3.2 Intoduction to probability 3.3 Probability rules. Sections 3.2 and 3.3. Elementary Statistics for the Biological and Life Sciences (Stat 205)

Transcription

1 3.2 Intoduction to probability Sections 3.2 and 3.3 Elementary Statistics for the Biological and Life Sciences (Stat 205) 1 / 47

2 Probability 3.2 Intoduction to probability The probability of an event E occurring is the long-run proportion of times it will occur in repeated experiments. Denoted Pr{E}. 0 Pr{E} 1. Pr{E} = 1 means E always occurs; Pr{E} = 0 means E never occurs; Example E = tails on fair coin toss, Pr{E} = 0.5. Half the tosses will be tails. 2 / 47

3 3.2 Intoduction to probability Probability and sampling a population Consider a population with proportion p of a characteristic. Randomly choose one member from the population. Let E = randomly chosen member has characteristic. Then Pr{E} = p. Example Large population of Drosophila melanogaster (fruit fly) kept in lab. Proportion that are black is p = 0.3; proportion gray is 1 p = 0.7. Say E = randomly sampled fruit fly is black. Pr{E} = / 47

4 3.2 Intoduction to probability Rolling a fair 6-sided die Say I roll one fair 6-sided die. E = roll a 7.5. Pr{E}? E = roll a number between zero and ten. Pr{E}? E = roll a 6, i.e. E = {6}. Pr{E}? E = roll a 1 or a 6, i.e. E = {1, 6}. Pr{E}? E = roll an even number, i.e. E = {2, 4, 6}. Pr{E}? If I roll the die 100,000 times and find the sample proportion of times I rolled an even number, what would this sample proportion be close to? 4 / 47

5 3.2 Intoduction to probability Frequency interpretation in more detail What do I (and the book) mean by Pr{E} is the long-run proportion of times E occurs in repeated experiments? We will show shortly that the probability of sampling two flies of the same color is = Let s look at what happens when we try repeating the experiment randomly sample two flies over and over and over and over again... After each sample we will update the sample proportion of times two flies of the same color are sampled. 5 / 47

6 Cumulatively estimating ˆp

7 Plotting ˆp versus number of experiments

8 What is happening? 3.2 Intoduction to probability As more and more information (data!) are collected, we can estimate the probability p = 0.58 almost perfectly. In some textbooks, probability is defined as a limit # times E occurs out of n experiments Pr{E} = lim = lim n n ˆp. n This is long-run proportion. The previous two slides allow n to get really large, but not infinite. We see that as n gets large, ˆp Pr{E} ( gets arbitrarily close to ). We can replace by = only at n =. 8 / 47

9 3.2 Intoduction to probability Section 1 Rule (1) 0 Pr{E} 1 for any event E. 2 Rule (2) If E 1, E 2,..., E k are all possible experimental outcomes (smallest events possible), then k Pr{E i } = Pr{E 1 } + Pr{E 2 } + + Pr{E k } = 1. i=1 3 Rule (3) The probability that an event does not happen, E C is Pr{E C } = 1 Pr{E}. 9 / 47

10 3.2 Intoduction to probability Probability of any event Let all experimental outcomes be listed as the smallest events E 1, E 2, E 3,..., E k. We can make new events from these, e.g. A = {E 2, E 4 }. The probability of any event is the sum of the probabilities of the experimental outcomes in the event Pr{A} = Pr{E i }. in A Computing probabilities involves a lot of counting and summing. E i 10 / 47

11 3.2 Intoduction to probability Example Blood type The smallest events possible are the individual experimental outcomes O, A, B, and AB. The proportions in the U.S. are Pr{O} = 0.44, Pr{A} = 0.42, Pr{B} = 0.10, and Pr{AB} = All of these are between 0 and 1. Pr{O} + Pr{A} + Pr{B} + Pr{AB} = 1. The probability that a randomly selected individual does not have type AB is Pr{AB C } = 1 Pr{AB} = = The probability of either A or AB is Pr{A, AB} = Pr{A} + Pr{AB} = = / 47

12 3.2 Intoduction to probability Union & intersection We can make a new event by taking the union or intersection of two events E 1 & E 2. The union of two events combines the experimental outcomes in both, written E 1 or E 2 (either can happen). The union happens if either E 1 or E 2 or both occur. The intersection of two events are the experimental outcomes that are only in both at the same time, written E 1 and E 2 (both must happen). The intersection of two events is the event that both occur. 12 / 47

13 Rolling a 6-sided die 3.2 Intoduction to probability The experimental outcomes (smallest events possible) are 1, 2, 3, 4, 5, and 6. Each has probability 1 6. Let E 1 = {1, 3, 5} be the event that an odd number is rolled. Let E 2 = {1, 2, 3} be the event that the roll is 3 or less. Union: E 1 or E 2 = Intersection: E 1 and E 2 = {1, 3}. Pr{E 1 or E 2 } = 4 6. Pr{E 1 and E 2 } = / 47

14 Disjoint events 3.2 Intoduction to probability If two events have no experimental outcomes in common, they are said to be disjoint. For the 6-sided die, E 1 = {1, 3, 5} and E 2 = {2, 4, 6} are disjoint. Question if E 1 and E 2 are disjoint, what is their intersection? 14 / 47

15 Disjoint events 3.2 Intoduction to probability 15 / 47

16 3.2 Intoduction to probability Union and intersection 16 / 47

17 Addition rules 3.2 Intoduction to probability Rule (4) If E 1 and E 2 are disjoint then Pr{E 1 or E 2 } = Pr{E 1 } + Pr{E 2 }. Rule (5) For any two events (disjoint or not) Pr{E 1 or E 2 } = Pr{E 1 } + Pr{E 2 } Pr{E 1 and E 2 }. 17 / 47

18 Hair and eye color 3.2 Intoduction to probability Relationship between hair and eye color for 1770 German men. 18 / 47

19 3.2 Intoduction to probability Hair color Brown Black Red Total Eye color Brown Blue Total 1770 Pr{Black hair} = = = / 47

20 3.2 Intoduction to probability Hair color Brown Black Red Total Eye color Brown Blue Total 1770 Pr{Blue eyes} = = = / 47

21 3.2 Intoduction to probability Hair color Brown Black Red Total Eye color Brown Blue Total 1770 Pr{Black hair and blue eyes} = = / 47

22 3.2 Intoduction to probability Hair color Brown Black Red Total Eye color Brown Blue Total 1770 Pr{Black hair or blue eyes} = = = / 47

23 Using Rule Intoduction to probability Rule (5) gives Pr{Black hair or blue eyes} = Pr{Black hair} + Pr{blue eyes} Pr{Black hair and blue eyes} = = = / 47

24 Conditional probability Have Pets Do not have pets Total Male Female Total What is the probability a randomly selected person is male, given that they own a pet?

25 Conditional probability Have Pets Do not have pets Total Male Female Total Pr( male given that they own a pet) = Pr( male own a pet) Male and own a pet = own a pet = % =

26 3.2 Intoduction to probability Conditional probability Sometimes we have information before we compute a probability. For example we might know an individual from the table has blue eyes. Knowing this fact reduces the population to only those who have blue eyes. This can change the probability of having black hair. The conditional probability of E 2 given E 1 is defined to be Pr{E 2 E 1 } = Pr{E 2 and E 1 }. Pr{E 1 } provded that Pr{E 1 } > 0. In terms of computing conditional probabilities from a table, it s easiest just to ignore the part of the table that E 1 didn t happen! 26 / 47

27 Bayes Theorem 3.2 Intoduction to probability Named after the Reverend Thomas Bayes. Published after his death. The conditional probability of E 2 given E 1 is Pr{E 2 E 1 } = Pr{E 1 E 2 }Pr{E 2 }. Pr{E 1 } It is the cornerstone of the statistical methods known as Bayesian Statistics. 27 / 47

28 3.2 Intoduction to probability Eye color Hair color Brown Black Red Total Brown Blue Total Pr{Black hair blue eyes} = = / 47

29 3.2 Intoduction to probability Hair color Brown Black Red Total Eye color Brown 20 Blue 50 Total 70 Pr{Black eyes red hair} = = / 47

30 3.2 Intoduction to probability Hair color Brown Black Red Total Eye color Brown Blue Total Pr{Not red hair brown eyes} = = = / 47

31 3.2 Intoduction to probability In biological sciences The hair/eyes example is not that interesting (to us at least). More relavent examples might be Pr{heart attack this year cholesterol > 190} Pr{cavities don t floss} Pr{low stress exercise} Pr{pertussis vaccinated} 31 / 47

32 Review: Probability Rules 1 Rule (1) 0 Pr{E} 1 for any event E. 2 Rule (2) If E 1, E 2,..., E k are all possible experimental outcomes (smallest events possible), then k Pr{E i } = Pr{E 1 } + Pr{E 2 } + + Pr{E k } = 1. i=1 3 Rule (3) The probability that an event does not happen, E C is Pr{E C } = 1 Pr{E}.

33 Review: Union and Intersection Let E 1 = {1, 3, 5} be the event that an odd number is rolled. Let E 2 = {1, 2, 3} be the event that the roll is 3 or less. Union: {E 1 or E 2 }= {1, 2, 3, 5}. Intersection: {E 1 and E 2 } = {1, 3}.

34 Review: Union and Intersection

35 Review: Probability Rules Rule (4) If E 1 and E 2 are disjoint then Pr{E 1 or E 2 } = Pr{E 1 } + Pr{E 2 }. Rule (5) For any two events (disjoint or not) Pr{E 1 or E 2 } = Pr{E 1 } + Pr{E 2 } Pr{E 1 and E 2 }.

36 Review: Conditional Probability The conditional probability of E 2 given E 1 is defined to be provded that Pr{E 1 } > 0. Pr{E 2 E 1 } = Pr{E 2 and E 1 }. Pr{E 1 }

37 Rule (6): Multiplicative Rule Independent events: One event DOES NOT affect the likelihood the next event will occur. Rule (6): If two events E 1 and E 2 are independent, then Pr{E 1 and E 2 } = Pr{E 1 } Pr{E 2 }. Example 1: flip a coin and draw a card from a random deck Pr{head and A} = Example 2: draw two cards with replacement from a random deck Pr{ A and A} =

38 Dependent Events 3.2 Intoduction to probability Dependent events: One event DOES affect the likelihood the next event will occur. Pr{E 1 and E 2 } =Pr{E 1 } Pr{E 2 }. Example 3: draw two cards without replacement from a random deck Pr{ A and 10} = / 47

39 Independent events 3.2 Intoduction to probability Pr{E 2 E 1 } = Pr{E 2 and E 1 } Pr{E 1 } = Pr{E 1} Pr{E 2 } Pr{E 1 } = Pr{E 2 }. Two events E 1 and E 2 are independent if knowing one has occured does not change the probability of the other, Pr{E 2 E 1 } = Pr{E 2 }. Recall that Pr{Black hair} = 0.28 and Pr{Black hair blue eyes} = Hence black hair and blue eyes are dependent events. 39 / 47

40 Probability trees 3.2 Intoduction to probability The probabilities from probability trees are given by Rule (7): Pr{E 1 and E 2 } = Pr{E 1 }Pr{E 2 E 1 }. This is the definition of conditional probability with both sides multiplied by Pr{E 1 }. 40 / 47

41 Probability trees 3.2 Intoduction to probability If our event E is the result of two (or more) smaller experiments, probability trees provide a convenient way to find the probability of all possible outcomes. After the first experiment, a second experiment is carried out resulting in either C or D. The four possible events are AC, AD, BC, or BD and their probabilities are obtained from a probability tree by multiplying numbers going down the appropriate branch. 41 / 47

42 3.2 Intoduction to probability Example Two fair coin tosses Pr{Heads,Heads} = = 0.25 Pr{Heads,Tails} = = 0.25 Pr{Tails,Heads} = = 0.25 Pr{Tails,Tails} = = 0.25 What is the probability of obtaining heads on both coin flips? 42 / 47

43 3.2 Intoduction to probability Example sampling two flies Pr{Black,Black} = = 0.09 Pr{Black,Gray} = = 0.21 Pr{Gray,Black} = = 0.21 Pr{Gay,Gray} = = 0.49 What is the probability of sampling the same color both times? 43 / 47

44 3.2 Intoduction to probability Example nitric oxide example Pr{Treatment,Positive} = = Pr{Treatment,Negative} = = Pr{Control,Positive} = = Pr{Control,Negative} = = What is the probability of a negative outcome? 44 / 47

45 3.2 Intoduction to probability Example medical testing example Pr{Disease,Test positive} = = Pr{Disease,Test negative} = = Pr{No disease,test positive} = = Pr{No disease,test negative} = = What is the probability of testing positive? 45 / 47

46 3.2 Intoduction to probability Example A more relavent question... Given that my test comes up positive, what is the probability that I have the disease? The definition of conditional probability is Pr{disease test positive} = Pr{disease and test positive} Pr{test positive} = = = / 47

47 Rule of Total Probability Rule (8): For any two events E 1 and E 2, Pr{E 1 } = Pr{E 2 } Pr{E 1 E 2 } + Pr{E2 C } Pr{E 1 E2 C }. Pr{E 1 } = Pr{E 2 } Pr{E 1 E 2 } + Pr{E2 C } Pr{E 1 E2 C } = Pr{E 1 and E 2 } + Pr{E 1 and E2 C }.