Consider the case of a manufacturer who has an automatic machine that produces an important part. Past records indicate that at the beginning of the data the machine is set up correctly 70 percent of the time. Past experience also shows that if the machine is set up correctly it will produce good parts 90 percent of the time. If it is set up incorrectly, it will produce good parts 40 percent of the time. Since the machine will produce 60 percent bad parts, the manufacturer is considering using a testing procedure. If the machine is set up and produces a good part, what is the revised probability that it is set up correctly? [U]Determine our events:[/U] [LIST] [*]C = Correctly Set Machine = 0.7 [*]C|G = Correctly Set Machine And Good Part = 0.9 [*]I = Incorrectly Set Machine = 1 - 0.7 = 0.3 [*]I|G = Incorrectly Set Machine And Good Part = 0.4 [*]B< = BAD PARTS = 0.60 [/LIST] P[correctly set & part ok] = P(C) * P(C|G) P[correctly set & part ok] = 70% * 90% = 63% P[correctly set & part ok] = P(I) * P(I|G) P[incorrectly set & part ok] = 30% *40% = 12% P[correctly set | part ok] = P[correctly set & part ok]/(P[correctly set & part ok] + P[incorrectly set & part ok]) P[correctly set | part ok] = 63/(63+12) = [B]0.84 or 84%[/B]

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When five people are playing a game called hearts, each person is dealt ten cards and the two remaining cards are put face down on a table. Because of the rules of the game, it is very important to know the probability of either of the two cards being a heart. What is the probability that at least one card is a heart? Probability that first card is not a heart is 3/4 since 4 suits in the deck, hearts are 1/4 of the deck. Since we don't replace cards, the probability of the next card drawn without a heart is (13*3 - 1)/51 = 38/51 Probability of both cards not being hearts is found by multiplying both individual probabilities: 3/4 * 38/51 = 114/204 Having at least one heart is found by subtracting this from 1 which is 204/204: 204/204 - 114/204 = 90/204 [URL='https://www.mathcelebrity.com/search.php?q=90%2F204&x=0&y=0']This reduces to[/URL] [B]15/34[/B]