216 MATHEMATICAL ASSOCIATION OF AMERICA Logged in as wins 16 w5 3 HW5: Problem 3

Question

216 MATHEMATICAL ASSOCIATION OF AMERICA Logged in as wins 16 w5 3 HW5: Problem 3 Prev Up Next (1 pt) Scientific research on popular beverages consisted conducted with no corporate ties or those that were any sponsored by the food ndustry, 14 % of the partiapants found the products 24 % were neutral, and 62 % tund the products favorable or those that had no ndustry were neutral, and 47 % found the products tavorable What is the probability that a participant selected at random found the products tavorable? of 60 studies that were fully sponsored by the food industry, and 40 studies that were unfavorable, trding, 35 % found the products unfavorable, 18 % it a randomly selected participant tound the product favorable, what is the probablity that the study was sponsored by the tood industry? if a randomiy selected participant found the product unfavorable, what is the probability that the study had no inoustry tunding? Note: You can eam partial credit on mis You have attempted this problem 0 times You have unimited attempts remaining

Explanation / Answer

a)

Let

S = sponsored
U, N, F = unfavorable, neutral, favorable

As there are 100 samples here,

P(S) = 0.60
P(S') = 0.40

Hence, by Bayes' Rule,

P(F) = P(S) P(F|S) + P(S') P(F|S') = 0.60*0.62 + 0.40*0.47 = 0.56 [ANSWER]

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b)

P(S|F) = P(S) P(F|S)/P(F) = 0.60*0.62/0.56 = 0.664285714 [ANSWER]

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c)

Here,

P(U) = P(S) P(U|S) + P(S') P(U|S') = 0.60*0.14 + 0.40*0.35 = 0.224

Hence,

P(S'|U) = P(S') P(U|S')/P(U) = 0.40*0.35/0.224 = 0.625 [ANSWER]

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