Download Summary of Chapter 5 Probability Modelsa

Summary of Chapter 5 Probability Models
Statistics
Section 5.1
A sample space is a complete list or description of possible outcomes of a
chance process. The possible outcomes must be disjoint (mutually exclusive),
meaning that one outcome cannot be another outcome. An event is a subset of a
sample space.
For example, if we roll a six-sided dice, the sample space is the numbers 1, 2, 3,
4, 5, and 6 (complete list of possible outcomes). Each element of the sample
space is disjoint (rolling a 2 is not the same as rolling a 3). An example of an
event for this sample space might be rolling a 2. Another example of an event
for this sample space might be rolling an even number (2, 4, or 6). The event
rolling a 2 and the event rolling an even number are not disjoint (mutually
exclusive) because rolling a 2 is also rolling an even number. However, the
event rolling a 2 and the event rolling an odd number are mutually exclusive
because rolling a 2 is not rolling an odd number.
Probability is a numerical expression of how likely it is for an event to occur.
Numerically, it is equal to
Number of outcomes of event
Number of equally likely outcomes
The probability of an event, A, occurring, P(A), is always 0 ≤ P(A) ≤ 1.
A probability distribution gives a list a complete set of possible events from a
sample space and the probability of each event. Since the list is complete, the
sum of all the probabilities is equal to 1. For example, a two-way table for the
sums of rolling two four-sided die and the associated probability distribution are
shown below:
+ 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
If a sampling is random, the Law of Large Numbers says as the sample size
becomes large, the probability of Event A for the sample approaches what the
probability of Event A is for the population. For example, if a fair 6 sided dice
is tossed 10 times, the proportion of 5’s might be quite a bit higher or lower
than 1/6. However, as the number of times the dice is tossed increases to a very
large value, the proportion of 5’s would approach 1/6.
The total number of possible combinations for a number of separate possibilities
(stages) can be computed by the Fundamental Principle of Counting, which says
that “if there are k stages, with ni possible outcomes for stage i, then the number
of possible outcomes for all k stages taken together is n1n2n3 . . . nk.
For example, say for a meal we have a choice of 2 main dishes, 3 drinks, and 2
desserts. Each type of choice (main dish, drink, dessert) would be a stage. The
total number of possible combinations is 2 x 3 x 2 = 12. These 12 possible
combinations can be shown with a tree diagram:
B1
M1
B2
B3
B1
B2
M2
B3
D1
M1B1D1
D2
M1B1D2
D1
M1B2D1
D2
M1B2D2
D1
M1B3D1
D2
M1B3D2
D1
M2B1D1
D2
M2B1D2
D1
M2B2D1
D2
M2B2D2
D1
M2B3D1
D2
M2B3D2
Section 5.2 Simulations
A simulation is a procedure where a model of a chance process is used to copy
(simulate) a real situation. Simulations are useful to determine the probability
of something occurring or, if something has occurred already, to determine how
likely that event would be if it was controlled by chance alone. Digits are used
to represent situations.
The steps involved in running a simulation are:
1.) State the assumptions being made about the real-life situation.
2.) Describe the model to be used to represent the real-life situation. The
model should include instructions for how to perform one run.
3.) Repetition. Repeat the model run many times, recording the summary
statistic (what you are recording from each run) in a frequency table.
4.) Write a conclusion in the context of the situation. Part of the
conclusion should include an estimated probability.
For example, if a team has a record of winning only 10% of its games, you
could assume that the team’s chance of winning any game at random is 10%.
Say you want to determine the probability that this team will win 2 out of every
five games played. To do this you could let the digit 0 represent a win and the
digits 1 to 9 represent a loss. Using a random digit table or a random digit
generator (such as RandInt on the calculator) a run would consist of 5 digits,
each digit representing one of the five games. If 2 0’s are present, the run is a
“success” (meaning that the team won 2 of the 5 games). If there are not 2 0’s,
the run is a “failure” (meaning the team did not win 2 of the 5 games).
Repeat this run many times, keeping track of the number of “successes” and
“failures”. The simulation runs can be stopped when the proportion value stays
essentially constant.
Once the runs are completed, you can write a conclusion, which in this case
would include your estimated probability of the team winning 2 out of 5 games.
Section 5.3 The Addition Rule and Disjoint Events
In statistics, the phrase “A or B” can mean A or B or Both. If 2 events are
disjoint (mutually exclusive) they cannot occur at the same time. A necessary
condition for disjoint events, then, is that
P(A and B) = 0
(case for disjoint events)
If 2 events are disjoint, the probability of 1 event or the other happening is just
the sum of the two probabilities:
P(A or B) = P(A) + P(B)
(addition rule for disjoint events)
(in more general terms, if there are more than 2 events the probability of one of
them happening would be the sum of the probabilities for each).
For example, in rolling a six sided die, the events rolling a 1, 2, 3, 4, 5, 6 are all
mutually exclusive. Each has a probability of 1/6. The probability of rolling a
1, 4, or 5 is (1/6 + 1/6 + 1/6) = 3/6 = ½.
If events are not disjoint (for example, rolling a 2 or rolling an even number)
when computing the probability for P(A or B) the portions of the two events that
overlap will be accounted for twice, so it is necessary to subtract out the portion
that is treated twice. The addition rule for events that are not disjoint, then, is:
P(A or B) = P(A) + P(B) – P(A and B)
(Addition Rule)
Note that this rule is very general, as it will also work for disjoint events,
because for disjoint events P(A and B) = 0.
As presented in a two-way table form we can calculate the P(A or B) in two
ways.
B
Yes No
TOTAL
Yes
a
b
a+b
A
No
c
d
c+d
TOTAL a + c b + d a + b + c + d
Using P(A or B) = P(A) + P(B) – P(A and B) we have:
ab
ac
a
abc



abcd abcd abcd abcd
Using P(A or B) as the sum of three inner cells we have:
a
b
c
abc



.
abcd abcd abcd abcd
Section 5.4 Conditional Probability
Probabilities that involve the terms “given that” are called conditional
probabilities. In conditional probabilities, the probability is potentially affected
by some additional restriction or information about the situation.
One situation where conditional probabilities are involved is when one deals
with draws in a small population without replacement. For example, if we have
3 black socks and 4 white socks, the color of sock we draw for the first draw has
an effect on the probability of what is drawn on the second. At the first draw,
we have a 3/7 probability of drawing a black sock. If a black sock is drawn
(and not replaced) we have a 2/6 = 1/3 probability of drawing a black sock on
the second draw. The probability of the second draw is conditioned by what has
happened on the first draw. In this situation, we might say “The probability of
drawing a black sock on the second draw given that the first draw was black is
2/6”. In probability notation, this is written as
P(black 2nd draw | black 1st draw) = 2/6.
When you see the symbol “|” think of the words “given that”.
Using the two-way table from above,
B
Yes No
TOTAL
Yes
a
b
a+b
A
No
c
d
c+d
TOTAL a + c b + d a + b + c + d
P(A) =
ab
a
, but P(A | B) =
(the probability A given that B is
abcd
ac
yes).
Using a tree diagram, such as the titanic example on page 327 of the text, it can
be shown that:
P(A and B) = P(A)  P(B | A) or
P(A and B) = P(B)  P(A | B)
This is called the Multiplication Rule.
Rearranging the Multiplication Rule gives the Definition of Conditional
Probability:
P A and B 
.
P A | B  
P B 
Section 5.5 Independent Events
Two events are independent if the result of one event does not have an effect on
the outcome of the other event. Using the notation for conditional events of the
last section (where what happens in one event can have an effect on the
outcome of another event) we have the Definition of Independent Events:
Assume P(A) > 0 and P(B) > 0. Events A and B are
independent events if and only if:
P(A | B) = P(A)
P(B | A) = P(B)
(given that B occurred has no effect on A)
or
(given that B occurred has no effect on A).
For independent events, the multiplication rule, P(A and B) = P(B)  P(A | B), is
simplified to:
P(A and B) = P(B)  P(A)
because for independent events P(A | B) = P(A). For more than 2 independent
events, the multiplication rule becomes:
P(A1 and A2 and A3 and . . . and An) = P(A1)  P(A1)  P(A1) . . .  P(An)
A type of problem often encountered with independent events is the situation of
at least one. This might involve a large number of possible situations, but the
only situation it will not involve is that of none. To find the probability for at
least one scenario, simply compute its equivalent, 1 – P(none).