Repeated Trials and Chi

Repeated Trials and Chi - BICD 100 Genetics Winter 2008...

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BICD 100 Genetics Winter 2008 1 Probability for repeated trials and the Chi-squared test 1. The binomial distribution (Pierce page 57-58) --outcome of a trial is binary choice (heads or tails in coin flipping, rolling a six or not a six in a die) --you know the probability of one outcome p (the other outcome = 1-p = q) --trials are independent and p is the same for each trial. The binomial expansion (p + q) k gives you the probabilities of each possible outcome of k trials, e.g. for the family of four children there are five possible outcomes (4 girls, 3 girls 1 boy, 2 girls 2 boys, 1 girl 3 boys, and 4 boys). The probability of each outcome is given by (p+q) 4 . P(4 boys) = p 4 and so on. To find the probability of a specific outcome (e.g. 2 boys 2 girls) without having to expand the entire series, calculate the probability of an outcome and multiply but the number of permutations that will give the outcome. For k children with n boys and (k-n) girls, the probability of a single permutation is p n q (k-n) . How many permutations will give a specific outcome? The total number of permutations for k trials = k! (k factorial, 1 x 2 x 3 x . ..k) Total number of permutations for boys = n! and for girls = (k-n)! Divide total # permutations by number of relevant permutations: k!/[n! (k-n)!] So for 2 boys 2 girls there are 4!/[2! x 2!] = 24/4 = 6 possible permutations. There are six ways fo getting 2 boys and 2 girls. Now multiply probability of a single permutation x number of permutations:
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Repeated Trials and Chi - BICD 100 Genetics Winter 2008...

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