Chapter 3. Probability

Chapter 3: Probability 7115. Probability and Guessing10 true/false questions, thus P(correct by chance per item)= 0.5a. P(first 7 out of 10 correct, next three incorrect)=P(C) ∗ P(C) ∗ P(C) ∗ P(C) ∗ P(C) ∗ P(C) ∗ P(C) ∗ P(C )∗ P(C )∗ P(C )=0.5 ∗ 0.5 ∗ 0.5 ∗ 0.5 ∗ 0.5 ∗ 0.5 ∗ 0.5 ∗ 0.5 ∗ 0.5 ∗ 0.5= 0.5 10 = 0.000977b. No, since there would be more than just this pattern of getting 7 out of 10 correct and these other patternswould have to be included in the determination of the overall probability of getting 7 correct out of 1016. Poll Confidence LevelP(all five out of five polls are 95% accurate)= 0.95 ∗ 0.95 ∗ 0.95 ∗ 0.95 ∗ 0.95=0.95 5 = 0.774The results of the five polls would be independent of each other. We could expect that if the same confidencelevel is used in all polls, 95% of them would provide accurate results and 5% would provide inaccurate results.17. Testing Effectiveness of Gender-Selection MethodP(all ten babies are girls)= 0.5 10 =0.000977Yes, since the probability of getting all ten babies as girls is 0.000977 (or about 1 time out of 1000) if chancealone is operating, the actual outcome of getting all ten girls would indicate the gender selection method iseffective since chance clearly would not be a viable reason for this outcome.18. Flat Tire ExcuseP(all correct in selecting the same tire as being flat)= 0.25 4 = 0.00391 (about 4 times out of 1000)19. Quality ControlWould expect 2% of the monitors to have a defect.P(none are defective out of 15 sampled monitors)= 0.98 15 = 0.739No, since this result could be attributed to chance or random variation; would happen almost three out of everyfour tests of 15 sampled monitors.20. Redundancya. P(alarm clock will not work)= 1 – P(alarm clock will work)= 1 – 0.975= 0.025b. P(that two alarm clocks would fail)= P(one will fail) ∗ P(other will fail)=0.025 ∗ 0.025= 0.000625c. P(being awakened from two alarm clocks)= P(alarm 1 works or alarm 2 works)=1 – P(both would fail)= 1 – 0.000625= 0.99937521. Stocking FishIdentify all possible events: MMM, MMF, MFM, MFF, FFF, FMM, FFM, FMFOf these combinations, 6/8 have both Male and Female, P= 0.7503-5 Multiplication Rule: Beyond the BasicsDescribing Complements. In exercises 1 – 4, provide a written description of the complement of the given event.1. Blood TestingIf it is not true that at least one of the 10 students has Group A blood, then none of them has Group A blood.The complement of an event where least one student out of 10 has group A blood is 0 or none.2. Quality ControlIf it is not true that all of the units are free of defect, then at least one of them is defective. When 50electrograph units that are free of defects are shipped, the probability of not having a defect is 50 out of 50 or aprobability of 1.000, thus the probability of shipping 50 electrograph units that are all defective is 1 minus theprobability of being free of defects or 1 minus 1.000 or 0.000.