Pearson Unit 6 Topic 15: Probability 15-6: Conditional Probability Formulas Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007
TEKS Focus: (13)(D) Apply conditional probability in contextual problems. (1)(A) Apply mathematics to problems arising in everyday life, society, and the workplace. (1)(E) Create and use representations to organize, record, and communicate mathematical ideas. (13)(C) Identify whether two events are independent and compute the probability of the two events occurring together with or without replacement.
Example 1: Suppose the probability of rain on Saturday is 40%, what is the probability that you will clean the garage on Saturday? 10 60 50 P(rains on Saturday) = 40% = 0.4 P(rains and you clean garage) = 0.4 0.9 = 0.36 P(does not rain on Saturday) = 100% - 40% = 60% P(does not rain and you clean garage) = 0.6 0.5 = 0.3 P (clean garage on Saturday) = 0.36 + 0.30 = 0.66 = 66%
Example 2: In a study designed to test the effectiveness of a new drug, half of the volunteers receive the drug, while the other half receive a placebo, a tablet or pill containing no medication. The probability of a volunteer receiving the drug and getting well was 45%. What is the probability of getting well, given that, he receives the drug? P(B/A) = P(getting well, taking new drug) P(A) = P(taking new drug) = ½ = 50% = 0.5 P(A and B) = P(taking new drug and getting well) = 45% = 0.45 P(B/A) = 0.45 = 0.9 = 90% 0.5
Example 3: In a survey given at Madison High School last year, 45% of students owned a dog, 27% owned a cat and 12% owned a dog and a cat. What is the conditional probability that a dog owner also owns a cat? What is the conditional probability that a cat owner owns a dog? P(cat/dog) = P(owns cat and dog) = 0.12 ≈ 0.267 ≈ 26.7% P (owns dog) 0.45 P(dog/cat) = P(owns cat and dog) = 0.12 ≈ 0.444 ≈ 44.4% P (owns cat) 0.27
Example 4: If you choose a tile at random from the tiles shown at the right and then replace the first tile you choose before choosing again, what is the probability that you choose a dotted tile and then a dragon tile? P(dotted) = 4 P (dragon) = 3 = 1 15 15 5 P(dotted and dragon) = 4 1 = 4 ≈ .053 ≈ 5.3% 15 5 75
Example 5: If you choose a tile at random from the tiles shown at the right and then replace the first tile you choose before choosing again, what is the probability that you choose a bird tile and then after replacing it a flower tile? Round to the nearest hundredth of a percent. P(bird) = 2 P (flower) = 1 15 15 P(bird and flower) = 2 1 = 2 ≈ .89% 15 15 225
Example 6: If you choose a tile at random from the tiles shown at the right and do NOT replace the first tile you choose before choosing again, what is the probability that you choose a dotted tile and then a dragon tile? Round to the nearest tenth of a percent. P(dotted) = 4 P (dragon) = 3 15 14 P(dotted and dragon) = 4 3 = 12 = 2 ≈ 5.7% 15 14 210 35
Example 7: If you choose a tile at random, what is the probability that you choose a flower tie and then without replacing it you choose a bird tile? P(flower) = 1 P (bird) = 2 = 1 15 14 7 P(flower and bird) = 1 1 = 1 ≈ .95% 15 7 105
Example 8: A college reported that 70% of its freshmen had attended a public school. 60% of those who had attended public schools graduated within 5 years while 80% of other freshman graduated within 5 years. What percent of freshmen graduated within 5 years? P(public & graduate) = P(public) P(graduate/public) = 0.7 0.6 = 0.42 = 42% P(other & graduate) = P(other) P(graduate/other) = 0.3 0.8 = 0.24 = 24% P(graduate) = P(public & graduate) + P(other & graduate) = 0.42 + 0.24 = 0.66 = 66%
Example 9: Madison girl’s soccer team wins 65% of its games on muddy fields and 30% of their games on dry fields. The probability of the field being muddy for their next game is 70%. What is the probability the team will win the next game? P(muddy field & win) = P(win/muddy field) P(muddy field) = 0.65 0.7 = 0.455 P(dry field & win) = P(win/dry field) P(dry field) = 0.3 0.3 = 0.09 P (Win) = P(win/given muddy field) + P (win, given dry field) = 0.455 + 0.09 = 0.545 = 54.5%