Chapter 6 Exercise 1 2 3 4 5 & 6 Notes

Exercise 1: Suppose you have three bags each containing 100 marbles: Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles. Bag 3 has 45 red and 55 blue marbles. You choose one of the bags at random and then pick a marble from the chosen bag, also at random.   What is the probability that the chosen marble is red?   Suppose you observe that the chosen marble is red. What is the probability that bag 1 was chosen? Event R: marble is red Event Bi: which bag I choose [P(R|B_{1}) \cdot \frac{75}{100} = 0.75 = 75%] [P(R|B_{2}) \cdot \frac{60}{100} = 0.6 = 60%] [P(R|B_{3}) \cdot \frac{45}{100} = 0.45 = 45%] [P(R|B_{i})] Event R: given that we choose bag i, what is the probability that the marble is red? [P(R) = P(R|B_{1})P(B_{1}) + P(R|B_{2})P(B_{2}) + P(R|B_{3})P(B_{3}) = 0.75 \cdot \frac{1}{3} + 0.6 \cdot \frac{1}{3} + 0.45 \cdot \frac{1}{3}] [= \frac{75}{100} \cdot \frac{1}{3} + \frac{60}{100} \cdot \frac{1}{3} + \frac{45}{100} \cdot \frac{1}{3}] [= \frac{1}{4} + \frac{2}{10} + \frac{3}{20} = \frac{5+4+3}{20} = \frac{12}{20} = \frac{3}{5} = 0.6] [P(B_{1}|R) = \frac{P(R|B_{1})P(B_{1})}{P(R)} = \frac{0.75 \cdot \frac{1}{3}}{0.6} = \frac{\frac{75}{100} \cdot \frac{1}{3}}{\frac{6}{10}} = \frac{75}{100} \cdot \frac{1}{3} \cdot \frac{10}{6} = \frac{75}{360} = 0.21] Exercise 2. An airline commissioned a local television the advertising of a new service offered to customers. For each customer it is then detected whether s/he has seen the advertisement and whether s/he bought the new service. Let us denote with B the event "the customer saw the advertisement" and with A the event "the customer purchased the new service". Known that: P(A) = 0.6 P(B|A) = 0.7 P(B|A') = 0.2 Can you affirm that the advertisement has been efficient? Saying that the advertisement has been efficient is equivalent to affirm that it is more likely that a customer will purchase the new service if she has seen the advertisement. Then we need to evaluate if P(A|B) > P(A|B'). P(A|B) = P(A ∩ B) / P(B) = (P(A) * P(B|A)) / P(B) = (0.6 * 0.7) / (0.6 * 0.7 + 0.4 * 0.2) = 0.84 We also need to compute P(A|B') to assess if the advertisement has been efficient: P(A|B') = P(A ∩ B') / P(B') = (P(A) * P(B'|A)) / (1 - P(B)) = (0.6 * 0.3) / (1 - 0.5) = 0.36 Since P(A|B) > P(A|B'), we can conclude that the advertisement was successful. Exercise 3. In evaluating a sales training program, a company found that out of 50 salespersons making a bonus last year, 20 had attended a special sales training program. The company employs 200 people, 40% of whom have attended the special training program. Find the probability that: a. A salesperson attended the special training program, given that he makes a bonus. b. A salesperson makes a bonus and has attended the special training program. c. A salesperson makes a bonus given that the salesperson attended the special training program. a. P(S|B) = 20/50 = 0.4 b. P(B ∩ S) = P(S) * P(B|S) = 0.4 * 0.25 = 0.1 c. P(B|S) = P(S) * P(B|S) / P(S) = 0.4 * 0.25 / 0.4 = 0.25 Note that P(B|S) = P(B), so there is no difference in the results if the special training program is attended. Exercise 4. A patient goes to see a doctor. The doctor performs a test with 99% reliability—that is, 99% of people who are sick test positive and 99% of the healthy people test negative. The doctor knows that only 1% of the people in the country are sick. If the patient tests positive, what are the chances the patient is sick? Suppose that the population consists of 10000 people. A: a person is sick B: the test is positive P(A|B) = P(A ∩ B) / P(B) = (P(A) * P(B|A)) / P(B) P(A) = 0.01 = 1% P(B|A) = 0.99 = 99% P(B) = 9999 / 10000 P(A|B) = (0.01 * 0.99) / (9999 / 10000) ≈ 0.0099 Exercise 5. Three factories produce light bulbs to supply the market. Factory A produces 20%, 50% of the tools are produced in factory B and 30% in factory C.   2% of the bulbs produced in factory A, 1% of the bulbs produced in factory B and 3% of the bulbs produced in factory C are defective. A bulb is selected at random in the market and found to be defective. What is the probability that this bulb was produced by factory B?   P(A) = 20% P(B) = 50% P(C) = 30% P(D) = the probability that a defective bulb is selected P(D|A) = 2% P(D|B) = 1% P(D|C) = 3% To compute the conditional probability that the defective bulb was produced by factory B (P(B|D)) we use the Bayes' theorem: P(B|D) = (P(D|B) * P(B)) / P(D) = (0.01 * 0.5) / 0.018 = 5/18 ≈ 0.28 Exercise 6. A consumer group conducts a nationwide survey of charity shops to ascertain the quality of the items on sale. They find, amongst other things, that: Only 90% of the CDs on sale play perfectly. 1% of the books on sale are missing a page. 65% of the books on sale have visible imperfections. (You may assume that the qualities of items in this shop are independent of each other.) William and his friend Kate go into a charity shop on their local High Street. a. A book is considered acceptable if it has no missing pages and no visible imperfections. Kate buys 1 book. Find the probability that the item is acceptable. b. William buys five CDs. Find the probability that exactly 4 of them will play perfectly. a. A = the book has missing pages B = the book has visible imperfections The probability that the item is acceptable is given by the probability that it has no missing pages and no visible imperfections. Thus, we can compute this probability by using the complement rule and the additional rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 65/100 + 1/100 - (65/100 * 1/100) = 0.6535 Employing the complement rule, we can conclude that the probability that the book bought by Kate is acceptable is 1 - 0.6535 = 0.3465. b. Let A₁, A₂, A₃, A₄ be the events that the first four CDs bought by William play well and A₅ the event that the last CD doesn't play well. P(A₁ ∩ A₂ ∩ A₃ ∩ A₄ ∩ A₅') = P(A₁) * P(A₂) * P(A₃) * P(A₄) * P(A₅') = 90/100 * 90/100 * 90/100 * 90/100 * 10/100 = 0.06561 Since each of the 5 CDs has the same probability of not playing perfectly, we retrieve the probability that exactly 4 of them will play perfectly by multiplying by 5 the previous result. Thus the answer is 5 * 0.06561 = 0.3281