# Bernoulli Binomial Distribution

• Destroxia
In summary, the Bernoulli binomial distribution is derived by generalizing the probability of a coin flip and taking into account the different outcomes that can occur. The probability of winning k times in a row is represented by p^k, and the probability of losing the remaining (n-k) trials is represented by (p-1)^(n-k). The combination of these two probabilities, multiplied by the number of possible ways to achieve k wins and (n-k) losses, gives us the probability of obtaining the desired outcome. This distribution is useful in scenarios where there are a fixed number of independent trials with only two possible outcomes
Destroxia

## Homework Statement

Derive the bernoulli binomial distribution by generalizing the probability of a coin flip.

## P(k, n) = \binom{n}{k}p^{k}q^{(n-k)} ##, q = p - 1

## Homework Equations

Combination: ## \binom{n}{k} = \frac {n!} {k!(n-k)!} ##

Prob. of coin flip: ## \frac {\binom{n}{k}} {2^n} ##

3. The Attempt at a Solution

I don't have much background in statistics, so I'm trying to figure out how these distributions were derived and such.

Looking at it in terms of a game, like flipping a coin, where heads is a win, tails is a lose, I don't understand things like why the probability of getting a win is ## p^{k} ## , or why the probability of losing is ## (p-1)^{n-k} ##. Then after that, why is the ## \binom{n}{k} ## even included in the equation.

I've been reading books, and articles, and I'm just not really wrapping my head around how multiplying the combination of ## \binom{n}{k} ## by the probability of winning by the probability of losing, even gives us the probability of the outcome.

If p is the probability of a win, then p^k is the probability of winning k times in a row.
If you have n trials and only win k times, then you lose the rest (n-k) of te trials. So the probability of winning the first k and then losing the rest would be ##p^k(1-p)^{n-k}##.
Now comes the combination part.
As I said, the probability of winning the first k and losing the rest is that piece of the formula. But the probability would be the same if you lost the first n-k and won the last k.
In all, you have to add up all the possible ways to win k times out of n. Each way has the same probability, so the total probability of winning k times is the probability of one of the ways times the number of combinations of k wins and n-k losses.

RyanTAsher said:

## Homework Statement

Derive the bernoulli binomial distribution by generalizing the probability of a coin flip.

## P(k, n) = \binom{n}{k}p^{k}q^{(n-k)} ##, q = p - 1

## Homework Equations

Combination: ## \binom{n}{k} = \frac {n!} {k!(n-k)!} ##

Prob. of coin flip: ## \frac {\binom{n}{k}} {2^n} ##

3. The Attempt at a Solution

I don't have much background in statistics, so I'm trying to figure out how these distributions were derived and such.

Looking at it in terms of a game, like flipping a coin, where heads is a win, tails is a lose, I don't understand things like why the probability of getting a win is ## p^{k} ## , or why the probability of losing is ## (p-1)^{n-k} ##. Then after that, why is the ## \binom{n}{k} ## even included in the equation.

I've been reading books, and articles, and I'm just not really wrapping my head around how multiplying the combination of ## \binom{n}{k} ## by the probability of winning by the probability of losing, even gives us the probability of the outcome.

Suppose n = 5, for example, and k = 2; so, we toss a coin 5 times and are interested in the event {2 heads, 3 tails}. We can list all the possible outcomes in order from toss 1, toss 2, ..., toss 5:
(1) HHTTT
(2) HTHTT
(3) HTTHT
(4) HTHTT
(5) HTTTH
(6) TTTHH
(7) TTHTH
(8) TTHHT
(9) THHTT
(10) THTHT
If p = probability of H in one toss (and 1 = 1-p = probability of T), ask yourself: what is P(outcome 1)? What is P(outcome 2)? ... P(outcome 10)? Rember, we are assuming independent tosses, so getting H or T on one toss does not affect in any way the probability of H or T in any other toss.

Now put it all together: what is P(2 H and 3 T)?

RUber said:
If p is the probability of a win, then p^k is the probability of winning k times in a row.
If you have n trials and only win k times, then you lose the rest (n-k) of te trials. So the probability of winning the first k and then losing the rest would be ##p^k(1-p)^{n-k}##.
Now comes the combination part.
As I said, the probability of winning the first k and losing the rest is that piece of the formula. But the probability would be the same if you lost the first n-k and won the last k.
In all, you have to add up all the possible ways to win k times out of n. Each way has the same probability, so the total probability of winning k times is the probability of one of the ways times the number of combinations of k wins and n-k losses.

Thank you, this outline of all the constituents really helped, I was able to visualized how it was put together here.

Ray Vickson said:
Suppose n = 5, for example, and k = 2; so, we toss a coin 5 times and are interested in the event {2 heads, 3 tails}. We can list all the possible outcomes in order from toss 1, toss 2, ..., toss 5:
(1) HHTTT
(2) HTHTT
(3) HTTHT
(4) HTHTT
(5) HTTTH
(6) TTTHH
(7) TTHTH
(8) TTHHT
(9) THHTT
(10) THTHT
If p = probability of H in one toss (and 1 = 1-p = probability of T), ask yourself: what is P(outcome 1)? What is P(outcome 2)? ... P(outcome 10)? Rember, we are assuming independent tosses, so getting H or T on one toss does not affect in any way the probability of H or T in any other toss.

Now put it all together: what is P(2 H and 3 T)?

Okay, this exercise helped, a lot. I think a lot of my confusion was stemming from the fact that the game of flipping a coin, both the win, and lose have equal probability, so I kept just writing the probability as ## p^{n} ##, which would work in the case of a coin, but since we are generalizing it for other scenarios, we need to take into account the fact that the probability of losing and winning could be different, and it really showcased the need for that combination as well.

So, I ended up doing outcome 1, and outcome 2, that was where I realized the point of the ## 1 - p = p(T) ##.

## P(O_1) = p * p * (1-p) * (1-p) * (1-p) = p^{2}(p-1)^{3} ##
## P(O_2) = p * (1-p) * p * (1-p) * (1-p) = p^{2}(p-1)^{3} ##

That's kinda when I realized you could generalize it as ## p^{k}(1-p)^{(n-k)} ##...

Then, to take into account all the different combinations, you would have to use the ## \binom{n}{k} ##, to take into account the probability of all those outcomes together.

## P(k,n) = \binom{n}{k} p^{k}(1-p)^{(n-k)} ##

To solve the problem...

## P(2H and 3T) = \binom{5}{2} p^{2}(1-p)^{3} = \binom{5}{2} (\frac 1 2)^{5} = (10)(\frac {1} {32}) = .3125 = 31.25 \% ##

RUber

## What is Bernoulli Binomial Distribution?

Bernoulli Binomial Distribution is a probability distribution that describes the number of successes in a sequence of independent binary experiments, where each experiment has only two possible outcomes - success or failure.

## What is the difference between Bernoulli and Binomial Distribution?

Bernoulli Distribution is a special case of Binomial Distribution where there is only one trial or experiment, while Binomial Distribution involves multiple independent trials or experiments with the same probability of success.

## How is Bernoulli Binomial Distribution used in real life?

Bernoulli Binomial Distribution is commonly used in statistics and data analysis to model and predict the number of successes or failures in a given set of binary experiments, such as the success rate of a marketing campaign or the likelihood of a coin toss resulting in heads or tails.

## What are the assumptions of Bernoulli Binomial Distribution?

The assumptions of Bernoulli Binomial Distribution include that each trial or experiment is independent, the probability of success remains constant throughout all trials, and there are only two possible outcomes - success or failure.

## How is Bernoulli Binomial Distribution related to the Central Limit Theorem?

Bernoulli Binomial Distribution is a discrete probability distribution, while the Central Limit Theorem states that the sampling distribution of the sample mean from a large sample size approximates a normal distribution regardless of the underlying distribution. However, for a large enough sample size, the Binomial Distribution can also approximate a normal distribution, making it useful in applications related to the Central Limit Theorem.

• Calculus and Beyond Homework Help
Replies
14
Views
2K
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Quantum Interpretations and Foundations
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Set Theory, Logic, Probability, Statistics
Replies
10
Views
991
• Calculus and Beyond Homework Help
Replies
33
Views
3K
• Calculus and Beyond Homework Help
Replies
8
Views
1K