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OMG !!! I am sorry to say this but my paper went awesome :/
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S1 paper6 how was it????
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almost everyone in our centre did well :/
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How did it go for you people ?
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sse2010 said:
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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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 !!
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 !!
ashhadazam said:It was pure ownageMe onwing the paper
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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Umm.. Quoting someone else as me ? !
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Well thats the part i am confused about and thats the part i haev don 1 mark mistake.. !!
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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About the ducks question the PDF was :
0 - 1/7
1 - 4/7
2 - 2/7
0 - 1/7
1 - 4/7
2 - 2/7
ashhadazam said:
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".
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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(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
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.............
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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well I am damn sure A wud be on 40