Qualification > Math
Urgent: Maths CIE Stats
cs:
The permutation and combination for p63 was hard.. 10 marks for that, i only manage to do the first question, 8 marks gone.. other question was quite okay, except the first one, which distribution is used to model weight of students..? i answered stem and leaf, most of my friends answered normal distribution cos of the "distribution" on the question, 2 marks for that..
mousa:
--- Quote from: $!$RatJumper$!$ on November 05, 2010, 06:52:38 am ---These were my answers:
i) (10C4*9C1*14C1) + (10C4*9C2) + (10C5*9C1) = 28728
ii) (9C4*8C0) + (9C3*8C1) / 28728 = 1/36
iii) (9C3*8C1*13C1) + (9C3*8C2) + (9C4*8C1) + (8C1*9C4) + (8C0*9C5) = 13230
iv) 6!-(5!*2!) / 6! = 2/3
Can anyone confirm my answers :)
--- End quote ---
i dunno realy why you've done it like that;;;;; since there must be 6 ppl in total and they said a min. of 4 m and 1 w.. I have done it like that:
possible combinations are// 5 men and 1 women OR 4 men and 2 Women
therefore the answer would be :
10C5*9C1 + 10C4 * 9C2 =9828 ways...so i dunno know which one of us is correct.
$!$RatJumper$!$:
--- Quote from: mousa on November 05, 2010, 07:27:54 am ---i dunno realy why you've done it like that;;;;; since there must be 6 ppl in total and they said a min. of 4 m and 1 w.. I have done it like that:
possible combinations are// 5 men and 1 women OR 4 men and 2 Women
therefore the answer would be :
10C5*9C1 + 10C4 * 9C2 =9828 ways...so i dunno know which one of us is correct.
--- End quote ---
ur talking about part i right? if u are, i agree what u did but if u see i also added + (10C4*9C1*14C1). This was because u can ALSO have 4 men, 1 woman and any 1 other, either man or woman.
Deadly_king:
Hmm............am not sure myself.
Just wait for some time. I'll ask Astar to clear it up for us. ;)
$!$RatJumper$!$:
ok thankx.. do let us know :) and By the way DK, as i asked b4, what do u think my grade wil be based on the marks i gave you.
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