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S1 paper6 how was it????

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sse2010 said:
yes they were mutually exclusive bcuz P(QandR)=0...but they were not independent

no...it wasn't mutually exclusive..
P(Q)...was probability of getting the product of 2 no.s 24
and..
P(R)...was.. one or both the numbers greater than 8 (am I correct??? =s...??)
so in that case when we get 12 and 2....we get the product 24 which is satisfied by P(Q) as well as P(R)
so i dont think they were mutually exclusive....!
 
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anamsohail said:
sse2010 said:
yes they were mutually exclusive bcuz P(QandR)=0...but they were not independent

no...it wasn't mutually exclusive..
P(Q)...was probability of getting the product of 2 no.s 24
and..
P(R)...was.. one or both the numbers greater than 8 (am I correct??? =s...??)
so in that case when we get 12 and 2....we get the product 24 which is satisfied by P(Q) as well as P(R)
so i dont think they were mutually exclusive....!

P(R) was BOTH numbers greater than 8. So YES, they were mutually exclusive. lol, editted that, sorry my bad.
 
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Well no they were mutually exclusive... Mutualy exclusive means having no outcome in common and that is what was there. !!!
P(Q) was product 24 which u cud get by (2,12) (3,8) and (4,6) . None of which is present in P(R) which said both the number should be greater than 8 !!
 
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ashhadazam said:
It was pure ownage :D Me onwing the paper :p !!! Behtreen paper hua yar !! and almost everyone in our centre did well :/

what was your answer to the last question's last part? Thats the only part thats causing me some confusion, ppl got such diff answers!
 
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It was not that bad for me.......an "OK" paper I must say (pure maths was worse) .......I proved the E(X) - the probabilities were sometin like 1/7......can't remember exactly.......The only bit I'm not that sure bout is the last question - permutation n combination.......I remember doin sometin like 1*8C1.......sometin - not sure........
 
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ashhadazam said:
ANd they were independent too !!! Because event 2 wasnt affected by event 1

how were they independant! If they were, P(Q) x P(R) would equal P(Q intersection R) which you found is zero (mutually exclusive). This was obviously not the case, P(Q) x P(R) was giving a normal probability, not zero. Hehe sorry to burst your bubble, ashhadazam, but methinks your paper is slowly starting to become "un-great". :p
 
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yeah man no need to confirm the 3 values in the prob dist table in that duck question. well yeah they were wat ur getting. if u had E(x)= 8\7 proved ur values r rite.
 
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ahahah.. lol !! even if i get that wrong it would still be great :mrgreen: :mrgreen:

P(Q and R) = P(R) x P(Q)

Mutually exclusive mai P(Q|R) = 0 thats y we get P(Q and R) = 0 !!

so i stick with the probablities being independent :p
 
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MashALLAH my paper wnt really goood......... was a bit different but mine went awesome.... did it in 45 mins.........

they were mutually exclusive.......... not independant........... last ques last part was 14........ bcz if there were three card and th pink and green were not together then they should be on the sides........ and it could start with green or with pink ..... and the choice for the remaining one would be 7C1........ so the ans was : 2*(7C1)= 14.............
 
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Kaer123 said:
Here are my answers or part of them:



5) Not independent not mutually exclusive



ANYONE PLZ CONFIRM :D
it was mutually exclusive for sure
 
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ajmaeen said:
MashALLAH my paper wnt really goood......... was a bit different but mine went awesome.... did it in 45 mins.........

they were mutually exclusive.......... not independant........... last ques last part was 14........ bcz if there were three card and th pink and green were not together then they should be on the sides........ and it could start with green or with pink ..... and the choice for the remaining one would be 7C1........ so the ans was : 2*(7C1)= 14.............
y r u using C? v had to arrange them too.... but nvm, 2 marks only!
 
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