Lecture 14
Nancy Pfenning Stats 1000
Chapter 7: Probability
Last time we established some basic definitions and rules of probability:
Rule 1:
P
(
A
C
) = 1

P
(
A
).
Rule 2:
In general, the probability of one event or another occurring is
P
(
A
or
B
) =
P
(
A
) +
P
(
B
)

P
(
A
and
B
)
If the events are mutually exclusive, then
P
(
A
and
B
) = 0 and so
P
(
A
or
B
) =
P
(
A
) +
P
(
B
).
Rule 3:
In general, the probability of one event and another is
P
(
A
and
B
) =
P
(
A
)
P
(
B

A
)
which we can reexpress as
P
(
B

A
) =
P
(
A
and
B
)
P
(
A
)
If the events are independent,
P
(
B

A
) =
P
(
B
) and so
P
(
A
and
B
) =
P
(
A
)
P
(
B
).
This time we utilize these rules to solve some more complicated problems, and discuss why at times
probabilities can be counterintuitive.
Tree diagrams
are often helpful in understanding conditional probability problems.
Example
(Game Show) Suppose a prize is hidden behind one of three doors A, B, or C. after the contestant
picks one of the three, the host reveals one of the remaining two doors, showing no prize. He gives
the guest a chance to switch to the door he or she did not select originally. Is the probability of
winning the prize higher with the “keep” or “switch” strategy? Use a probability tree to calculate
the probability of winning using each strategy.
When events occur in stages, we normally are interested in the probability of the second, given that
the first has occurred. At times, however, we may want to know the probability of an earlier event having
occurred, given that a later event ultimately occurred.
Example
Suppose that the proportion of people infected with AIDS in a large population is .01. If AIDS is
present, a certain medical test is positive with probability .997 (called the
sensitivity
of the test),
and negative with probability .003. If AIDS is not present, the test is positive with probability
.015, negative with probability .985 (called the
specificity
fo the test).
Does a positive test
mean a person almost certainly has the disease? Let’s figure out the following: if a person tests
positive, what is the probability of having AIDS?
58
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A tree diagram is very helpful for this type of problem. We’ll let
A
and not
A
denote the events
of having AIDS or not,
T
and not
T
denote the events of testing positive or not.
First we’ll find the overall probability of testing positive. Either a person has AIDS and tests
positive
or
he/she does not have AIDS and tests positive:
P
(
T
) =
P
(
A
and
T
) +
P
(not
A
and
T
)
According to the multiplication rule, it follows that
P
(
T
) =
P
(
A
)
P
(
T

A
) +
P
(not
A
)
P
(
T

not
A
)
=
.
01(
.
997) +
.
99(
.
015) =
.
00997 +
.
01485 =
.
02482
Now, we can apply the definition of conditional probability to find the probability we seek (prob
ability of having AIDS, given that a person tested positive):
P
(
A

T
) =
P
(
A
and
T
)
P
(
T
)
=
.
00997
.
02482
=
.
40
Thus, even if a person tests positive, he/she is more likely
not
to have the disease!
.
01
.
99
A
not
A
.
997
.
003
.
985
.
015
T
not
T
T
not
T
P
(
A
and
T
) =
.
01(
.
997) =
.
00997
P
(not
A
and
T
) =
.
99(
.
015) =
.
01485
Most students—and even most physicians—would have expected the probability to be much higher.
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 Fall '06
 taeyoungpark
 Normal Distribution, Probability, Probability theory, r.v.

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