Question

A company produces only one type of light bulbs. The light bulbs are being manufactured two factories: Factory A and Factory B. Factory A produces 10, 000 bulbs every year, while Factory B produces 2, 000. Also 2% of bulbs from Factory A are defective while 1% of bulbs from Factory B are. After production all bulbs are brought to the same facility and shipped to retail stores.
Required:
a. What is the probability that a bulb bought from a store is defective?
b. Given that the bulb is defective, what is the probability that it was produced by Factory A?
c. Given that the bulb is functional, what is the probability that it was produced by Factory A?

Answer 1

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given values:

$P(\frac{D}{A})=0.02 \\\\P(\frac{D}{B})=0.01\\\\P(A)=\frac{10,000}{10,000+2000} =\frac{10}{12} =\frac{5}{6} \\\\P(B)=\frac{1}{6}$

For point a:

$P(D) =p(D\cap A) +P(D \cap B)$

         $=P(\frac{D}{A}) \times P(A) + P(\frac{D}{B}) \times P(B)\\\\=0.02 \times \frac{5}{6} + 0.01 \times \frac{1}{6}\\\\=\frac{0.1}{6} + \frac{0.01}{6}\\\\=\frac{0.1+0.01}{6} \\\\=\frac{0.11}{6} \\\\=\frac{11}{600} \\\\=0.0183$

For point b:

$P(\frac{A}{D})= \frac{P(\frac{D}{A}) \times P(A)}{P(D)}$

         $= \frac{0.02 \times \frac{5}{6}}{\frac{11}{600}}\\\\= \frac{0.02 \times \frac{5}{6}}{0.0183333333}\\\\= \frac{0.02 \times 0.833333333}{0.0183333333}\\\\= \frac{0.0166666667}{0.018}\\\\=0.9090 \approx 0.0991$

For point c:

$P(\frac{A}{D^C})=\frac{P(\frac{D^c}{A}) \times P(A)}{P(D^c)}$

           $= \frac{1-0.02 \times \frac{5}{6}}{1- \frac{11}{600}}\\\\= \frac{0.98 \times \frac{5}{6}}{0.98}\\\\=\frac{5}{6}\\\\=0.833$