9211 WordsApr 16, 201137 Pages

4: Probability and Probability Distributions
4.1 a This experiment involves tossing a single die and observing the outcome. The sample space for this experiment consists of the following simple events: E1: Observe a 1 E4: Observe a 4 E2: Observe a 2 E5: Observe a 5 E3: Observe a 3 E6: Observe a 6
b Events A through F are compound events and are composed in the following manner: A: (E2) D: (E 2) B: (E 2, E 4, E 6) E: (E 2, E 4, E6) C: (E 3, E 4, E 5, E 6) F: contains no simple events
c Since the simple events Ei, i = 1, 2, 3, …, 6 are equally likely, [pic]. d To find the probability of an event, we sum the probabilities assigned to the*…show more content…*

d-e Sum the probabilities of the appropriate simple events: [pic]
4.12 a Similar to Exercise 4.7. List the six jurors as M1, M2, M3, M4, F1 and F2. The simple events can be found using a tree diagram and are shown below: M1M2 M1M3 M1M4 M1F1 M1F2 M2M3 M2M4 M2F1 M2F2 M3M4 M3F1 M3F2 M4F1 M4F2 F1 F2 b Sum the probabilities of the appropriate simple events: [pic]
4.13 a Experiment: A taster tastes and ranks three varieties of tea A, B, and C, according to preference. b Simple events in S are in triplet form. [pic] Here the most desirable is in the first position, the next most desirable is in the second position, and the least desirable is in third position. c Define the events D: variety A is ranked first F: variety A is ranked third Then [pic]
The probability that A is least desirable is [pic]
4.14 a Visualize a tree diagram with four stages – selecting the runner who places first, second, third and fourth, respectively. There are four choices at the first stage, three choices (branches) at the second stage, two choices (branches) at the third stage, and only one choice (branch) for the last runner. The total number of simple events is [pic]. The simple events are listed below: JBED JEBD JEDB JBDE JDEB JDBE BJED BEJD BEDJ BJDE BDEJ BDJE EBJD EJBD EJDB EBDJ EDJB EDBJ DBEJ

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