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Stat v. 62

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The paper was manageable :)
What did you get the values for the probability distribution table?
 
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Total trees 15.
Hibiscus:4, Oleander: 2 J-something: 9

(i) selection of 12 trees where there has to be a minimum of two of each kind.
A: 2O, 2H, 8J = 2C2*4C2*9C8 = 54
2O,3H, 7J = 2C2*4C3*9C7 = 144
2O, 4H, 6J =2C2* 4C4*9C6= 84
total = 54+144+ 84 =282

(ii) The company picks 6J, 4H and 2O. How can they be arranged if the trees of the same type are are kept together.
A: 3!*4!*6!*2!= 207360

(iii) Arrange the 12 trees so that the hibiscus plants are not together.
A 12! -(9!*4!)

syed1995
saifookhan

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)

Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a
different design. Find in how many ways these 9 mugs can be arranged in a row if the china
mugs are all separated from each other.

littlecloud11

Check http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_61.pdf

Question 6 part ii...

Solve this for me..

Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a
different design. Find in how many ways these 9 mugs can be arranged in a row if the china
mugs are all separated from each other.

Using your method first. and then check the MS.
 
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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 8!*4!*9C4 = 1.22*10^8 Answer (3sf) (Which can otherwise be written as 8!*9P4)

Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a
different design. Find in how many ways these 9 mugs can be arranged in a row if the china
mugs are all separated from each other.

littlecloud11

Check http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_61.pdf

Question 6 part ii...

Solve this for me..

Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a
different design. Find in how many ways these 9 mugs can be arranged in a row if the china
mugs are all separated from each other.

Using your method first. and then check the MS.
this or 8!*9*8*7*6 same thing
 
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syed1995 littlecloud11

Do you guys remember the answers to the probability distribution table?

one of them was 4/27 I think.. I don't remember them sorry.. I had a chemistry practical today.. so maths is totally out of my mind... My total sum of probabilities was coming as 1 so I did that one right. :D

Btw how did your practical go? Do you remember yours.. if you post them maybe I will be able to help better.. I wrote them all in fraction.. They won't penalize me would they?
 
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Q6 iii) it was 8!x9P4 not 12!-(9!x4!)
yes i did the same
8! *9*8*7*6
Consider the following
_P_p_p_p_p_p_p_p_
Where _ = a place each hibiscus plant canb e to be seprated from each other AND p= other plants (non hibcsu ones)
For first plant there are 9 places, for 2ndplant 8.3rd platn 7 and 4th plant 6
And as for remaining 8 non hibiscus plants ( that j and oleander thingy) there are 8! arrangements :)

EDIT - o lol didnt see one persone xplained it above my post xD
 
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one of them was 4/27 I think.. I don't remember them sorry.. I had a chemistry practical today.. so maths is totally out of my mind... My total sum of probabilities was coming as 1 so I did that one right. :D

Btw how did your practical go? Do you remember yours.. if you post them maybe I will be able to help better.. I wrote them all in fraction.. They won't penalize me would they?
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 :)
 
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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 :)

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