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Chapter 3

Chapter 3. Section 3.7 Independence. Independent Events

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Chapter 3

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  1. Chapter 3 Section 3.7 Independence

  2. Independent Events Two events A and B are called independent if the chance of one occurring does not change if the other has occurred. In terms of probability this means the conditional probability is the same as the original (i.e. P(A) = P(A|B)). Example: Experiment: Flip 3 coins S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} F : The event the first coin is a head. A: The event all coins are the same. P(A) =The chance all coins are the same = S A F HTT TTT HHH THH HHT THT HTH TTH These two events are independent the fact that one happens does not change the chance the other will happen. The chance all are the same given or assuming the first is a head. P(A|F) =

  3. Example A man is tested and finds that he has a 70% chance of fathering a girl (G) each time he has a child. If he decides he will father 2 children and each time he has a child his chance of fathering a girl is independent of the previous time, draw a probability tree that shows all the possibilities for having two children along with their probabilities. P(GG) = (.7)(.7) = .49 = 49% G .7 Notice that if you add up all the probabilities at the end you will get 1 if you use decimals and 100% if you use percentages. G .7 .3 P(GB) = (.7)(.3) = .21 = 21% B P(BG) = (.3)(.7) = .21 = 21% G .7 .3 B .3 P(BB) = (.3)(.3) = .09 = 9% B What is the probability both children will be the same sex? 58% (i.e. 49%+9%) 91% (i.e. 49%+21%+21%) What is the probability at least one is a girl?

  4. Example Two baseball teams A and B will play a "best of 3" series. Team A wins over team B 60% of the time and each game they play is independent of the previous games. Draw a probability tree that shows all possible outcomes of the series. P(AA) = (.6)(.6) = .36 = 36% A .6 P(ABA) = (.6)(.4)(.6) = .144 = 14.4% A A .6 .6 .4 B P(ABB) = (.6)(.4)(.4) = .096 = 9.6% B .4 P(BAA) = (.4)(.6)(.6) = .144 = 14.4% A .6 A .6 B P(BAB) = (.4)(.6)(.4) = .096 = 9.6% .4 .4 B .4 P(BB) = (.4)(.4) = .16 = 16% B What is the probability A wins the series? P(AA) + P(ABA) + P(BAA) = 36% + 14.4% + 14.4% = 64.8% What is the probability the series is a sweep? (i.e. only goes 2 games) P(AA) + P(BB) = 36% + 16% = 52%

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