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

Jan 25, 2017

The answer is $= 1 - {\left(1 - p\right)}^{n}$

Explanation:

The probability is

$P \left(X = k\right) = \left(\begin{matrix}n \\ k\end{matrix}\right) {p}^{k} {q}^{n - k}$

where,

$q = 1 - p$

Therefore,

$P \left(X = k\right) = \left(\begin{matrix}n \\ k\end{matrix}\right) {p}^{k} {\left(1 - p\right)}^{n - k}$

In our case,

$k \ge 1$

So,

$P \left(X = 0\right) = \left(\begin{matrix}n \\ 0\end{matrix}\right) {p}^{0} {q}^{n - 0}$

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

$P \left(X = 0\right) = {\left(1 - p\right)}^{n}$

$P \left(X \ge 1\right) = 1 - P \left(X = 0\right) = 1 - {\left(1 - p\right)}^{n}$

Jan 25, 2017

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:

$P \left(k \ge 1\right) = 1 - P \left(k = 0\right) = 1 - {\left(1 - p\right)}^{n}$