Answer to Question #107555 in Statistics and Probability for Emmanuel Daniels

Question

The proportion of defective items in a certain manufacturing process is 0.20. If 10
items are selected from the process, Find
a. The expected number of good items
b. Probability that none is defective
c. Probability that at least 3 are defective

Answer 1

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

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given that $p=0.20, n=10$

a. The expected number of good items

$$E(X^C)=n(1-p)=10(1-0.2)=8$$

b. Probability that none is defective 

$$P(X=0)=\binom{10}{0}0.2^0(1-0.2)^{10-0}=0.8^{10}=0.1073741824$$

c. Probability that at least 3 are defective

$$P(X\geq3)=1-P(X=0)-P(X=1)-P(X=2)=$$

$$=1-\binom{10}{0}0.2^0(1-0.2)^{10-2}-\binom{10}{1}0.2^1(1-0.2)^{10-1}-$$

$$-\binom{10}{2}0.2^2(1-0.2)^{10-2}=1-0.1073741824-$$

$$-0.268435456-0.301989888=0.3222004736\approx$$

$$\approx0.3222$$