4 you do 1/9 x 1/8 x 1/7 x 1/6 = 1/3024 then x by 4! because they can be in any order. Then you do the same for 9,8,6,5 so you x by 2. That should give you 48/3024 or 0.0159

I also got 2/3cm too.

For the question where you had the 9 numbered discs, what did people get?
I got 4! for the first part because it said in the question Andrea picks 4 discs and puts them down in a random order. So I thought it was how many arrangements can you be made from the 4 that Andrea picked and not how many different combinations of 4 she could pick...

However, everyone else I've talked to did something like 9C4 instead so I'm thinking I'm probably wrong and got confused reading the question..

For question 7 (that 60% of voters being women one)...was that a geometric distribution?

For question 7 (that 60% of voters being women one)...was that a geometric distribution?

Yes it was

Well it may seem that way but I'm just a concerned parent.

If you don't trust me then fine - all you need to answer is the first bit - what marks were the last couple of questions worth so I can work out the potential impact on her score.

I've no need to cheat to get a score on these exams as I'm fairly confident with a few days prep I'd get 100% in them all if I needed to take them but with a First Class honours degree in microelectronic systems design - I'll give that a pass and not waste my time.

If you are stuck on a questions just ask me.

more like an overconcerned obssessive parent? shouldnt your daughter be asking the questions if she cares?
also theres no need to boast about these exams

YOU DIDNT TAKE THE EXAM YOUR DAUGHTER DID SO LEAVE


also what did it mean by one unit is 2cm? was one unit one of those tiny squares, or 1 unit made up from a 10x10 of the little squares

also what did it mean by one unit is 2cm? was one unit one of those tiny squares, or 1 unit made up from a 10x10 of the little squares

Anyone else notice the mistake. For the question where you had to say which graph was the geometric distribution there was a value for 0 which is not possible. Although it wouldn't have prevented anyone from getting that question right they may have then got the next one wrong as they may have thought that it returns the number of failures rather than the number of trials upto and including the success.

I think I got 100 UMS just not sure whether it was 100% or not. Might have made a silly mistake somewhere.

It meant 10x10 little squares cos it was referring to the F.D units, they're worded stupidly

thanks, what was the answer?

thanks, what was the answer?

Sorry, I'll try to explain better.
Last part of question 8 you have to find probability that there are more red than blue (or the other way around I can't remember). For the part before that you had to work out the probability that there are 11 of each. so you do P(R=11) 22C11 x 0.5^11 x 0.5^11. Which comes to about 0.168.
The only possibilities are that Red and Blue have the same number, there are more red, or there are more blue. Because p = 0.5 the probability that there are more red or more blue are the same. So you do 1 - 0.168, then divide by two because the other two outcomes are equally likely.

And for the last part of question 9, the only possible combinations of making 28 and 9,8,7,4 and 9,8,6,5. so to work out out the probability that it's 9,8,7,4 you do 1/9 x 1/8 x 1/7 x 1/6 = 1/3024 then x by 4! because they can be in any order. Then you do the same for 9,8,6,5 so you x by 2. That should give you 48/3024 or 0.0159