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Randomization in a study of TV commercials. Gonzaga University professors conducted a study of more than 1,500 television commercials and published their results in the Journal of Sociology, Social Work, and Social Welfare (Vol. 2, 2008). Commercials from eight networks—ABC, FAM, FOX, MTV, ESPN, CBS, CNN, and NBC—were sampled for 8 days, with one network randomly selected each day. The table below shows the actual order determined by random draw:

ABC—July 6 (Wed)

FAM—July 7 (Thr)

FOX—July 9 (Sat)

MTV—July 10 (Sun)

ESPN—July 11 (Mon)

CBS—July 12 (Tue)

CNN—July 16 (Sat)

NBC—July 17 (Sun)

a. What is the probability that ESPN was selected on Monday, July 11?

b. Consider the four networks chosen for the weekends (Saturday and Sunday). How many ways could the researchers select four networks from the eight for the weekend analysis of commercials? (Assume that the assignment order for the four weekend days was immaterial to the analysis.)

c. Knowing that the networks were selected at random, what is the probability that ESPN was one of the four networks selected for the weekend analysis of commercials?

Short Answer

Expert verified
  1. 0.125
  2. 70
  3. 2

Step by step solution

01

Step-by-Step SolutionStep 1: The probabilities that ESPN was chosen on Monday, July 11

The mathematical tool of likelihood is employed to look at unpredictability. It's involved with the chance (probability) of an occurrence occurring. The likelihood formula is as follows:

P(E)=NumberoffavourableOutcomesTotalNumberofOutcomes

Here, we have

Totalnumberofdates=8

P(ESPN)=18=0.125

Hence, the probability that ESPN will be chosen is 0.125.

02

Identify the researchers who choose four out of eight for the weekend commercial analysis.

No.ofways=C48=8!4!×4!=8×7×6×5×4×3×2×14×3×2×1×4×3×2×1=40,320576=70

Hence, the probability is 70.

03

The probability that ESPN was one of the four networks chosen for the weekend commercial analysis

ESPN is one of the four networks; therefore, we must select 3 of the remaining 7.

Probability=C37C48=7!3!×3!8!4!×4!=7×6×5×4×3×2×13×2×1×3×2×18×7×6×5×4×3×2×14×3×2×1×4×3×2×1=5,0403640,320576=14070

Hence, the probability is2.

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