- #1
Pushoam
- 962
- 52
Homework Statement
Derive the bernoulli binomial distribution.
Homework Equations
The Attempt at a Solution
Each bernoulii trial could have only two possible outcomes .
Let’s name one outcome as success and another outcome as failure.
Let’s denote the probability of getting success and failure in each Bernoulli trial by p and q respectively.
Clearly, q = 1-p
For n independent Bernoulli trials, let’s denote the probability of getting k successes by P (k, n).
The probability of getting a success in each trial is p.
So, the probability of getting k sucesses in k trials is ## p^k##.
The probability of getting k sucesses in n independent Bernoulli trials is = probability of getting k sucesses and n-k failures in n independent Bernoulli trials.
The probability of getting k sucesses in k independent Bernoulli trials and getting n – k failures in n-k independent Bernoulli trials are ## p^k~ and~ (1-p)^{(n-k)} ## respectively.
Let’s consider the following events.
Event A : getting k successes in the 1st k trials of n independent Bernoulli trials
Event B : getting n- k failures in the next n- k trials of n independent Bernoulli trials
Now, the probability of getting both events A and B is ## p^k (1-p)^(n-k) ##
But, according to the problem the k successes could be in any k trials of the n independent Bernoulli trials. It is not necessary that these k trials should be the 1st k trials of the n independent Bernoulli trials.
So, the events corresponding to the problem is,
Event C: getting k successes in any of the k trials of n independent Bernoulli trials
Event D: : getting n- k failures in the rest of the n- k trials of n independent Bernoulli trials
Now, the probability of getting both events C and D is what the question is asking.For event C,
In how many ways can I get any of the k trials out of n independent trials?
This is ##\binom n k## i.e. choosing k boxes out of n boxes.
After being chosen a way of getting k trials out of n independent Bernoulli trials , the probability of getting k successes in these chosen k trials is ## p^k##.
After being chosen the k trials, there is only one way of getting n – k trials.
So, after being chosen the k trials, the probability of getting k sucesses in these trials and n-k failures in the rest of n-k trials is ## p^k (1-p)^{(n-k)} ##.
Since, there are ##\binom n k## of choosing k trials and for each choice, the probability of getting k sucesses in these trials and n-k failures in the rest of n-k trials is ## p^k (1-p)^{(n-k) } ##,
the probability of getting k sucesses in any of the k trials of n independent Bernoulli trials and n-k failures in the rest of n-k trials is = sum of the probability of getting k sucesses in these trials and n-k failures in the rest of n-k trials i.e.## p^k (1-p)^{(n-k)} ## for ##\binom n k## times.
Hence, the probability of getting k sucesses out of n independent Bernoulli trials is =P(k,n) = ##\binom n k p^k (1-p)^{(n-k)}##Is this correct?