Probability and Volume Questions (1 Viewer)

[Jase]

A party of 10 people is divided at random into 5 groups of 2 people.
Find the probablity of 2 particular people being in the same group.

now, bear in mind that I dont really understand anything about probablity..
but is there a formulaic way to determine this (i.e using combinations/permutations or factorials) or do you have to think it all out and divide all the possibilties into the sample space manually?

also.. heres a volumes question:

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A symmetrical pier of height 5m has an elliptical base with eqaution:
x^2 / 25 + y^2 / 4 = 1
and slopes to a parallel top with equation
x^2 / 9 + y^2 = 1

If cross sections taken parallel to the base are also elliptical, find the volume of the pier given that the area of an ellipse with the semi major axis a and semi-minor axis b is ( Pi a b)
-----------------------------------------------

well.. im quite sure youd use the cross sections method right?
would be wise to use a erm.. semi cone as the base.. and a semi ellipse as the cross section. then multiply it by two? im quite confused. i dont understand how to find values for a and b..

thanks

 

[turtle_2468]

EDIT: dbl post that didn't work.

 

[turtle_2468]

A party of 10 people is divided at random into 5 groups of 2 people.
Find the probablity of 2 particular people being in the same group.

now, bear in mind that I dont really understand anything about probablity..
but is there a formulaic way to determine this (i.e using combinations/permutations or factorials) or do you have to think it all out and divide all the possibilties into the sample space manually?

also.. heres a volumes question:

-----------------------------------------------
A symmetrical pier of height 5m has an elliptical base with eqaution:
x^2 / 25 + y^2 / 4 = 1
and slopes to a parallel top with equation
x^2 / 9 + y^2 = 1

If cross sections taken parallel to the base are also elliptical, find the volume of the pier given that the area of an ellipse with the semi major axis a and semi-minor axis b is ( Pi a b)
-----------------------------------------------

well.. im quite sure youd use the cross sections method right?
would be wise to use a erm.. semi cone as the base.. and a semi ellipse as the cross section. then multiply it by two? im quite confused. i dont understand how to find values for a and b..

thanks

GRRR my third time writing this: This had better work

Probability: method 1. Let 2 people be A and B.
Let the group A belongs to be the first group. B is then randomly allocated to one of the other slots. There are 4 slots left in first group, 5 in second group, so prob of them being together is 4/9.
Method 2: There are 5*4 ways to arrange A and B in first group, and 5*4 in second, so 40 together
but 10*9=90 ways of arranging them altogether. Hence 4/9.

Conics: 1) Find equation of elliptic cross-section at height h in terms of h
2) find a and b in terms of h, and hence area in terms of h
3) integrate

More on finding a and b:
Recall standard formuila for conic x^2/a^2+y^2/b^2=1.
Hence if you have the equation of the base, a^2=25, a=5, similarly b=2. So area = 10*Pi.

 

[ngai]

its 5 groups of 2...not 2 groups of 5?

i get 1/9:
if no restrictions, then its (10C2*8C2*...*2C2)/5! = 945
for both ppl in same group
lets say all teh groups are named
so then choose a group for them two: 5C1 = 5
order the rest: 8C2*6C2*4C2*2C2 = 2520
now remove the ordering between groups: /5!
so u have (5*2520)/5! = 105
P = 105/945 = 1/9

or u can do it by having all groups named, then no restrictions becomes 113400
and don't remove ordering between groups, so #ways = 2520*5 = 12600

 

[Archman]

this may be a cheap shot using wat ngai got but how about:
call the 2 ppl A and B, now it doesnt matter which order the groups are picked in or the order which 2 ppl from each group are picked in. so say we pick A first
now there are 9 ppl left to pick to be in the same group as A, so B has a 1/9 chance.

 

[Jase]

yes, its 5 groups of 2 lol.

and yep. its 1/9.
could you perhaps give me a crash course on probablity?
i mean, what does "(10C2*8C2*...*2C2)/5!" = 945 actaully mean?
is it the total number of possible arrangements?
argh sorry for being dull or wasting your time. we never covered this much, and i didnt pay enough attention during 3u !

thanks for the hints on the conics, i tried using trapeziums , but it got too confusing. so i used another method and got the values for a and b in terms of h.

thanks for your time.

 

[coca cola]

could you perhaps give me a crash course on probablity?
i mean, what does "(10C2*8C2*...*2C2)/5!" = 945 actaully mean?

it means the total number ways without order (since divide 5!), that the five different groups can be selected.

i.e. first group can be selected in 10C2 ways, second 8C2, and so on...

 

[coca cola]

another version to this solution, think about it this way as in 10 slots. if A is one person and is at one slot in a particular group. B can be at 9 other different slots, but one of them is in the same group as A. therefore

Bayes theorem:

P(A|B) = n(A intersect B)/ n(S intersect B), S denotes the universe.

n(A intersect B) = 1, since there can only be one way they can be together in this particular group.

n(S intersect B) = 9, since there are 9 places B can be at without being at the same position of A.

therefore P(A|B) = 1/9.

 

[turtle_2468]

lol I can't read..

 

[siavash_s_s]

um... never miind