Question

Scientific research on popular beverages consisted of 70 studies that were fully sponsored by the food industry, and 30 studies that were conducted with no corporate ties. Of those that were fully sponsored by the food industry, 10 % of the participants found the products unfavorable, 22 % were neutral, and 68 % found the products favorable. Of those that had no industry funding, 35 % found the products unfavorable, 16 % were neutral, and 49 % found the products favorable
a. What is the probability that a participant selected at random found the products favorable?

b. If a randomly selected participant found the product favorable, what is the probability that the study was sponsored by the food industry?

c. If a randomly selected participant found the product unfavorable, what is the probability that the study had no industry funding?

Answer 1

Answer:

a) 0.623 (62.3%)

b) 0.764 (76.4%)

c) 0.6 (60%)

Step-by-step explanation:

a) defining the event F= the product is favorable , then the probability is

P(F) = probability that the study was sponsored by food industry * probability that the product is favorable given that it was sponsored + probability that the study had no corporate ties * probability that the product is favorable given that it was not sponsored = 0.70 * 0.68 + 0.30 * 0.49 = 0.623

b) for conditional probability we use the theorem of Bayes , then defining the event S= the study was sponsored by the food industry , we have:

P(S/F)=P(S∩F)/P(F)= 0.70 * 0.68/0.623 = 0.764 (76.4%)

where

P(S∩F)=probability that the study was sponsored by food industry and the product was found favorable

P(S/F)=probability that the study was sponsored by food industry given that the product was found favorable

c) for U= the product was found unfavourable , doing the same procedure as in a)

P(U)= 0.70 * 0.10 + 0.30 * 0.35 = 0.175

and the corresponding conditional probability is

P(N/U)=P(N∩U)/P(U)= 0.30 * 0.35 / 0.175 = 0.6 (60%)

where N represents the event = the study had no industry funding