How is a binomial distribution different from a Bernoulli distribution?

Nov 22, 2016

the Binomial is just the aggregate of many Bernoulli trials

Explanation:

if the Bernoulli is $f \left(x\right) = p \left(x\right) 1 - p \left(x\right)$ and we wanted to know what would happen if we had two trails then there are 4 total possibilities that can happen.

1= success , 2= success: $p {\left(x\right)}^{2} {\left(1 - p \left(x\right)\right)}^{0}$
1= fail , 2= success: $p \left(x\right) \left(1 - p \left(x\right)\right)$
1= success , 2= fail: $p \left(x\right) \left(1 - p \left(x\right)\right)$
1= fail , 2= fail: $p {\left(x\right)}^{0} {\left(1 - p \left(x\right)\right)}^{2}$

now if we wanted to know the probability of exactly 1 success out of 2 trials we know that

$2 \cdot p \left(x\right) \left(1 - p \left(x\right)\right)$ or (""_1^2)p(x)^1(1-p(x))^(2-1)

if we want exactly 2 success out of 2 trials then

$p {\left(x\right)}^{2} {\left(1 - p \left(x\right)\right)}^{0}$ or (""_2^2)p(x)^2(1-p(x))^(2-2)

it turns out that if you continue like this you end up with the binomial form
f(x,k,n) = (""_k^n)p(x)^k(1-p(x))^(n-k)