BICD 100 Genetics Winter 2008
1
Probability for repeated trials and the Chisquared test
1.
The binomial distribution (Pierce page 5758)
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 = 1p = 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 (kn) girls, the probability of a single permutation is p
n
q
(kn)
.
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 = (kn)!
Divide total # permutations by number of relevant permutations: k!/[n! (kn)!]
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.
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 Spring '08
 Nehring
 Genetics, Probability, Probability theory, Binomial distribution, Poisson

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