HOWS UR STATISTICS PAPER 62??????

[monty_993]

no u had to use permutations and most ppl were getting 4 hundred and sth i got 301
but u cant use combination because it said ARRANGEMENTS and tht means permutation
yea ur rite abt exclusive but i wrote independent cuz p(q)*p(r) was not equal to p(q and r):S

 

[umer khan]

i messed up in the last 3 parts of the paper :"( the permutation question

 

[dadevil92]

myn was the worst eva...myt end up wid a B...evn been able to score 70+ in p1...nd yea..i do agree wid laviva...i also drew dat stupid table nd messed up at the end as i did not hve tym...worst...

 

[umer khan]

what u guys think of the percentile ................. remember nov 09 went down to 37......

 

[supermagic]

i think the percnetile would be between 40and 43 but not likely 40

and yeah, i used combination to anwser very last question and i got 14 !!! (1C1*7C1*1C1*2! = 14)

and i think i wrote P(Q)andP(R) are exclusice but not independent since p(Q and R ) is not same as p(Q) * p(R)

how about very first question??? the mean and s.d. i think i got 9.97 or 9.79 for s.d.

 

[NightmarE]

Q and R were mutually exclusive as well as independant ....
R was if both dice gave more than 8 ...
if both dice wud giv more than 8 then u can nevr get da product 24 .....
if u get the product 24 then its impossible for both dice 2 gv no.s greater than 8 ...
they are mutually exclusive ....
and ..
suppose u throw both dice 1nce ...
u satisfy the condition of R ...
now ths wont affect da probability of satisfying Q in da next turn .... or vice versa ...
so yes, Q nd R were independant ....

 

[sse2010]

lol nightmare! is dis ur own theory about independent eventz? LMAO!!!!!
two eventz can only n only be independent if P(QandR)=P(Q)xP(R)...this wasnt da case here cuz P(QandR) was zero..so they were not independent!!!

 

[Buffer Solution]

It was tricky, that permutations ques and also the 1/3U ques. I think they were the distinguishers between good and the best candidates.

 

[ammadb]

i found the mutually exlcusve que easy.. although i drew up the complete table, i was able to end up in time =)..
but i got stuck in the relation que..
i think it was mu= -1.571s.d
is it correct??
n abt the last que.. prmutations were to be used..?? n how we had to do them?
wat was the arrngements of arrngng 3 cards??
if it shud have pink??
n if pink n green were not together..??

 

[sse2010]

we can use permutation n combination both fr last q...if using permutation, it wud be 9P3..n if usin combination then 9C3x3!

 

[a2man]

@sse2010 i think nightmare is right cuz thats wht is exactly mentioned in the CIE S1 book too, they were definitely independent too since 1st outcome is not affected by the second or vice versa...and the pretty much raised my total lost marks to 10!!!

 

[sse2010]

u both r wrong! For independent events, that is events which have no influence on one another, the rule simplifies to:

That is, the probability of the joint events A and B is equal to the product of the individual probabilities for the two events.

 

[a2man]

Dude whining isn't going to change the answers , i too thought (and wrote) that they were not independent , only mutually exclusive but i asked my math teacher and he said much the same thing ...

 

[sse2010]

LMAO!!!! jus wait fr resultz then dudE!! anyway bst ov luck!!

 

[hamizahmed]

yar I want to pull my hairs off..... I did the most foolish mistake of ma lyf, used binomial in prob distrib table.. WTF

 

[Jazib]

here, i'll go with the sse guy! cuz they were not independent at all!

 

[sse2010]

yo buddy!!!!!

 

[princesszahra]

chill ppl
its over!

 

[hamizahmed]

although its over bt one of the worst papers!

 

[NightmarE]

lolzz ... :-D
finally i ndrstd y they were not independant ... :-D
in smple wrds ... Q can only happen when R is not happening or viceversa ..
.therefore they r dependant ... :-D