<Language> Binomial Distribution and Ringe

[H. Mark Hubey]

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H. Mark Hubey wrote:

>

>

> Here they are: they are related to each other.

>

> 1. A marksman takes shots at a target so far away that we are

> not sure that he will hit a bullseye. Suppose his probability of

> hitting the bullseye is p. (Now, p is just a number here, a number

> greater than or equal to 0 and less than or equal to 1. We normally

> would write this as 0 <= p <= 1. In books the < sign and the =

> sign are put on top of each other so that it doesn't look like

> we have a representation of an arrow in <=)

>

> a) What is the number of shots that he will [probably have to] take

> before he hits the target (bullyseye)?



Here is the solution to this part. It gives us a new probability

density called the geometric density. The continuous version of

it is the exponential density. We are thinking of a series of

experiments of the type;



	S

	FS

	FFS

	...

	FFFFFFFS



Note that there is only one S (i.e. success) and all the other

tosses before it are F (failure). So this is the "time to success"

density (or distribution). Any of these can happen if the marksman

takes shots. The question is which is more likely to happen?

To compute that we have to average all these possibilities. But before

we do that we should write the probability density in the form

that one usually sees it;



	P(S)=((1-p)^x) * (p)



That is x failures (of probability 1-p each) followed by a single

success of probability p.





> b) What is the probability that out of 10 shots he will hit 6?



This is related to the above. Superficially it seems like we want

to change the above formula to something like



	P(S)= ((1-p)^4) * (p)^6



so that we have 4 failures and 6 successes out of 10. But this is

only one possibility of obtaining 6 successes and 4 failures. IOW,

this is something like FFFFSSSSSS. But we have to consider other

possibilities such as SSSSSSFFFF. To do this we have to find all

possible ways of obtaining 6 successes and 4 failures. This brings

up the problem of permutations and combinations.



To do that let's start with ways of arranging N objects along a line.

For example if we want to find arrangements of {a,b,c} by trial and

error we can see that the possibilities are abc,acb,bac,bca,cab,cba

which totals to 6. In general we can total the number of permutations

by noting that for the 1st position we have N possibilities. After

having made that choice we have N-1 possibilities left, and so on. So

for N objects the number can be computed via



	N*(N-1)*(N-2)....3*2*1



This is read as N-factorial and written as N!.

For N=3, we obtain 3*2*1 which is 6.



The more difficult aspect of this is to compute the number of

possibilities if we arrange N objects R at a time. To do that we

proceed similarly. For the 1st place we have N choices, for the

2nd place we have N-1 choices, etc. However, we have to stop this

before when we get to the Rth place, so the formula becomes



	N*(N-1)*(N-2)*...*(N-R+1)



There is an easier way to write this, in terms of factorials.



                 N!

	P(N,R)=----------

                (N-R)!


Now a combination C(N,R) can be obtained from P(N,R) by dividing by
R! because in the definition of combination the order is not
significant. And finally we get to the famous Binomial Distribution
which is

f(S)= C(N,S)*(p^S)(1-p)^(N-S)


p^S is the probability of S successes. (1-p)^(N-S) is the probability of
(N-S) failures out of N tries (i.e. S successes and N-S failures) and
the C(N,S) is the combination that distributes the S successes out of N
tries into the various possibilities. This is the formula used by Ringe
in his work. This is the formula that is used by Rosenfelder on his
website.



--
Best Regards,
Mark
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hubeyh at montclair.edu =-=-=-= http://www.csam.montclair.edu/~hubey
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