Chapter_2__selected

# Chapter_2__selected - PROBABILITY 2-12 a P(exploring the...

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CHAPTER 2 PROBABILITY 2-12. a) P(exploring the products) = 1790 / 2385 = 0.7505 b) P(purchase) = 387 / 2385 = 0.1623 c) P(purchase | explored the products) = 387 / 1790 = 0.2162 d) (a) approximately 75 percent of the visitors go beyond the homepage. 2-14. Based on murder statistics for a given time period: total murders per 100,000 population. 2-16. More likely to occur than not to occur. 2-18. P(first shopper detected) + P(second detected) - P(both detected) = 0.98 + 0.94 - 0.93 = 0.99 2-20. P(F) + P(>50) – P(F & >50) = 12/20 + 2/20 – 2/20 = 0.6 P(< 30) = 2/20 = 0.1 2-22. ) ( ) ( ) ( ) ( B S P B P S P B S P - + = = 0.85 + 0.33 - 0.28 = 0.90 2-24. a. 1268 / 2074, found by: 597 / 2074 + 962 / 2074 – 291 / 2074 = 1268 / 2074 b. 230 / 2074 c. 419 / 2074, found by: 230 / 2074 + 189 / 2074 = 419 / 2074 d. 306 / 1112 e. 189 / 427 f. 773 / 2074, found by: 310 / 2074 + 291 / 2074 + 172 / 2074 g. 773 / 1647, found by: (310 + 291 + 172) / (648 + 597 + 402) = 773 / 1647 2-26. 30%: The three foreign banks make up approximately 10% of the market and SBC Warburg has approximately a 3% share. 3 / 10 = .30 2-28. P(M | R) = 0.32 Given that R) P(M = 0.80 and P(R) = 0.4, P(M | R) = P(M / R) P(R) = (0.80)(0.40) = 0.32 2-30. 2.5% of the packages are late. Given that P(N | D) = 0.25 and P(D) = 0.10, D) P(N = P(N | D) P(D) = (.25)(.10) = 0.025 2-32. 61.1% successfully completed. Given that P(A | H) = 0.94 and P(H) = 0.65, H) P(A = P(A | H) P(H) = (.94)(.65) = 0.611 2-34. Let E,S denote the events: top Executive made over \$1M, Shareholders made money, respectively. Then: a. P(E) = 3 / 10 = 0.30 b. ) S P( = 3 / 10 = 0.30 c. ) S | P(E = ) S )/P( S P(I = (2 / 10) / (3 / 10) = 2/3 = 0.667 d. P(S | E) = E)/P(E) P(S = (1/10)/ (3/10) = 1/3 = 0.333 2-36. For any single issue, the probability that it is foreign underwritten is 10% and the probability that it is not foreign underwritten is 90%. The probability that none of the six issues is foreign underwritten would be (.90) 6 = .5314. The probability that at least one is foreign underwritten is:

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