What is the probability of 4 successes for a binomial experiment consisting of 11 trials with probability 0.4 of success on each trial?

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Round to the nearest percent.

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Answer 1 · 2017-01-16T05:06:49

The probability is about 23.7%

#X~b(n,p)# where #n=11# and #p=0.4#

and we are looking for #P(X=4)#

Since

#X~b(n,p) => P(X=k)=(C_(n,k)) p^k(1-p)^(n-k)#

where #C_(n,k)=(n!)/(k!(n-k)!)#

we have to calculate

#P(X=4)=(11!)/(4!(11-4)!)(0.4)^4(0.6)^7#

#=(11!)/(4!(7!))(0.4)^4(0.6)^7=((8)(9)(10)(11))/(4!)(.0256)(.0279936)#

#=(2(9)(10)(11))/(3!)(.0256)(.0279936)=((2)(3)(10)(11))/(2!)(.0256)(.0279936)#

#=330(.0256)(.0279936)approx.237#