251, Math October 16, 2003, Quiz 4 Names: ANSWERS . 1. The following represents the outcomes of a flu vaccine study. Got the Flu No Flu Shot Given Flu Shot Column Total 223 446 669 Did not get Flu 777 1554 2331 Row Total 1000 2000 3000
Let F represent the event the person caught the flu, let V represent the even the person was vaccinated, let H represent the event the person remained healthy (didn't catch the flu), and let N represent the even that the person was not vaccinated. (a) Compute: P(F), P(V), P(H), P(N), P(F given V) and P(V given F), P(V and F), P(V or F). Ans: P(F) = .223, P(V) = 2/3, P(H)= .777, P(N) = 1/3, P(F given V) = .223, P(V given F) = 446/669=2/3 P(V and F) = 446/3000 .149 or, by the multiplication rule we get the same answer: P(V and F) = P(F) P(V given F) = (.223) (2/3) .149 P(V or F) = (223+446+1554)/3000 = .741, or by the addition rule: P(V or F) = P(V) + P(F) P(V and F) .667+.223 - .149 = .741 (b) Are the events V and F mutually exclusive? Are the events and V F independent? Explain your answers. Ans: V and F are not mutually exclusive because they can both occur together. Another way of saying this is they are not mutually exclusive because P(V and F) > 0. (As a contrast, F and H are mutually exclusive because they cannot occur together.) V and F are independent because P(V given F) = P(V), and P(F given V) = P(F).
2. (Chapter Review Problem #8, p. 199) Class records at Rockwood College indicate that a student selected at random has a probability 0.77 of passing French 101. For the student who passes French 101, the probability is 0...