Pr(M I) Pr(I) Pr(M I) Pr(I) + Pr(M G) Pr(G)

The Prosecutor s Fallacy: A large factory has 000 male employees. An assault on a female worker takes place. DNA profiling of a suspect gives rise to a rare trait, X, such that Pr(X occurs in general population)
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The Prosecutor s Fallacy: A large factory has 000 male employees. An assault on a female worker takes place. DNA profiling of a suspect gives rise to a rare trait, X, such that Pr(X occurs in general population) 0000 which is equivalent to Pr(Match Innocent) 0000 This is of no interest!! We want Pr(I M) Pr(M I) Pr(I) Pr(M I) Pr(I) + Pr(M G) Pr(G) Thus this evidence alone does not prove guilt beyond reasonable doubt. Note that the more employees there are, the less evidence of guilt is provided by the DNA profiling on its own. Independence We have seen that knowledge of one event can change the probability of another event. If this is so, the two events are said to be dependent. Definition: Two events A and B are said to be independent if: Pr(A B) Pr(A) i.e. they are independent if knowing that B has occurred tells us nothing about A. We saw earlier that for any events A and B: For independent events, this simplifies to: Pr(A AND B) Pr(A B) Pr(B) Pr(A AND B) Pr(A) Pr(B) To test if two events are independent, one can check if Pr(A B) Pr(A). Alternatively, if data are given as a frequency table, calculate: (Frequency of A) (Frequency of B) Overall Total If this is the same as the frequency of A AND B, the two events are independent. Example: One hundred people were asked about their political views with the following results: Con Lab SNP LD M F An interviewee is chosen at random. In other words, each of the 00 people is equally likely to be chosen. Note that: Pr(Con) Pr(Male) Pr(Con AND Male) Pr(Con Male) Pr(Con AND Male) Pr(Male) Thus Pr(Con) Pr(Con Male) 0.2 and so the events Con and Male are independent. An alternative calculation is: (Freq. of Con) (Freq. of Male) Overall Total This is the same as the number of Con and Male individuals, so the events are independent. However, Lab and Male are not independent. Pr(Lab) 0.4 Pr(Lab AND Male) Pr(Lab Male) Pr(Male) (23/00) (50/00) This is not to be confused with: Pr(Male AND Lab) Pr(Male Lab) Pr(Lab) (23/00) (40/00) Example Venn diagrams for independence: Independent E F Not Independent E F Not Independent E F Example: Using an unfair coin You have are given an unfair coin which comes up heads with probability 0.7 and tails with probability 0.3. Can you use this coin to producea fair result (i.e. the probability of heads being 0.5)? Yes, using the following algorithm:. Flip the coin twice and record the results in order, for example HT for a head followed by a tail or TH for tail followed by head. 2. If the coin comes up the same on both flips (HH or TT), start again. 3. If the coin is different both times, declare the result as the outcome of the second flip, e.g. HT gives the result tail and TH gives the result heads. So Pr(The result is heads) Pr(Second flip comes up heads both flips are different) Pr(Second flip comes up heads and both flips are different) Pr(both flips are different) Pr(TH) Pr(HT) + Pr(TH) Note: we used independence of coin flips when calculation Pr(TH) and Pr(HT). Example: Roads There are two roads from A to B and two roads from B to C. Each of the four roads is blocked by snow with probability 0.2, independently of the others. What is the probability of an open route from A to C? If there is no open routes from A to C what is the probability there is an open route A to B? For there to be an open route A to C there has to be an open route A to B and an open route B to C. We label the roads a,b for those A to B and c, d for thos B to C. Pr(Open route A to B) Pr(Road a open OR Road b open) Pr(Road a open) + Pr(Road b open) Pr(Road a open AND Road b open) Since all the roads have the same probability of being blocked we have So Pr(Open route A to B) Pr(Open route C to D). Pr(Open route A to C) Pr(Open route A to B AND Open route C to D) Pr(Open route A to B) Pr(Open route C to D) Recap: Suppose E, F and C are events.. 0 Pr(E) 2. If outcomes are equally likely, then: Favourable outcomes Pr(E) Total outcomes 3. Pr(E) Pr( not E) 4. Pr(E or F) Pr(E) + Pr(F) Pr(E and F) Pr(E and C) 5. Pr(E C) Pr(C) Pr(E and F) Pr(E) Pr(F E). If Pr(E C) Pr(E) then, E and C are independent events. 7. If Pr(E and F) Pr(E) Pr(F) then, E and F are independent events. Random Variables (Rees.) Probability theory can be used to build models of the way in which data can arise. Example: In a large population of oysters, 0% of the oysters contain a pearl. A random sample of oysters is opened until the first pearl is found. Let R be the oyster in which the first pearl is found. R can take the values, 2, 3,.... Pr(R is ) p() 0. Pr(R is 2) p(2) [failure then success] Pr(R is 3) p(3) [2 failures then success] Pr(R is 4) p(4) [3 failures then success] and in general Pr(R is r) p(r) 0.9 r 0. This is an example of a discrete random variable. The above has assumed that the population is large enough that the probability of a success is not affected by the number of preceding failures. Definitions: A random variable is a variable that takes random numerical values. The random variable is discrete if the values that it takes can be listed. Notation: Capital letters will be used for random variables: X,Y,... Lower case letters will be used for the possible values of random variables: x,y,... The probability that X takes the value x is written Pr(X is x) p(x) and is called the probability function of X. Two properties of the probability function are 0 p(x) ; p(x). Here are some examples of random variables: S Number of Seeds that germinate Y Yield in a barley trial R Number of days with Rain in February Q Score on an IQ test M Manganese concentration in water Definitions: The mean of a discrete random variable is: and the standard deviation is: µ x p(x) mu σ x2 p(x) µ 2 sigma Note: These definitions can be extended to the case of random variables that are not discrete. Interpretation: Suppose that we take a random sample from a population which has mean µ and standard deviation σ. We can calculate the sample mean x and the sample standard deviation s. Then, unless n is small, we will find that: x µ s In practice, µ and σ are often unknown, and so we estimate them by x and s, which are calculated from the data. Example: The number of bank customers queueing to speak to a teller follows the following distribution: Number Probability The mean of this discrete random variable is: σ µ For the standard deviation we need: x 2 p(x) So σ Notice that the calculation of these statistics is similar to the calculations used for sample data in a frequency table. Example: Let X be the face value on the single roll of a die. What are the mean and standard deviation of X? Experiment: roll the die 20 times: and use the data: Face (x) Freq. (f) n f 20; xf ; x 2 f 274 and so x s xf f ( f ) 9 ( ) ( x2 f ( ) xf) 2 f From the probability function X Pr(X x) µ x p(x) x; σ x2 p(x) µ ( ) s. Permutations How many ways are there to arrange n different objects in order? Example: 3 letters A, B, C possibilities: ABC, ACB, BAC, BCA, CAB, CBA There are 3 choices for the first position, 2 for the next and for the last. So there are: 3! 3 2 ways (called: 3 factorial) Definition: n factorial is: n! n (n )... 2 for all n 0! More generally: How many ways are there to arrange r objects in order, if they are chosen from n different objects? There are n choices for first position, (n ) for the next position, down to (n r + ) for the last. n! Thus there are (n r)! ways. Example: Can arrange 3 items chosen from 5 in 0 different ways. Definition: The Binomial Coefficient is ( ) n n! r r! (n r)! and is the number of ways of choosing r items out of n. The New Cambridge Statistical Tables tabulate values of this for n 30. Note: Some books use a different notation for the binomial coefficient. For example: C n r or C(n,r). One way to think of the Binomial coefficient is that it is the number of ways of permuting the original n objects divided by the number of ways of permuting both the r chosen objects and the (n r) objects that were not chosen. Example: The number of ways of choosing two items out of five is: ( ) 5 5! 2 2! 3! (2 ) (3 2 ) 0 To see why the permutations and combinations might be useful, consider the following: Example: A fair die is tossed 5 times. Let Y be the number of sixes. What is the probability that Y 2? First, consider one suitable outcome: Pr({,, not, not, not }) ( ) 2 ( 5 ) 3 There are other outcomes for which Y 2. Each such outcome has the same probability of occurring. We can just list the combinations there are 0 of them (do it!). So: ( ) 2 Pr(Y is 2) 0 0. ( ) 3 5 In practice, there is no need to list combinations; we use the binomial coefficient instead. Suppose there are n independent trials Binomial Distribution (Rees.2.7) Each trial has only two outcomes (designated success and failure ) P(Success) is the same value p for each trial. The random variable of interest X is number of successes from the n trials. The probability of getting r successes is: Pr(X r) ( ) n p r ( p) n r r We write: X B (n, p) This is called a Binomial (n, p) distribution. Note: The symbol means has the distribution. Although it is usual to talk about The Binomial Distribution, it is really a family of distributions. The values of n and p determine which member of the family is being considered. They are said to be the parameters of the distribution. Example: The probability that a patient recovers from a certain disease is 0.3. Twelve patients are selected. What is the probability that: (i) exactly 4? (ii) between 2 and 4? (iii) 2 or more patients recover? We assume that the patients are independent and equate Success with Recover. We have a Binomial situation with n 2 and p 0.3. If X is the number of people that recover then Substituting r 4 in the formula gives: X B(2, 0.3) (i) Pr(X 4) ( ) Note that we can use cancelling to calculate ( ) 2 4 : ( ) 2 2! 4 4! 8! and so (ii) We can use Pr(X 4) Pr(2 X 4) p(2) + p(3) + p(4) and work out each term separately. Exercise: Check that: p(2) 0.8 p(3) p(4) 0.23 and so ( ) 0.39 (iii) To find Pr( 2) directly involves a lot of work as there are probabilities to calculate. It is much easier to consider the complementary event! Pr(X 2) Pr(X 2) p(0) p() A simple alternative is to use the Lindley & Scott statistical tables (NCST). Table (pp 4 23) gives the Binomial distribution for n up to 20 and for p a multiple of 0.0. The probabilities in these tables are all cumulative probabilities. So, using Table, with n 2,r 4,p 0.3: Pr(X 4) All of the questions above can be solved by looking up one or two values in the tables. ( ) (i) Pr(X 4) Pr(X 4) Pr(X 3) (ii) Pr(2 X 4) Pr(X 4) Pr(X ) (iii) Pr(X 2) Pr(X ) See also Rees.5; Table C. of Rees (4th ed.) is a short version of NCST Table. Example: Suppose we change the last example just a little and make the probability of recovering from the disease 0.. Twelve patients are selected. What is the probability that (i) exactly 4 recover? (ii) between 4 and 9 patients recover? (iii) less than 5 recover? Now the appropriate model is: X B(2, 0.) The first question can be done directly from the formula, but the other two parts really require the use of tables. We will see how to use the tables for all three parts. The NCST Table only gives values of p 0.5. We must use a little trick before we can apply Table to this problem. We let X be the number who recover and Y be the number who do not recover. Pr(Recover) 0. Pr(Not cured) 0.4. Saying that 4 patients recover out of 2 implies that 8 patients do not recover. Hence where Pr(X 4) Pr(Y 8) Y B(2, 0.4). The example asked 3 questions about X. The three equivalent questions about Y are: (i) exactly 8 do not recover? (ii) between 3 and 8 patients do not recover? (iii) more than 7 do not recover? Hint: When the labels of Success and Failure are switched, as here, we always need to rephrase the questions. When the question involves a range of values, consider the equivalent questions on the boundaries. (i) Pr(Y 8) Pr(Y 8) Pr(Y 7) (ii) Pr(3 Y 8) Pr(Y 8) Pr(Y 2) (iii) Pr(Y 7) Pr(Y 7) These answers about Y will be the same as the answers to the original questions about X. Some properties of the Binomial distribution it is symmetrical if the probability of success is 0.5 and skewed otherwise. it becomes more skewed as the probability gets closer to zero or one. it becomes less skewed as the sample size gets larger. The plot shows the probability function of Y B(2, 0.4): Note that any value between 2 and 8 is likely to occur and more extreme values are possible! The theoretical mean and standard deviation of a Binomial random variable are (Rees.): For Y B(2, 0.4): µ np µ np σ np( p) σ np( p) Rules of thumb that were given in the descriptive statistics section can be applied to distributions. Thus, approximately 95% of the probability will be within 2 standard deviations of the mean. For this example, the interval is: (.4, 8.2)
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