91Ó°ÊÓ

Bayes' theorem relates the conditional probability of two events occurring. In this case, we have the probability that a TV is of a specific brand, given that it is still working after 30,000 hours. The formula for Bayes' theorem is: \(P(A|B) = \frac{P(B|A) * P(A)}{P(B)}\) Where: - \(P(A|B)\) is the probability of event A occurring given that event B has occurred. - \(P(B|A)\) is the probability of event B occurring given that event A has occurred. - \(P(A)\) is the probability of event A occurring. - \(P(B)\) is the probability of event B occurring.

First, we need to find the probability that a TV sold by Ace that year is of each brand: - The probability that a TV is Brand-A, \(P(A)\), is 1000/4500 = 2/9 - The probability that a TV is Brand-B, \(P(B)\), is 1500/4500 = 1/3 - The probability that a TV is Brand-C, \(P(C)\), is 2000/4500 = 4/9

For this step, we need to find the probability a TV is of Brand-A given that it is still working after 30,000 hours, denoted as \(P(\mathrm{Brand{-}A}| \mathrm{Working})\). We have: - \(P(\mathrm{Working}| \mathrm{Brand{-} A}) = 0.90\) - \(P(\mathrm{Brand{-}A}) = 2/9\) We also need to find the total probability that a TV is working after 30,000 hours, \(P(\mathrm{Working})\). Using the law of total probability: \(P(\mathrm{Working}) = P(\mathrm{Working}| \mathrm{Brand{-}A}) * P(\mathrm{Brand{-}A}) + P(\mathrm{Working}| \mathrm{Brand{-}B}) * P(\mathrm{Brand{-}B}) + P(\mathrm{Working}| \mathrm{Brand{-}C}) * P(\mathrm{Brand{-}C})\) Plugging in the given values and the probabilities calculated in step 2: \(P(\mathrm{Working}) = (0.90 * \frac{2}{9}) + (0.85 * \frac{1}{3}) + (0.80 * \frac{4}{9})\) \(P(\mathrm{Working}) = \frac{72}{45}\) Now, apply Bayes' theorem: \(P(\mathrm{Brand{-}A}|\mathrm{Working}) = \frac{P(\mathrm{Working}| \mathrm{Brand{-}A}) * P(\mathrm{Brand{-}A})}{P(\mathrm{Working})}\) \(P(\mathrm{Brand{-}A}|\mathrm{Working}) = \frac{(0.90 * \frac{2}{9})}{\frac{72}{45}}\) \(P(\mathrm{Brand{-}A}|\mathrm{Working}) = \frac{1}{3}\) So, the probability that the randomly chosen working TV is a Brand-A TV is 1/3.

For this part, we need to find the probability that the randomly chosen working TV is not a Brand-A TV. We can find this by subtracting the previous result from 1: \(P(\mathrm{Not \ Brand{-}A}|\mathrm{Working}) = 1 - P(\mathrm{Brand{-}A}|\mathrm{Working})\) \(P(\mathrm{Not \ Brand{-}A}|\mathrm{Working}) = 1 - \frac{1}{3}\) \(P(\mathrm{Not \ Brand{-}A}|\mathrm{Working}) = \frac{2}{3}\) The probability that the randomly chosen working TV is not a Brand-A TV is 2/3.