Thread: [SOLVED] questions are about sampling.

Can someone help me with the following:

see the attachment.

I am really stuck and do not know were to start

Thanks

Maths_Brain

Originally Posted by maths_brain

Can someone help me with the following:

see the attachment.

I am really stuck and do not know were to start

Thanks

Maths_Brain

You might find that many members are reluctant to open attached files (I'm one of those members). So you might get a better response if you take the time and type out your questions.

Originally Posted by mr fantastic

You might find that many members are reluctant to open attached files (I'm one of those members). So you might get a better response if you take the time and type out your questions.

Having been stupid enough to open said attachment what do I find? It's not a word document its a wrapper for picture!! Grrrr...

CB

(I'm not vulnerable to macro virus's but I don't like opening word and excel attachments because it takes a long time to boot up OOO on my slow computers (I won't open an office document on the my work computers as a matter of principle)

Originally Posted by maths_brain

Can someone help me with the following:

see the attachment.

I am really stuck and do not know were to start

Thanks

Maths_Brain

The third part is just algebra, write out the right hand side in terms of factorials and simplify.

$\displaystyle
{n-1 \choose r}+{n-1 \choose r-1}=\frac{(n-1)!}{r!(n-1-r)!} + \frac{(n-1)!}{(r-1)!(n-r)!}
$

Alternativly one may observe that it is the sum of the number of ways of choosing a subset of $\displaystyle r$ elements from the given set which does not include $\displaystyle 1$, and the number of ways that does include $\displaystyle 1$.

CB

Originally Posted by maths_brain

Can someone help me with the following:

see the attachment.

I am really stuck and do not know were to start

Thanks

Maths_Brain

When choosing a subset consisting of $\displaystyle r$ elements from a set of $\displaystyle n$ elements the first element may be choosen in $\displaystyle n$ ways the second in $\displaystyle (n-1)$ ways and so on to the $\displaystyle r$-th which may be choosen in $\displaystyle (n-r+1)$ ways. So the number of ways that the set may be choosen is $\displaystyle n(n-1)...(n-r+1)$, but in this number every permutation of each distinct choice appears, so to get the number of choices independent of order we divide by the number of permutations of $\displaystyle r$ distinct objects. Hence the number of ways of choosing the subset is:

$\displaystyle N(r,n)=\frac{n(n-1)...(n-r+1)}{r!}=\frac{n!}{r!(n-r)!}$

CB

Originally Posted by maths_brain

Can someone help me with the following:

see the attachment.

I am really stuck and do not know were to start

Thanks

Maths_Brain

When choosing a subset consisting of $\displaystyle r$ elements from a set of $\displaystyle n$ elements the first element may be choosen in $\displaystyle n$ ways the second in $\displaystyle (n-1)$ ways and so on to the $\displaystyle r$-th which may be choosen in $\displaystyle (n-r+1)$ ways. So the number of ways that the set may be choosen is $\displaystyle n(n-1)...(n-r+1)$, but in this number every permutation of each distinct choice appears, so to get the number of choices independent of order we divide by the number of permutations of $\displaystyle r$ distinct objects. Hence the number of ways of choosing the subset is:

$\displaystyle N(r,n)=\frac{n(n-1)...(n-r+1)}{r!}=\frac{n!}{r!(n-r)!}$

Which is what we expect.

Now the number of ways of choosing a subset that does not contain $\displaystyle 1$ is $\displaystyle N(r,n-1)$, and so the probability required is:

$\displaystyle p=\frac{N(r,n-1)}{N(r,n)}$

Similarly if $\displaystyle 1$ is in the set the number of ways of choosing it is $\displaystyle N((r-1),(n-1))$

CB

Thanks for the help Captin Black.

Just wondering is the probaility of it containing 1 in the set equal to N((r-1),(n-1))

and the number of ways of choosing a subset that does not contain is N(r,n-1)

Thanks

Maths Brain

Originally Posted by maths_brain

Thanks for the help Captin Black.

Just wondering is the probaility of it containing 1 in the set equal to N((r-1),(n-1))

and the number of ways of choosing a subset that does not contain is N(r,n-1)

Thanks

Maths Brain

If you read what I said earlier you will realise that N((r-1),(n-1)) is the number of ways of choosing the set with a 1 in it, to get the probability you have to divide by the total number of sets (N(r,n))

(same for the other one)

CB