Question

find the n that maximizes P(A)

Answer 1

$$P(A) = {{\left( {\matrix{
{10} \cr
4 \cr

} } \right)\left( {\matrix{
{n - 10} \cr
{16} \cr

} } \right)} \over {\left( {\matrix{
n \cr
{20} \cr

} } \right)}}$$

Answer 2

maximum P(A) when (n,20 ) minimum right ? like when its 1 for example ?

how ever that wont work since n-10 must be >16

Answer 3

how are you going by solving this?

Answer 4

wait thinking

Answer 5

@ganeshie8

Answer 6

This is what we're talking about right?\[\Large \left(\begin{matrix}p \\ k\end{matrix}\right)=\frac{p!}{(p-k)! \ k!}\]

Answer 7

yep

Answer 8

wanna expand ??

Answer 9

there is a rule how to get the largest(n,k) value by controling k
to chose k in the middle to get the middle triangle number which also the largest

Answer 10

but that would be like trial /error if we did it

Answer 11

So these are all little pascal's triangles. That means the middle one will be the largest since we know the outside is all 1's right?

Since we can only have a factorial of 0 or greater, that puts some limits on what n can be. Since we have n-16 above a 10 then we absolutely have to have at least n>=26

Answer 12

sowe would have least n of 22 ?

Answer 13

Expanding it out makes it a little clearer. \[\Large \frac{10!}{(10-4)!*4!} \frac{(n-10)!}{(n-26)!*16!} \frac{(n-20)!*20!}{n!}\]

Answer 14

\[\frac{ f(n+1) }{ f(n) }>1; \frac{ f(n+2) }{ f(n+1) }<1\]
\[\frac{ f(n+1) }{ f(n) }=\frac{ \left(\begin{matrix}n-19 \\ 16\end{matrix}\right) }\left(\begin{matrix}n \\ 20\end{matrix}\right){ ?\left(\begin{matrix}n+1 \\ 20\end{matrix}\left(\begin{matrix}n-10 \\ 16\end{matrix}\right)\right) }\]

Answer 15

fkkkkkkk

Answer 16

i feel it should be 21 or 22 mmm

Answer 17

$${{f(n + 1)} \over {f(n)}} = {{\left( {\matrix{
{n - 19} \cr
{16} \cr

} } \right)\left( {\matrix{
n \cr
{20} \cr

} } \right)} \over {\left( {\matrix{
{n + 1} \cr
{20} \cr

} } \right)\left( {\matrix{
{n - 10} \cr
{16} \cr

} } \right)}} = {{{\textstyle{{(n - 9)!} \over {16!(n - 25)!}}}{\textstyle{{n!} \over {(n - 20)!20!}}}} \over {{\textstyle{{(n - 10)!} \over {16!(n - 26)!}}}{\textstyle{{(n + 1)!} \over {(n - 19)!20!}}}}}$$

Answer 18

what ur trying to do nin ?

Answer 19

committing a mathematical abomination? laughing out loud IDK

Answer 20

a few cancelations can be done and rewriting

Answer 21

16! 20! cancel out

Answer 22

ok ill do trial error /last option lol
ok so we need maximum (n-10,16)
and minimum (n,20)
n>=26

for max 16*2=32 so n=42 mmmm

Answer 23

\[=\frac{ (n-9)!n! (n-26)!(n-19)! }{ (n-10)!(n+1)!(n-25)!(n-20)! }\]

Answer 24

It's not that complicated really if we just think of these two parts separately with the n's in them.

Since just simply looking at it shows us n has a lower possible bound of 26, let's just do a change of variable to help us look at what we really need:

n=26+k

\[\Large \frac{10!}{6!*4!} \frac{(k+13)!}{(k)!*16!} \frac{(k+6)!*20!}{(26+k)!}\]

This looks more manageable. Let's make it more manageable and scoot all the stuff without k's to the left and the stuff with k's to the right. \[\Large \frac{10!*20!}{6!*4!*16!}*\frac{(k+6)!*(k+13)!}{k!(26+k)!}\]Alright so what's going to maximize the right part? The rest are constants so it really doesn't matter. It looks like we can simplify some of these down to get: \[\Large \frac{(k+1)(...)(k+6)}{(k+14)(...)(k+26)}\]

Yeah now take the derivative, set it equal to zero and uhhhhh... lul idk

Answer 25

laughing out loud kai

Answer 26

so maximize function ?@ganeshie8

Answer 27

ok ok 10 and 4 there for a reason
what course ur taking @nincompoop is it calcules or descrete math ?

Answer 28

\[= \frac{ (n-9) (n+9)}{ (n+1)(n-25) }\rightarrow (n-9)(n-19)>(n+1)(n-25)\]

Answer 29

\[49>n\]

Answer 30

so I was dreaming about not knowing what I am doing in math, and I think this was it

Answer 31

Wait! I just think I figured it out. So remember if we set k=0 then that's the lowest possible number our thing can have, so that will give us: \[\Large \frac{6!*13!}{26!}\]

Now just try every value from here on up haha. I'm going to bed, good luck everyone, may the force be with you etc... lol.

Answer 32

wait can we make usfull of this identity?
(n,k) (k,r)=(n,r)(n-r, k-r)

Answer 33

go ahead @ikram002p

Answer 34

mmmm see this :-\
http://www.wolframalpha.com/input/?i=maimize+%28 \binom{10}{4}*\binom{n-10}{16}%29%2F\binom{n}{20}

Answer 35

also table from 26 to 42 gives rational numbers mmm

Answer 36

check this
http://prntscr.com/42hkwz

Answer 37

Example 5 here might help http://math.arizona.edu/~jwatkins/o-mle.pdf

Answer 38

as ic in the graph n btw 48 to 50

Answer 39

according to that result, the max should be 50, which looks like it corresponds with the graph @ikram002p showed

Answer 40

50 to 48 but not sure mmmm lets see value of it when n>100

Answer 41

49

Answer 42

48, but n+1, so 49

Answer 43

that was only trial error :-\
just tell what course u have ?

Answer 44

probability and statistics

Answer 45

http://math.arizona.edu/~jwatkins/o-mle.pdf shows a method of maximum likelihood estimation (example 5)

Answer 46

supposedly, an intro

Answer 47

shiz that's with an R software, finally! laughing out loud

Answer 48

I suppose you can just use R yeah to find the max. They do go through a proof though and they show
\[ \hat{n}=\lfloor \frac{10(20)}{4}\rfloor\]

Answer 49

I don't know how to use it yet, but I am learning this stuff and subject for the first time

Answer 50

nice !

Answer 51

Hm lemme see if I can get a code that will work.
Hm that is a difficult question for a first time o.o

Answer 52

I tried this in R:
> optimize(fn, interval=c(40, 60), maximum=TRUE)
$maximum
[1] 49.49498

$objective
[1] 0.2801753

I'm not sure which algorithm R uses though for this function.

Answer 53

@nincompoop

Answer 54

oh and of course,
fn<-(choose(10,4)*choose(x-10,16))/(choose(x,20))