A Bernoulli trial (or binomial trial) is a random experiment with exactly two possible outcomes, “success” and “failure”, in which the probability of success is the same every time the experiment is conducted. The mathematical formalization of the Bernoulli trial is known as the Bernoulli process.

**Multinomial Distribution**

If in every try we have three and only three outcomes (A B C) and the probability of A Pa and B is Pb and for C is Pc.

We may want to know what is the chance that after exactly n trials we get exactly Sa number of As and Sb number of B and Sc number of C.

All different ways that we can select s things out of n things is good in this case.

We want 1 A and 2 B and and 3 C in 6 trials:

ABBCCC

BBACCC

BBCCCA

ABCBCC

ABCCBC

ABCCCB

………

Set of possibilities are {1A` 1B 2A 2B 3A 3B 4A 4B 5A 5B 6A 6B} from which three we want to select

In **Binomial Distribution** we want the probability of getting exactly s successes in exactly n Bernoulli trials.

All different ways that we can select s things out of n things is good in this case.

We want 3 s out of 5 trials:

Set of possibilities are {1s 2s 3 s 4s 5s} from which three we want to select —- —– —-

The possibilities are

1s 2s 3s

the other is

2s 1s 3s

But the above two events mean the same

but are a different event than

1s 2s 4s

……

so the question is: “in how many different ways we can select three 3s from this set of 5?”

The answer is c=nCs.

The probability of each outcome is p^s*q^(n-s).

so the probability that we get c successes

P(1s2s3s or 1s2s4s or ……)

=nCs*p^s*q^(n-s)

p is the probability of success

q is the probability of failure

n is total number of trials

and s is total number of successes

And we are interested in all different ways that we can select s things out of n things

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If we want to find out what is the probability that we have got s successes when we get a success in the nth trial.

means we need s-1 successes have happened when we we have performed the (n-1)th trial (no matter how) and then another success at nth trial which makes our total number of successes to s.

p(s-1 successes when we get have performed the (n-1)th trials)=binomial=n-1Cs-1*p^(s-1)*q^(n-1-s+1)=n-1Cs-1*p^(s-1)*q^(n-s)

But we need the nth trial to be a success and that has a probability of p

P(total s successes when we have got a success in the nth trial)=n-1Cs-1*p^(s-1)*q^(n-s) * p=n-1Cs-1*p^(s)*q^(n-s)

This is called **negative binomial distribution** or **Pascal distribution** or **Pólya distribution**

(c) 2015 Amir H. Ghaseminejad

References:

http://mathworld.wolfram.com/NegativeBinomialDistribution.html

http://en.wikipedia.org/wiki/Bernoulli_trial