Question

relationship between bernoulli and binomial distribution

Free throw basketball: we have p=0.6 to make a shot successfully, 0.4 miss the shot.

Bernoulli: P(X=1)=0.6, P(X=0)=0.4

now we shoot the ball 5 times, so we have 5 random variable X1, X2, ,,,,,,X5

P(X1=Miss,X2=Success,X3=X4=Miss,X5=Success,X6=Miss)=p^2(1-p)^4

this is not binomial because we have the certain order, however, who can tell me the difference?

becasue everyone said Bernoulli only n=1, but here n=5 and it is still bernoulli and it isnot binomial becasue it doesn't have 5 choose 2

Answer 1

Yes you are right, i is not Binomial distribution.

Because Binomial distribution is sum of n independent Bernoulli trials. But given expression is the product of Bernoulli trials.

For Bernoulli distribution,

P(X=x) = p^x * (1-p)^1-x ; x = 0,1

Now

P(X=miss) = P(X=0) = p⁰ * (1-p)¹

P(X=success) = P(X=1) = p¹ * (1-p)⁰

Therefore,

P(X1=Miss,X2=Success,X3=X4=Miss,X5=Success,X6=Miss) = {p⁰ * (1-p)¹ } * { p¹ * (1-p)⁰ } * { p⁰ * (1-p)¹ } * { p⁰ * (1-p)¹ } *

{p¹ * (1-p)⁰} * { p⁰ * (1-p)¹}

= p² * (1-p)⁴