Question

Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. n =12, x= 5, p=0.25?

Answer 1

Probability of `5` successes among `12` trials is `0.10324`

Explanation:

Binomial distribution gives the probability of `x` successes among `n` trials as `C_x^np^xq^((n-x))`,

where `p` is probability of success in one trial and `q` is probability of failure i.e. `q=1-p`

In the given case, we have `n=12`, `x=5` and `p=0.25` i.e. `q=1-0.25=0.75`

Hence probability of `5` successes among `12` trials is

`C_5^12(0.25)^5(0.75)^((12-5))`

= `(12*11*10*9*8)/(1*2*3*4*5) * (0.25)^5*(0.75)^7`

= `11*72*0.25^5*0.75^7`

= `792*0.000976562*0.13348388671875`

`~=0.10324`