Angio_talk

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(Stat49N - April 7, 2004)
MULTIPLICATION LAW. Let A and B be events and
assume P(B) ? 0. Then P(A ? B) P(A B)
P(B) The multiplication law is often useful in
finding the probabilities of intersections, as
the following examples illustrate.
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Example A. An urn contains three red balls and
one blue ball. Two balls are selected without
replacement. What is the probability that they
are both red? Let R1 and R2 denote the events
that a red ball is drawn on the first trial and
on the second trial, respectively. From the
multiplication law, P(R1 ? R2) P(R1) P(R2
R1) P(R1) is clearly ¾ , and if a red ball has
been removed on the first trial, there are two
red balls and one blue ball left. Therefore P(R1
? R2) 2/3. Thus P(R1 ? R2) ½ .
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LAW OF TOTAL PROBABILITY. Let B1, B2, Bn be
such that U Bi ? and Bi ? Bj ? for i ?
j, with P(Bi) gt 0 for all i. Then, for any
event A, P(A) ? P(A Bi) P(Bi)
n i1
n i1
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EXAMPLE C. Referring to Example A, what is the
probability that a red ball is selected on the
second draw? The answer may or may not be
intuitively obvious that depends on your
intuition. On the one hand, you could argue that
it is clear from symmetry that P(R2) P(R1)
¾. On the other hand, you could say that it is
obvious that a red ball is likely to be selected
on the first draw, leaving fewer red balls for
the second draw, so that P(R2) lt P(R1). The
answer can be derived easily by using the law of
total probability.
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P(R2) P(R2 R1) P(R1) P(R2 B1) P(B1)
2/3 x 3/4 1 x 1/4 3/4 where B1 denotes the
event that a blue ball is drawn on the first
trial. BAYES RULE. Let A and B1, , Bn be
events where the Bi are disjoint, U Bi ? ,
and P(Bi) gt 0 for all i. Then P(Bj A)
-----------------------------------
n i1
P(A Bj) P(Bj) ? P(A Bi) P(Bi)
n i1
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INDEPENDENCE. Intuitively, we would say that two
events, A and B, are independent if knowing that
one had occurred gives us no information about
whether the other had or will occur that is, P(A
B) P(A) and P(B A) P(B). Now, if P(A)
P(A B) --------------- then P(A ? B)
P(A) P(B) We will use this last relation as the
definition of independence. Note that it is
symmetric in A, and B and does not require the
existence of a conditional probability that is,
P(B) can be 0. DEFINITION. A and B are said to
be independent events if P(A ? B) P(A) P(B).
P(A ? B) P(B)
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EXAMPLE a. A card is selected randomly from a
deck. Let A denote the event that it is an ace
and D the event that it is a diamond. Knowing
that the card is an ace gives no information
about its suit. Checking formally that the
events are independent, we have P(A) 4/52
1/13 and P(D) 1/4. Also, A ? D is the event
that the card is the ace of diamonds and P(A ? D)
1/52. Since P(A) P(D) (1/4) x (1/13) 1/52,
the events are in fact independent.
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EXAMPLE c. A fair coin is tossed twice. Let A
denote the event of heads on the first toss, B
the event of heads on the second toss, and C the
event that exactly one head is thrown. A and B
are clearly independent, and P(A) P(B) P(C)
.5. To see that A and C are independent, we
observe that P(C A) .5. But P(A ? B ? C)
0 ? P(A) P(B) P(C) To encompass situations such
as that in Example c, we define a collection of
events, A1, A2, An, to be mutually independent
if for any subcollection, Ai , , Ai , P(Ai
? ? Ai ) P(Ai ) P(Ai )
m
1
1
m
m
1
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• Discrete Random Variables.
• A random variable is essentially a random number.
  We will be interested in random numbers that are
  determined by experiments. As motivation for a
  definition, let us consider an example. A coin
  is thrown three times, and the sequence of heads
  and tails is observed thus,
• ? hhh, hht, htt, hth, ttt, tth, thh, tht
• Examples of random variables associated with ?
  are
• the total number of heads,
• the total number of tails, and
• the number of heads minus the number of tails.
• Each of these is a real-valued function defined
  on ? that is,
• each is a rule that assigns a real number to
  every point ? ? ?. Since the outcome in ? is
  random, the corresponding number is random as
  well.

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In general, a random variable is a function from
? to the real numbers. Since the outcome of the
experiment for which ? is the sample space is
random, the number produced by the function is
random as well. It is conventional to denote
random variables by italic uppercase letters from
the end of the alphabet. For example, we might
define X to be the total number of heads in the
experiment described above. A discrete random
variable is a random variable that can take on
only a finite or at most a countably infinite
number of values. The random variable X just
defined is a discrete random variable since it
can take on only the values 0, 1, 2, and 3. For
an example of a random variable that can take on
a countably infinite number of values, consider
an experiment that consists of tossing a coin
until a head turns up and defining Y to be the
total number of tosses. The possible values of Y
are 0, 1, 2, 3, . In general, a countably
infinite set is one that can be put into
one-to-one correspondence with the integers.
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If the coin is fair, then each of the outcomes in
? above has probability 1/8, from which the
probabilities that X takes on the values 0, 1, 2,
and 3 can be easily computed P(X 0)
1/8 P(X 1) 3/8 P(X 2) 3/8 P(X 3)
1/8
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Generally, the probability measure on the sample
space determines the probabilities of the various
values of X if those values are denoted by x1,
x2, , then there is a function p such that p(xi)
P(X xi) and ?i p(xi) 1. This function is
called the probability mass function, or the
frequency function, of the random variable X.
Figure below shows a graph of p(x) for the coin
tossing experiment. The frequency function
describes completely the probability properties
of the random variable.
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In addition to the frequency function, it is
sometimes useful to use the cumulative
distribution function (cdf) of a random variable,
which is defined to be F(x) P(X ? x), -? lt x
lt ? Cumulative distribution functions are
usually denoted by uppercase letters and
frequency functions by lowercase letters. Figure
below is a graph of the cumulative distribution
function of the random variable X of the
preceding paragraph. Note that the cdf jumps
where p(x) gt 0 and that the jump at xi is p(xi).
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It is useful to define here the concept of
independence of random variables. In the case of
two discrete random variables X and Y, taking on
possible values x1, x2, and y1, y2, , X and Y
are said to be independent if, for all i and
j, P (X xi and Y yj) P(X xi) P(Y
yj) The definition is extended to collections of
more than two discrete random variables in the
obvious way for example, X, Y, and Z are said to
be mutually independent if, for all i, j, and
k, P (X xi ,Y yj , Z zk) P(X xi) P(Y
yj) P(Z zk) We next discuss some common
discrete distributions that arise in
applications.
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Bernoulli Random Variables. A Bernoulli random
variable takes on only two values 0 and 1, with
probabilities 1 p and p, respectively. Its
frequency function is thus p(1) p p(0) 1
p p(x) 0, if x ? 0 or x ? 1 An alternative
and sometimes useful representation of this
function is p(x) ?
px (1 p) 1-x , if x 0 or x 1 0,
otherwise
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If A is an event, then the indicator random
variable, IA, takes on the value 1 if A occurs
and the value 0 if A does not occur IA (?)
IA is a Bernoulli random variable. In
applications, Bernoulli random variables often
occur as indicators. A Bernoulli random variable
might take on the value 1 or 0 according to
whether a guess was a success or a failure.
1, if ? ? A 0, other wise
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The Binomial Distribution. Suppose that n
independent experiments, or trials, are
performed, where n is a fixed number, and that
each experiment results in a success with
probability p and a failure with probability 1
p. The total number of successes, X, is a
binomial random variable with parameters n and p.
For example, a coin is tossed 10 times and the
total number of heads is counted (head is
identified with success). The probability that
X k, or p(k), can be found in the following
way. Any particular sequence of k successes
occurs with probability pk (1 p)n-k, from the
multiplication principle. The total number of
such sequences is ( ) , since there are ( )
ways to assign k successes to n trials.
Thus, p(k) ( ) pk (1 p)n-k
n k
n k
n k
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Two binomial frequency functions are shown in
Figure 2-3. Note how the shape varies as a
function of p.
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EXAMPLE. Tay-Sachs disease is a rare but fatal
disease of genetic origin occurring chiefly in
infants and children, especially those of Jewish,
eastern European extraction. If a couple are
both carriers of Tay-Sachs disease, a child of
theirs has probability .25 of being born with the
disease. If such a couple has four children,
what is the frequency function for the number of
children that will have the disease?
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We assume that the four outcomes are independent
of each other, so, if X denotes the number of
children with the disease, p(k) ( ) .25k
.754-k, k 0, 1, 2, 3, 4 These probabilities
are given in the following table k p(k) ------
------------ 0 .316 1 .422 2 .211 3 .047 4 .0
04
4 k