Urgent: Maths CIE Stats

[$!$RatJumper$!$]

i have no idea if this is correct.. i would like someone who is sure about their answer to comment on if i did something wrong and i think the threshold should vary between 36-39 out of 50

[lana]

I predict between 35-38 i think everyone found it harder than any other year .. but I hope we all get an A 

[Deadly_king]

Am sorry...................i've done so badly in the last question that I can't confirm any answers in it.

Yeah it was kinda hard for everyone. So i expect the threshold to decrease upto 30-35.

I really hope to get an A in that paper though it's not very likely due to the number of mistakes.

[mousa]

As far as i remember, the info they gave was:
If there are 10 men and 9 women. What are the ways of choosing 6 people if there are atleast 4 men and 1 woman.

iii) If one particular man or one particular women are on the committee but not both.

Answer I got:

(9C3*8C1*13C1) + (9C3*8C2) + (9C4*8C1) + (8C1*9C4) + (8C0*9C5) = 13230

I realy dont remeber wt I got, but i am confident of my working out, By the way, for the 1st part regarding how many combinations for the committe having at least 4m and at least 1 w??
it meant that only possible combinatins are 5m and 1 w OR 4m and 2 w

so my answer was 9828 ways is that correct??

[mousa]

hey guyss how about Questions 5 and 6 any of you remember their answers???
which bitts you found tricky???

[Deadly_king]

hey guyss how about Questions 5 and 6 any of you remember their answers???
which bitts you found tricky???

That was the normal and binomial distribution, right?

Unfortunately I can't remember the values I got

[$!$RatJumper$!$]

do u guys remember the actual question for the last question includin all its parts?

[Deadly_king]

do u guys remember the actual question for the last question includin all its parts?

Here you are :

7. A committee of 6 people, which must contain at least 4 men and at least 1 women, is to be chosen from 10 men and 9 women.

i) Find the number of possible committee that can be chosen. [3]

ii) Find the probability that one particular man, Albert and one particualr woman, Tracey are both on the committee.[2]

iii) Find the number of possible committees that include either Albert or Tracey but not both. [3]

iv) The committee that is chosen of 4 men and 2 women. They queue up randomly in a line for refreshments. Find the probability that the women are not next to each other in the queue. [3]

[$!$RatJumper$!$]

These were my answers:

i) (10C4*9C1*14C1) + (10C4*9C2) + (10C5*9C1) = 36288

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

[$!$RatJumper$!$]

And i think i got 60/75 for P1 and 43/50 for S1 ... what do you think my grade will be, based on the threshold you guys consider to be used

[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]

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

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$!$]

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.

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.