Stat v. 62

[salvatore]

Yo if you can tell me the toffees / chocolates in each bag.. I can calculate it for you right here right now.

It was something like this.. 7 chocolates, 5 toffees in Syed's bag and 3 chocolates, 4 toffees, 2 boiled sweets in Tina's bag.
I just chose random names

 

[A star]

I got 0.25, 0.175 and 0.575 as the probabilities.. but I kinda feel I've messed up somewhere. Yeah, I don't think they would penalise you since the format wasn't mentioned in the question.

You're talking about AS chemistry practical right? I did that on 14th May! It was pretty lengthy but not hard

very low gt is expected this year near 25

 

[lionkill3r]

same. It was lengthy but not hard.. Do you remember the number of Toffees/Chocolates each of Suzan/Sarah or Ahmed had?

suzan - 5 chocolate 7 toffe
Ahmed- 3 chovolate, 2 boiled 4 tofee (before taking one from sarah ofcourse)
I remeber this coz i drew 2 bags and labbeled chcolate otffe and stuff xD (for convenience)

 

[mrmt]

I'm not too sure but from what I've seen in the mark schemes. 1 mark would have been for the answer, 1 for (9!*4!) for all the hibiscus being together, and the other for them not being together so 12!- (9!*4!). That's how they would have split the 3 marks.
So...

the ans ws 8! *9p4..
coz none should be together ... not 'not all should b together.!

 

[A star]

suzan - 5 chocolate 7 toffe
Ahmed- 3 chovolate, 2 boiled 4 tofee (before taking one from sarah ofcourse)
I remeber this coz i drew 2 bags and labbeled chcolate otffe and stuff xD (for convenience)

actually your first and last is same as mine but second is diff

 

[Ryan123]

I kinda have a doubt in the last part.. I think you did it wrong. I remember a similar question came in the papers previously.. where it said no two women are together...

we did something like ..

_*_*_*_*_*_*_*_*_ where * is one of the other trees.. and the _ is where the Hill billy would go..

I did it as arrangement of other trees * arrangement of the hillbilly trees * the selection of places where each hillbilly tree can go.

8!*4!*9C4 = 1.22*10^8 Answer (3sf) (Which can otherwise be written as 8!*9P4)

You are most likely right i did it the other way 12!- onee but thats wrong.
Abt the last one its
X 0 1 2
P(X) 7/24 19/40 7/30

i think toffes and choco in first bag was 7 and 5 respectively and in 2nd bag choco was 3 and the rest 6

 

[syed1995]

That's the tree diagram..

Now P(O) = 1C' * 2C`
P(1) = 1C * 2C` or 1C`*2C
P(2) = 1C * 2C

P(0) = 5/12 * 7/10 = 7/24 (0.292)
P(1) = 7/12 * 6/10 + 5/12 * 3/10 = 19/40 (0.475)
P(2) = 7/12 * 4/10 = 7/30 (0.233)

Total Of Probabilities = 1 either fraction/decimal..

That's What I did as far as I remember...

 

[salvatore]

very low gt is expected this year near 25

Yeah.. hope so

 

[lionkill3r]

That's the tree diagram..

Now P(O) = 1C' * 2C`
P(1) = 1C * 2C` or 1C`*2C
P(2) = 1C * 2C

P(0) = 5/12 * 7/10 = 7/24 (0.292)
P(1) = 7/12 * 6/10 + 5/12 * 3/10 = 19/40 (0.475)
P(2) = 7/12 * 4/10 = 7/30 (0.233)

Total Of Probabilities = 1 either fraction/decimal..

That's What I did as far as I remember...

I got the same answers !!!

 

[syed1995]

suzan - 5 chocolate 7 toffe
Ahmed- 3 chovolate, 2 boiled 4 tofee (before taking one from sarah ofcourse)
I remeber this coz i drew 2 bags and labbeled chcolate otffe and stuff xD (for convenience)

That's wicked.. Suzan has 5 chocolates and 7 toffees.. yet you take the one from sarah.. where did she come from ?

 

[A star]

That's the tree diagram..

Now P(O) = 1C' * 2C`
P(1) = 1C * 2C` or 1C`*2C
P(2) = 1C * 2C

P(0) = 5/12 * 7/10 = 7/24 (0.292)
P(1) = 7/12 * 6/10 + 5/12 * 3/10 = 19/40 (0.475)
P(2) = 7/12 * 4/10 = 7/30 (0.233)

Total Of Probabilities = 1 either fraction/decimal..

That's What I did as far as I remember...

you forgot B whatever it is in P(0)

 

[salvatore]

That's the tree diagram..

Now P(O) = 1C' * 2C`
P(1) = 1C * 2C` or 1C`*2C
P(2) = 1C * 2C

P(0) = 5/12 * 7/10 = 7/24 (0.292)
P(1) = 7/12 * 6/10 + 5/12 * 3/10 = 19/40 (0.475)
P(2) = 7/12 * 4/10 = 7/30 (0.233)

Total Of Probabilities = 1 either fraction/decimal..

That's What I did as far as I remember...

For P(0), isn't it supposed to be 5/12 * 6/10.
Coz there were 3 chocolates in Ahmad's bag, so the remainder is 6.

 

[syed1995]

Yeah.. hope so

Yea the GT will be low .. that Probability question / Permutation Question and that n and k question.. these 3 were done poorly.. that's 20 marks right there!!!

 

[syed1995]

you forgot B whatever it is in P(0)

Eh? I added the probabilities..

Doing (5/12 * 5/10 ) + (5/12 * 2/10 ) would give the same answer.. as 5/12 * 7/10 lol..

 

[salvatore]

Yea the GT will be low .. that Probability question / Permutation Question and that n and k question.. these 3 were done poorly.. that's 20 marks right there!!!

I got the value of n as 144 and k as 6. Is that right?

 

[A star]

I got the value of n as 144 and k as 6. Is that right?

yup

 

[syed1995]

I got the value of n as 144 and k as 6. Is that right?

I didn't do that question.

 

[A star]

Eh? I added the probabilities..

Doing (5/12 * 5/10 ) + (5/12 * 2/10 ) would give the same answer.. as 5/12 * 7/10 lol..

i took total candy ahmad has as 11 shit shit shit so it will cound as a genuine error and maks will only be deducted in one part right?

 

[lionkill3r]

I got the value of n as 144 and k as 6. Is that right?

yap just using u=np and sigma^2=npq wil get yu the answer ...
BTW suzan and sarah watever both sound the same

 

[A star]

I didn't do that question.

it was so easy why not it was simply
mean=n(k/9)
var=n(k/9)(1-K/9)