Question

How do you find the probability of at least one success when `n` independent Bernoulli trials are carried out with probability of success `p`?

Answer 1

The answer is `=1-(1-p)^n`

Explanation:

The probability is

`P(X=k)=((n),(k))p^kq^(n-k)`

where,

`q=1-p`

Therefore,

`P(X=k)=((n),(k))p^k(1-p)^(n-k)`

In our case,

`k>=1`

So,

`P(X=0)=((n),(0))p^0q^(n-0)`

`((n),(1))=(n!)/(1!(n-1)!)=n`

`P(X=0)=(1-p)^n`

`P(X>=1)=1-P(X=0)=1-(1-p)^n`

Answer 2

See explanation.

Explanation:

To calculate the probability of getting at least one success you use opposite event formula.

The event opposite to given is You got no success in `n` trials. Therfore the probability is:

`P(k=0)=(""_0^n)(p^0)(1-p)^n=(1-p)^n`

So the probability we are looking for is:

`P(k>=1)=1-P(k=0)=1-(1-p)^n`