2     Probability Laws

2.1     Axioms of Probability

Given a sample space S, we will assign probability values to events (subsets) which obey the following axioms:
1. P(A) >= 0 for every event A
2. P(S) = 1
3. If A1, A2, … are mutually exclusive events,

4. then
P(A1  A2  …)  =  P(A1) + (A2) + …        (addition rule)
Note:   When we assign probabilities to all of the subsets of a sample space, we create what is called a probability measure on the space; behaves very similarly to area.

Can think of as Venn diagram, where total area is 1 (and so areas of subsets are <= 1); then probability of subset is just its area.

ex:

of M & M's manufactured, 20% are red, 10% orange, 10% green, 10% blue, 20% yellow and 30% brown.  Thus if our experiment consists of selecting one M & M at random and considering its color, our sample space could be
S = {R, Or, G, Bl, Y, Br}
The logical assignment of probabilities to the single-element subsets of the sample space is
P({R})  = .20
P({Or}) = .10
P({G})  = .10
P({Bl}) = .10
P({Y})  = .20
P({Br}) = .30
(we would usually write these as just  P(R) = .20,  without using the set brackets, when we're dealing with single elements of the sample space, even though strictly speaking probability values are associated with subsets)

We would then get probabilities for other subsets by using the addition rule!
So the probability of getting a red or a green M&M would be
P({R,G})  =  P({R}) + P({G})  =  .20 + .10  =  .30

Note:   when a sample space is discrete, we usually assign probabilities to individual elements, then find probabilities of other subsets from these, as in the above example

Note:  if the outcomes in our sample space are equally likely, and there are N possible outcomes, then the probability of each outcome is 1/N and thus the probability of any event A is

P(A) =  n/N,  where
n = number of outcomes for which A occurs
N = total number of outcomes
this is just the classical approach to assigning probabilities!

Theorem 2.1.2           P(A') = 1 - P(A)
proof:

Since AA' = S, P(AA') = P(S) = 1
Since AA' = , we can use the addition rule to get  P(AA') = P(A) + P(A')
Thus   P(A) + P(A') = 1    or    P(A')  =  1 - P(A)
Theorem 2.1.1           P() = 0
proof:
= S',   so  P()  =  P(S')  =  1 - P(S)
Note:  sometimes much easier to find P(A') than P(A)!

From a Venn diagram, we can get an idea of how to find P(A1A2) when A1 & A2  aren't disjoint:

P(A1) + P(A2)   counts the overlap twice, so subtract   P(A1A2)
This suggests the General Addition Rule:
P(A1A2 )  =  P(A1) + P(A2) - P(A1A2)
This can be used when events A1 and A2 aren't mutually exclusive (disjoint), in which case the previous addition rule wouldn't apply.

ex:

Suppose that the probability that a child will have blue eyes is .25, the probability that a child will have blonde hair is .30, and the probability that a child will have both is .13.

What's the probability that a child selected at random will have either blue eyes or blond hair (or both)?
Let  E  = the event a child has blue eyes,  H = the event a child has blond hair. We want P(EH).
The general addition rule gives us
P(EH) = P(E) + P(H) - P(EH)
=  .25 + .30 - .13
=  .42

ex:
On a particular football team, the probability that a player plays offense is .60 and the probability that a player plays defense is .65. Everybody plays one or the other. What's the probability a player selected at random plays both?
Let  A  = the event a player plays offense,  B  = the event a player plays defense
We want the probability of  AB.
We can solve the general addition rule for P(AB) to get
P(AB) = P(A) + P(B) - P(AB)
= .60 + .65 - 1
= .25

Note:  Sometimes easiest to use a Venn diagram to compute probabilities; just find the probability associated with each separate region.

ex:

From the example above, what's the probability a child will have blonde hair but not blue eyes?
With E and H as above,  E'  = the event a child doesn't have blue eyes; then we want  P(E'H).
The Venn diagram for the problem is shown, with all of the pertinent regions labelled:
The shaded region is what we want; it's everything that's in H but not in E. From the diagram, we can see that
P(E'H)  =  P(H) - P(EH)
=    .30  -  .13
=   .27

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