Probexamp
- 1. More Probability Examples (Probability Reviewed) Example 1 Each of a sample of four home mortgages is classi…ed as …xed rate (F) or variable rate (V ). a. What are the 16 outcomes in S? Solution S = {FFFF, FFFV , FFV F, FV FF, FV FV , FV V F, FV V V , V FFF, V FFV , V FV F, V FV V , V V FF, V V FV , V V V F, V V V V }. b. Which outcomes are in the event that exactly three of the selected mortgages are a …xed rate? Solution A = {FFFV , FFV F, FV FF, V FFF}. c. Which outcomes are in the event that all four mortgages are of the same type? Solution B = {FFFF, V V V V }. d. Which outcomes are in the event that at most one of the four is a variable-rate mortgage? Solution C = {FFFF, FFFV , FFV F, FV FF, V FFF}. e. What are the union and intersection of the two events in parts b and d? Solution A [ C = {FFFV , FFV F, FV FF, V FFF, FFFF, V V V V }; A C = A: f. What are the union and intersection of the two events in parts c and d? Solution B [ C = {FFFV , FFV F, FV FF, V FFF, FFFF, V V V V }; B C = {FFFF}. Example 2 A construction …rm is working on two di¤erent projects. Let A be the event that the …rst one is completed by the contract date and de…ne B analogously for the second project. If P(A [ B) = 0:9 and P(A B) = 0:5, what is the probability that exactly one project is completed by the contract date? Solution The desired probability is: P(exactly one project is completed by the contract date) = P(A or B but not both) = P(A [ B) ¡ P(A B) = 0:9 ¡ 0:5 = 0:4: Example 3 The route used by a certain motorist in commuting to work contains two intersections with tra¢c signals. The probability that he must stop at the …rst signal is 0:4, the analogous probability for the second signal is 0:5, and the probability that he must stop at least once is 0:6. What is the probability that he must stop: 1
- 2. a. At both signals? Solution Let A be the event that he must stop at the …rst signal and B be the event that he must stop at the second signal. Then P(AB) = P(A)+P(B)¡P(A[B) = 0:4+0:5¡0:6 = 0:3. b. At the …rst signal but not at the second signal? Solution P(A B0 ) = P(A) ¡ P(A B) = 0:4 ¡ 0:3 = 0:1: c. At exactly one signal? Solution P(A [ B) ¡ P(A B) = 0:6 ¡ 0:3 = 0:3: Example 4 The probability that a grader will make a marking error on any particular question of a multiple-choice exam is 0:1. Assume there are ten questions and questions are marked independently. a. What is the probability that no errors are made? Solution (0:9)10 = 0:3487: b. What is the probability that at least one error is made? Solution 1 ¡ (0:9)10 = 0:6513: c. If there are n questions and the probability of a marking error is p rather than 0:1, give expressions for the probabilities in a and b. Solution P(no errors are made) = (1 ¡ p)n ; P(at least one error is made) = 1 ¡ (1 ¡ p)n : Example 5 Sixty percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the probability that: a. All of the next three vehicles inspected pass. Solution (0:6)3 = 0:2160: b. At least one of the next three vehicles inspected fails. Solution 1 ¡ (0:6)3 = 0:7840: c. Exactly one of the next three inspected passes. Solution Let A be the event that exactly one of the next three vehicles inspected passes. Let Ei be the event that the ith vehicle passes and the other two fail. Then A = E1 [E2 [E3. Then: P(A) = P(E1) + P(E2) + P(E3) = (0:6)(0:4)(0:4) + (0:4)(0:6)(0:4) + (0:4)(0:4)(0:6) = 0:2880: d. At most one of the next three vehicles inspected passes. 2
- 3. Solution P(exactly one of the next three vehicles passes) + P(all three fail) = 0:2880 + (0:4)3 = 0:3520: Example 6 A box contains the following four slips of paper, each having exactly the same di- mensions: (1) win prize one, (2) win prize two, (3) win prize 3, and (4) win prizes one, two, and three. One slip will be randomly selected. Let A1 = {win prize one}, A2 = {win prize two}, and A3 = {win prize three}. Show that A1 and A2 are independent, that A1 and A3 are independent, and that A2 and A3 are independent (that is, pairwise independence exists). However, show that P(A1 A2 A3) 6= P(A1) £ P(A2) £ P(A3), so that the three events are not mutually independent. Solution We show pairwise independence as follows: P(A1) = P(draw slip 1 or 4) = 1 2 P(A2) = P(draw slip 2 or 4) = 1 2 P(A3) = P(draw slip 3 or 4) = 1 2 P(A1 A2) = P(draw slip 4) = 1 4 P(A2 A3) = P(draw slip 4) = 1 4 P(A1 A3) = P(draw slip 4) = 1 4 Hence: P(A1) £ P(A2) = 1 2 £ 1 2 = 1 4 = P(A1 A2) P(A2) £ P(A3) = 1 2 £ 1 2 = 1 4 = P(A2 A3) P(A1) £ P(A3) = 1 2 £ 1 2 = 1 4 = P(A1 A3) Thus pairwise independence exists. But: P(A1 A2 A3) = P(draw slip 4) = 1 4 P(A1) £ P(A2) £ P(A3) = ³ 1 2 ´3 = 1 8 Hence: P(A1 A2 A3) 6= P(A1) £ P(A2) £ P(A3) Thus the events are not mutually independent. 3