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probability
 P(E) = m/n
 m= #desired outcomes
 n= total # of outcomes

Independence
 if the occurrence of one event does not effect the probability of the other
 ie: probability of first having a boy, then having a second child a boy (1/2)

Addition rule
 the "or" rule
 P(A or B) = P(A) + P(B)
 ie: probability of rolling a 5 or 6 on die (1/6 + 1/6) = 2/6 = 1/3

Multiplication rule
 the "and" rule
 for independent events..the probability of BOTH events occurring
 ie: probability of A_bb from a AaBb x AaBb dihybrid cross
 P(A_) * P(bb) = 3/4 * 1/4 = 3/16

Binomial expansion
 Used for calculating the probability of MULTIPLE outcomes occurring
 1. Calculate the # of outcomes that satisfy the question
 2. Probability of each permutation
 P(outcome) = (n!/a!*b!)p^a * q^b
 n = # of trials
 a = # of p results
 b = # of q results
 p = probability of a
 q = probability of b
 ie: if Aa x Aa have 4 kids, what is the prob of getting 3 A_ and 1 aa?
 P = (4!/3! * 1!) (3/4)^3 * (1/4)^1
 P = 27/64

Binomial Expansion w/MULTIPLE terms
 P = (n!/a!b!c!d!)p^a * q^b * r^c * s^d
 a+b+c+d = n
 q+r+s+t = 1

Outcomes
 formula for how many outcomes are possible (ie: how many combinations of 2 hats you can pick out of a set of 5)
 nCr = n!/(nr)! * r!
 n = total number
 r = # of selections
 outcomes = 5! / (52) * 2! = 10
 ALTERNATE is Pascal's triangle:
 go to the 5th row, and count from left to right starting from 0, 1, 2 (end at 2 since you are picking 2 hats)

At Least One
 probability of At least one occurrence
 ie: probability of eldest girl of 3 children having AT LEAST ONE brother
 P = P(1st brother) + P(2nd brother)  P(both being brothers)
 P = 1/2 + 1/2  (1/2*1/2)
 P = 1  1/4 = 3/4= 75%

Chi Squared test
 1. add all X^2 values (each value is a n)
 X^2 = (OE)^2/E
 2. calculate degrees of freedom
 df = (n1)
 3. look at chart and see if is Greater than 0.05
 4. If is greater than 0.05, then FAIL TO REJECT hypothesis (Ho)
 5. If is .05 or less, then REJECT Ho and ACCEPT Ha (alternate hypothesis)

