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AMS 311
Lecture 5
February 8, 2000
Coming next class: Review of Chapter one homework problems.
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Theorem 2.1 (Counting Principle)
If the set E contains n elements and the set F contains m elements, there are nm ways in
which we can choose, first, an element of E and then an element of F.
Theorem 2.2 (Generalized Counting Principle)
Let E1, E2, , Ek be sets with n1, n2, , nk elements, respectively. Then there n1 n2
 nk ways in which we can, first, choose an element of E1, then an element of E2, ,
and finally an element of Ek .
Theorem 2.3. A set with n elements has 2n subsets.
Theorem 2.4.
The number of distinguishable permutations of n objects of k types, where n1 are alike,
n2 are alike, , nk are alike, and n=n1+n2++nk is
n!
.
n1 ! n2 !  nk !
Theorem 2.5 (Binomial Expansion) For any integer n>0,
( x  y) n 
  x
n
i0
n
i
ni
yi .
Theorem 2.7. (Stirling’s Formula)
n! 
2 nn n e  n ,
where the sign  means
lim
n 
n!
 1.
2 nn n e  n
Chapter Three
Conditional Probability and Independence
Estimates of conditional probabilities are fundamental statistics in the health sciences.
For example, age-specific mortality rates are an example of estimates of conditional
probabilities.
Definition: If P(B)>0, the conditional probability of A given B, denoted by P(A|B), is
P( AB)
P( A | B) 
.
P( B)
Example 3.2. From the set of all families with two children, a family is selected at
random and found to have a girl. What is the probability that the other child of the family
is a girl? Assume that in a two-child family all sex distributions are equally probable.
Answer : 1/3.
Example 3.3. From the set of all families with two children, a child is selected at random
and is found to be a girl. What is the probability that the second child of this girl’s family
is also a girl? Assume that in a two-child family all sex distributions are equally probably.
Answer: ½.
Theorem 3.1.
Let S be the sample space of an experiment, and let B be an event of S with P(B)>0. Then
a) P(A|B)0, for any event A of S.
b) P(S|B)=1.
c) If A1, A2,  is a sequence of mutually exclusive events, then


i 1
i 1
P Ai | B)   P( Ai | B).
Proof?
Reducing the sample space to get a conditional probability.
Example 3.8. Let’s make a deal!
On a TV game show, there are three curtains. Behind two of the curtains there is nothing,
but behind the third there is a prize that the player might win. The probability that the
prize is behind a given curtain is 1/3. The game begins with the contestant randomly
guessing a curtain. The host of the show (master of ceremonies), who knows begind
which curtain the prize is, will then pull back a curtain other than the one chosen by the
player, and reveal the prize is not behind that curtain. The host will ot pull back the
curtain selected by the player, nor will he pull back the one with the prize, if different
from the player’s choice. At this point, the host gives the player the opportunity to change
his choice of curtain and select the other one. The question is whether the player should
change his choice. That is, has the probability of the prize being behind the curtain
chosen changed from 1/3 to ½ or is it still 1/3? If it is still 1/3, the contestant should
definitely change his choice. Otherwise, there is no point in doing so.