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?

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Answer 1 · 2018-05-07T06:43:21

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

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

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

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

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

= $792*0.000976562*0.13348388671875$

$~=0.10324$