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FALL 2015 MATH 141 - ASSIGNMENT (7.4-7.6) DUE: 11/11/15 Name All steps must be written clearly and neatly to get full credit. If you use your calculator for anything beyond an arithmetic calculation, please indicate how at the appropriate step. 1. (5pts) An unbiased coin is tossed five times. Find the probability that we get tails exactly twice. Solution The number of outcomes in the whole sample space is 25 = 32. The number of outcomes in the event that we get tails exactly twice is C(5, 2) = 10. So the probability that we get tails exactly twice is 10 . 32 2. (5pts) Electronic baseball games manufactured by Tempco Electronics are shipped in lots of 24. Before shipping, a quality-control inspector randomly selects a sample of 8 from each lot for testing. If the sample contains any defective games, the entire lot is rejected. What is the probability that a lot containing exactly 2 defective games will be shipped? Solution The number of outcomes in the whole sample space is C(24, 8) = 735, 471. The number of outcomes in the event that a lot containing exactly 2 defective games will be shipped is C(22, 8) = 319, 770. This is because the lot will be shipped if none of the 8 games in the sample is defective, so we can think of it as picking all 8 games from the 22 non-defective ones in the lot. So the probability is 319,770 . 735,741 3. (10pts) Four balls are selected at random without replacement from an urn containing three white balls, five blue balls, and four red balls. Find the probability that at least two of the balls selected are blue. Solution P (≥ 2 blue) = 1 − P (0 blue) − P (1 blue) C(7, 4) C(5, 1)C(7, 3) − C(12, 4) C(12, 4) 285 57 = = 495 99 =1− 1 2 ASSIGNMENT (7.4-7.6) 4. (15pts) A fair six-sided die is rolled twice. Let E be the event that the number landing uppermost from the first roll is a 3, and let F be the event that the absolute difference between the numbers from the two rolls (i.e. larger minus smaller) is 2. (i) Construct an appropriate tree diagram and fill in the probabilities on each branch. (See above.) (ii) Find P (F ). Solution We look for all the branches ending with F , multiply along each branch, then add. 8 P (F ) = 16 26 + 16 16 + 16 16 + 16 62 + 16 16 + 16 61 = 36 = 92 (iii) Determine whether E and F are independent events. Solution 1 P (E) = 16 and P (F ) = 29 so P (E)P (F ) = 27 . But P (E ∩ F ) = P (E ∩ F ) 6= P (E)P (F ), the events E and F are not independent. 12 66 = 1 . 18 Since 5. (5pts) A study of the faculty at U.S. medical schools in 2006 revealed that 32% of the faculty were women and 68% were men. Of the female faculty, 31% were full/associate professors, 47% were assistant professors, and 22% were instructors. Of the male faculty, 51% were full/associate professors, 37% were assistant professors, and 12% were instructors. If a faculty member at a U.S. medical school selected at random in 2006 held the rank of full/associate professor, what is the probability that the faculty member was female? Solution (Tree diagram omitted) ∩F A) (.32)(.31) P (W |F A) = P (W = (.32)(.31)+(.68)(.51) = .22 (2 dec. places) P (F A)