Download WORKING WITH NAMED PROBABILITY MODELS

WORKING WITH NAMED PROBABILITY MODELS
Hoffman
Purpose of Lesson: Find real-world events and check assumptions associated with known named
probability models to see if they can be used to predict outcomes.
Example: You and your mother and your sister all have horrible tooth aches and have dental appointments
and read on the Georgia Dental Board’s website that patients who present themselves to dentists and have
your condition have abbesses that require extraction 20% of the time. What is the expected number of you
who will need to have the tooth removed? What is the probability all of you will need a tooth removed?
Is it binomial? 1. Only two outcomes? Removal or non-removal Independent trials? You, sis and mom’s
teeth conditions influence each other? Not a lot, so, yes, independent. Probability of removal is the same
for each of you? Maybe mom being older has a greater than 20% chance, but let’s go with each of you has
a 20% chance. Question asks about ‘how many’? Yes.
We know (because of theorems from class that we ‘proved’ via simulation with our coins…and that was
the sole purpose of discussing all that abstract junk) the following:
Theorem 1: The expected value E(X) or mean or mu  of a Binomial random variable is given by n*p
where n is the number of trials and p is the probability of a success in each trial.
Example: 3 people = 3 trials = n and p = .20 (i.e. the probability of a success. Note the success is defined
by how the question is phrased). So we expect 3*.2 = .6 of you to get a tooth pulled. Whew, that is less
than one and somewhat comforting.
Theorem 2: The variance Var(X) or variance or sigma-squared of a Binomial random variable is given
by n*p*(1-p).
Example: For us, n = 3 and p = .2 so we have 3*.2*(1-.2) = 3*.2*.8=.48 and therefore our standard
deviation is the square root of .48 which is .69 . If I think that I want a range of possible numbers of people
(me, mom and sister) to know who will need extractions I might use the mean plus or minus the standard
deviation which in this case is .6 minus .69 or about 0 and .6 plus .69 which is better than 1. So it looks
like one of you might need a tooth extracted.
Theorem 3: The probability that a Binomial random variable will ‘come up’ with k successes in n trials is:
where kCn is the number of ways you can choose k things from n given
kCn * (p*p*…p)*(q*q*…*q)
\ k times/ \(n-k) times/
by the formula
n!/{k!*(n-k)!} where, for example, 5 !=5*4*3*2*1=120
Example: So the probability none of you will need to have a tooth removed is:
C3 *(.2*.2*.2)*(.8…multiplied by itself zero times is by definition 1) =
3
3!/{3!*0!} * .008 * 1 = .008 (note that by definition 0! is 1). So, you won’t all be commiserating together.