If 12% of a population of parts defective, what is the probability of randomly selecting 60 parts and finding that 8 more are defective?

Let $X=$ the number of defective parts: $X\sim Bin(n, p).$

Given $n=60, p=0.12, q=1-p=1-0.12=0.88$

$P(X>8)=1-P(X=0)-P(X=1)$

$-P(X=2)-P(X=3)-P(X=4)$

$-P(X=5)-P(X=6)-P(X=7)$

$-P(X=8)=1-\dbinom{60}{0}(0.12)^0(0.88)^{60-0}$

$-\dbinom{60}{1}(0.12)^1(0.88)^{60-1}-\dbinom{60}{2}(0.12)^2(0.88)^{60-2}$

$-\dbinom{60}{3}(0.12)^3(0.88)^{60-3}-\dbinom{60}{4}(0.12)^4(0.88)^{60-4}$

$-\dbinom{60}{5}(0.12)^5(0.88)^{60-5}-\dbinom{60}{6}(0.12)^6(0.88)^{60-6}$

$-\dbinom{60}{7}(0.12)^7(0.88)^{60-7}-\dbinom{60}{8}(0.12)^8(0.88)^{60-8}$

$=0.29007983737$

The probability of randomly selecting 60 parts and finding that 8 more are defective is 0.29008.