The binomial distribution is
[tex]\begin{gathered} P(X=k)=(nbinomialk)p^k(1-p)^{n-k} \\ n\rightarrow\text{ total number of trials} \\ k\rightarrow\text{ number of successful trials} \\ p\rightarrow\text{ probability of a successful trial} \end{gathered}[/tex]Therefore, in our case, the function is (Notice that 12%=0.12)
[tex]\begin{gathered} \Rightarrow P(X=k)=(8binomialk)(0.12)^k(1-0.12)^{8-k} \\ \Rightarrow P(X=k)=(8binomialk)(0.12)^k(0.88)^{8-k} \end{gathered}[/tex]Thus, we need to find the probability that fewer than 3 of the 8 parts are defective; this is,
[tex]P(X<3)=P(X=0)+P(X=1)+P(X=2)[/tex]Calculating P(0), P(1), and P(2),
[tex]\begin{gathered} P(X=0)=(8binomial0)(0.12)^0(0.88)^8=\frac{8!}{(8-0)!0!}(1)(0.88)^8=0.88^8=0.35963452... \\ P(X=1)=\frac{8!}{(8-1)!1!}(0.12)^1(0.88)^7=8(0.12)^1(0.88)^7=39.232857... \end{gathered}[/tex]and
[tex]P(X=2)=\frac{8!}{6!2!}(0.12)^2(0.88)^6=28(0.12)^2(0.88)^6=0.18724772...[/tex]Hence,
[tex]\Rightarrow P(X<3)\approx0.9392[/tex]