Respuesta :
In this problem, we have a binomial probability distribution
where
p=80%=0.80
q=1-p -----> q=0.20 probability of fail
the probability that more than 3 of them are successful is
P(X>3)=P(x=4)+P(X=5)+P(x=6)
step 1
Find out P(X=4)
[tex]\begin{gathered} P(X=4)=\frac{6!}{4!(6-4)!}*0.80^4*0.20^2 \\ P(X=4)=0.24576 \end{gathered}[/tex]step 2
Find out P(X=5)
[tex]\begin{gathered} P(X=5)=\frac{6!}{5!(6-5)!}\times0.80^5\times0.20^1 \\ P(X=5)=0.393216 \end{gathered}[/tex]step 3
Find out P(X=6)
[tex]\begin{gathered} P(X=6)=\frac{6!}{6!(6-6)!}\times0.80^6\times0.20^0 \\ P(X=6)=0.262144 \end{gathered}[/tex]step 4
Adds the probabilities
P(X>3)=0.24576+0.393216+0.262144
P(X>3)=0.90112
therefore
