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

2 Answers
Jan 25, 2017

Answer:

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#

Jan 25, 2017

Answer:

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#