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

The data follow a Binomial.

a. Expected number of good items

$E(X)=nP$

n= 10 and P= 0.8

$E(X)=10×0.8=8$

b . Probability of no defective.

$P(=0)$ can be obtained from =BINOM.DIST(0,10,0.2,FALSE) Excel formula which yields 0.107374.

Similarly, substituting to the Binomial distribution, we have $(1-P)^n=0.8^{10}=0.107374$.

c. $P(\ge3$ defectives).

The Excel formula =1-BINOM.DIST(2,10,0.2,TRUE) can be used which yields 0.3222.

Similarly from Binomial distribution, $1-(P(=0)+P(=1)+P(=2))$ can be obtained:

$P(=0)=0.107374$ from (b),

$P(=1)=\binom{10}{1}×0.2×0.8^9=0.268435$ ,

$P(=2)=\binom{10}{2}×0.2^2×0.8^8=0.30199$ .

Thus, $P(\ge3)=1-0.107374-0.268435-0.30199=0.3222$ .