Question

companies a, b, and c produce 10%, 20%, and 70%, respectively, of the major appliances sold in a certain area. in that area, 1% of the company a appliances, 4\frac{1}{2}% of the company b appliances, and 5% of the company c appliances need service within the first year. suppose a defective appliance is chosen at random; find the probability that it was manufactured by company b.
the probability that it came from company b is
(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Define the probabilities

Let $P(A)=0.1$, $P(B)=0.2$, $P(C)=0.7$ be the probabilities that an appliance is produced by company A, B, and C respectively. Let $P(D|A) = 0.01$, $P(D|B)=0.045$, $P(D|C)=0.05$ be the conditional - probabilities that an appliance needs service (is defective) given that it is produced by company A, B, and C respectively.

Step2: Use the law of total probability to find $P(D)$

By the law of total probability, $P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$.
\[

$$\begin{align*} P(D)&=(0.01\times0.1)+(0.045\times0.2)+(0.05\times0.7)\\ &=0.01\times0.1 + 0.045\times0.2+0.05\times0.7\\ &=0.001 + 0.009+0.035\\ &=0.045 \end{align*}$$

\]

Step3: Use Bayes' theorem to find $P(B|D)$

By Bayes' theorem, $P(B|D)=\frac{P(D|B)P(B)}{P(D)}$.
\[

$$\begin{align*} P(B|D)&=\frac{0.045\times0.2}{0.045}\\ &= 0.2000 \end{align*}$$

\]

Answer:

$0.2000$