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Welcome to MM150
Unit 7 Seminar
Definitions
Experiment - A controlled operation that
yields a set of results.
Outcomes - The possible results from an
experiment.
Event - A subcollection of the outcomes of
an experiment.
Definitions Continued
Empirical Probability - The relative
frequency of occurrence of an event
and is calculated by ACTUAL
OBSERVANCE of an experiment.
Theoretical Probability - The
determination of probability through
a STUDY OF POSSIBLE
OUTCOMES that can occur for the
experiment.
Definitions Continued
Empirical Probability
P(E) = number of times event E has occurred
total number of time exp. is performed
Theoretical Probability (Outcome equally likely)
P(E) = number of outcomes favorable to E
total number of possible outcomes
Example
Let’s use 10 flips of a fair coin, which has 2
outcomes: heads or tails.
Empirical: I toss the coin 10 times and flip
heads 7 times.
P(Head) = 7/10 = 0.7
Theoretical: There are 2 outcomes, 1 of
which is favorable to being a head.
P(Head) = 1/2 = 0.5
Law of Large
Numbers
Probability statements apply in practice to a
large number of trials, not to a single trial. It is
the relative frequency over the long run that is
accurately predictable, not individual events
or precise totals.
Is it a fair die? Consider a standard six sided die. What is the theoretical
probability of rolling an even number in a single toss? P(E) = 1/2
Experiment: Roll the die many times and record if the result is even or not.
Number
of Rolls
Number
that are
even
results
Empirical Probability
10
4
4/10 = 0.4
100
53
53/100 = 0.53
Law of Large Numbers: If
it is a fair die, then as we
repeat the experiment more
and more the empirical
probability should approach
the theoretical probability.
It is a fair die.
500
241
241/500 = 0.482
1000
511
511/1000 = 0.511
4 Important Facts
1. The probability of an event that cannot occur is 0. It is
said to be impossible.
2. The probability of an event that must occur is 1. It is
said to be certain.
3. Every probability is a number between 0 and 1
inclusive; that is, 0 ≤ P(E) ≤ 1.
4. The sum of the probabilities of all possible outcomes
of an experiment is 1. That is,
P(A) + P(not A) = 1
P(not A) = 1 - P(A)
Single Toss of Fair Coin
Outcomes - There are 2, heads or tails.
Fact 1: P(Yellow) = 0
Fact 2: P(Head or Tails) = 1
Fact 3: Cannot be more likely than 100% nor less likely
than never!
Fact 4:
P(Head) = 1/2 = 0.5
P(Tail) = 1/2 = 0.5
P(Head) + P(Tail) = ½+ ½
P(Tail) = 1 - P(Head) = 1 - ½ = ½
Odds
Odds against event
P(event fails to occur) = P(failure)
P(event occurs)
P(success)
Odds in favor of event
P(event occurs) = P(success)
P(event fails to occur) P(failure)
Odds Example
Look at page 293 of your text, #49.
Odds against A+ = P(person not having A+)
P(person having A+)
= 66/34 or 66:34
Now, what if the question was what are the odds for a
person having A+ blood? It would be 34/66 or 34:66.
Probability from Odds
P(A+ Blood) = # of outcomes favorable to A+
total # of outcomes
= 34/100 = 0.34
NOTE Odds for was 34:66; therefore
probability of A+ is 34/(34+66) = 34/100
Expected Value
E = P1*A1 + P2A2 + ... + PnAn
where P1 is the probability the first event
will occur and A1 is the net amount won
or lost if the first event occurs
P2 is the probability the second event will
occur and A2 is the net amount won or
lost if the second event occurs ...
Expected Value
Example
Page 301 #16 In July in Seattle, the grass grows 1/2 in. a day on a sunny day and
1/4 in. a day on a cloudy day. In Seattle in July, 75% of the days are sunny and
25% are cloudy.
B) Determine the expected total grass growth in the month of July in Seattle.
E = 0.75*1/2 + 0.25*1/4
= 0.375 + 0.0625
= 0.4375 This is the expected growth per day.
There are 31 days in July
(0.4375)(31) = 13.5625 inches is the expected growth in July in Seattle
Definitions
Sample Space - All possible outcomes of an
experiment.
Sample Point - Each individual outcome in the
sample space.
Counting Principle - If a first experiment can be
performed in M distinct ways and a second
experiment can be performed in N distinct
ways, then the two experiments in that specific
order can be performed in M*N distinct ways.
Tree Diagram Example
Mathy’s Pizza has a choice of:
3 crusts: hand-tossed, pan, thin
2 sauces: traditional, meat
3 toppings: pepperoni, sausage, green peppers
Construct a tree diagram and list the sample space.
What is the probability that a student selects meat sauce?
Crust
Sauce
r
h
m
r
p
m
r
t
m
Topping Sample Space
e
s
g
e
s
g
e
s
g
es
g
e
s
g
e
s
g
hre
hrs
hrg
hme
hms
hmg
pre
prs
prg
pme
pms
pmg
tre
trs
trg
tme
tms
tmg
With/Without Replacement
A bag with 5 red marbles, 2 blue marbles and 1 white
marble.
With Replacement
You pick one marble out of the bag, record it and
put it back in the bag. It is available to be picked
again.
Counting Principle: 8 * 8 = 64
There are 64 outcomes.
For example:
R1R1
R1R2
R1R3
R1R4
R1R5
R1B1
R1B2
R1W
R2R1
R2R2
R2R3
R2R4
R2R5
R2B1
R2B2
R2W
...
Without Replacment
You pick one marble out of the bag, record it and set
it aside (not back in the bag). It is not available to be
picked again.
Counting Principle: 8 * 7 = 56
There are 56 outcomes.
For example:
R1R2
R1R3
R1R4
R1R5
R1B1
R1B2
R1W
R2R1
R2R3
R2R4
R2R5
R2B1
R2B2
R2W
...
Note: doubles missing
Or Problems
P(A or B) = P(A) + P(B) - P(A ∩ B)
When A and B are mutually exclusive
P(A or B) = P(A) + P(B)
Or Problems
P(Heart or 6) = P(Heart) + P(6) --Now...think about this...There is a 6 of hearts that will be counted in the
hearts and with the 6s, so it is counted twice. We only want it to be counted
once. Therefore we have to subtract 1/52 from the equation above.
P(Heart or 6) = P(Heart) + P(6) - 1/52
= 13/52 + 4/52 - 1/52
= 16/52
NOTE: P(Heart ∩ 6) = 1/52, so
P(Heart or 6) = P(Heart) + P(6) - P(Heart ∩ 6)