A container holds 50 electronic components, of which 10 are defective. if 6 components are drawn at random from the container, the probability that at least 4 are not defective is . if 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is .

We know that there's a probability of [tex]p=\frac{40}{50}=\frac{4}{5}=0.8[/tex] of choosing an electronic component that is not defective and a probability of [tex]q=\frac{10}{50}=\frac{1}{5}=0.2[/tex] of choosing an electronic component that is defective:

[tex]\displaystyle P(X\geq4)=P(X=4)+P(X=5)+P(X=6)\\\\P(X\geq4)=\binom{6}{4}(0.8)^4(1-0.8)^{6-4}+\binom{6}{5}(0.8)^5(1-0.8)^{6-5}+\binom{6}{6}(0.8)^6(1-0.8)^{6-6}\\\\P(X\geq4)=\frac{6!}{4!(6-4)!}(0.8)^4(0.2)^2+\frac{6!}{5!(6-5)!}(0.8)^5(0.2)^1+\frac{6!}{6!(6-6)!}(0.8)^6(0.2)^0\\ \\P(X\geq4)=0.90112[/tex]

So, if 6 components are drawn at random from the container, the probability that at least 4 are not defective is 0.90112

Problem 2

[tex]\displaystyle P(X=3)=\binom{8}{3}(0.2)^3(1-0.2)^{8-3}\\\\P(X=3)=\frac{8!}{3!(8-3)!}(0.2)^3(0.8)^5\\ \\P(X=3)\approx0.1468[/tex]

So, if 8 components are drawn at random from the container, the probability that exactly 3 are defective is 0.1468