Question

1. A research on popular smartphones consists of 80 studies that were fully sponsored by the corporate industry, and 50 studies that were conducted with no corporate ties. Of those that were fully sponsored by the corporate industry, 15% of the participants found that the product unfavourable, 22% were neutral, and 63% found the product favourable. Of those that had no industry funding, 35% found the products unfavourable, 16% were neutral, and 49% found the product favourable. 1.1 What is the probability that a participant selected a randomly found product favourable? [15%] 1.2 If a randomly selected participant found the product favourable, what is the probability that the study was sponsored by the corporate industry? [15%] 1.3 If a randomly selected participant found the product unfavourable, what is the probability that the study had no industry funding? [20%]

Answer 1

0 sol. Given Given Data A Research 80 studies on popular smart phones consists of 50 Studies that were conducted with no Corporate no corporate les Industry ondo 5% of participants found that the product Unfavourable 22% were neutral, 66% Found that product favourable. No Industry Funding : 35% Found the products un favourable 16% were neutral; 49% found the product fancurable. une we have to find the probability that a pastice Ans! -pont Selected a randomly found product Favourable. P(FIS) - P(FOS) PCS) piens) P(FIS) pcs) - 0.63% 30 = 0.236 P(FOS) - P(FIS) * P13) = 0.49 x 50/30 0.106 We get from their combination PIP)= P(FOS) + P(POS) = 0.2 36 + 0.306 IPLE)= 0.542 We have to find the probability that the study was sponsored by the Food industry. PCSif) - PCF ns)/PCF) = 0.234 0.542 P(SP) = 0.4352 He get the mandomly selected participant Found the product favourable

1 Gle have to find the probability that the study had no industry Funding. Pluns) - P(VIS) x pcs) = 0.15 x 30 50.0562 plung) = P(1/3)* P(5) = 0.35% = 0.218 we get from the combination, then PLU) - Pluns) + PCUNS) : 0.0562 +0.218 Pe 0.2745 ) :p(510) a plsnu) PLU) 0.218 0.2749 = 0.7928 selected 1 P(510) randomly unfovorable ) participant a We get the product Found