What is the derivation for the formula p^y(1-p)^n-y.
I can understand for bernouli random variable
P(X=1) = p
P(X=0) = p-1
therefore , E[x] = p.
thus, var(x) = p*(1-p)…
Can somebody xplain the derivation for Binomial part…
You’re aware that if A & B are independent events, P(AB) = P(A)P(B), yes?
The binomial distribution formula gives the probability of getting y successes in n independent trials, where the probability of success on each trial is p.
Suppose that you had 10 trials, 3 successes and 7 failures, with P(success) = 0.4; the exact outcome, in order, was:
SFFSFFFFSF
So, the probability of getting SF on the first two trials – given that they’re independent – is:
P(SF) = P(S)P(F) = (0.4)(0.6)
The probability of getting SFF on the first three trials is:
P(SFF) = P(S)P(F)P(F) = (0.4)(0.6)²
The probability of getting SFFS on the first three trials is:
P(SFFS) = P(S)P(F)P(F)P(S) = (0.4)(0.6)²(0.4) = (0.4)²(0.6)²
And so on.
P(SFFSFFFFSF) = P(S)P(F)P(F)P(S)P(F)P(F)P(F)P(F)P(S) = [(0.4)^(3)][(0.6)^(7)]