**Unformatted text preview: **Probability is the relative frequency with which an event occurs. These measures help managers make educ
their companies. Learning Objectives: Describe the types of probability:
1) Empirical Probability is determined by observation. For example, the number of defects in 10,000 widgets
be 25. So, P(defective) = 25/10,000.
2) Subjective Probability - Subjective probability measures the likelihood of an event subjectively
● Rationally derived from knowledge and experience,
● A "personal belief" allowing substantial variation in estimates. Based on my experience and conditions of
that the probability of candidate A winning the election is about 55%. 3) Classical Probability - Classical probability involves enumerating all possible outcomes, counting the num
of interest, and computing probability of the outcomes of interest = number of the outcomes of interest / th
of outcomes. If there are 40 even numbers and 60 odd numbers randomly distributed in a box, the probabil
even number, P(even) = 0.40.
Learning Objective: What are the probability rules?
● Probability of an outcome can take any value from 0(0%, impossible) to 1(100%, certainty).
● The probability of all possible outcomes is 100%.
● An event consists of 1 or more outcomes.
● The probability of an event occurring is 1 minus the probability that it doesn't occur. Question #
1 Highlight the values which are NOT VALID for probabilities
● 0.50
● -0.10
● 1.10 The Law of Large Numbers means that as the number of observations of an event becomes larger, the runn
relative frequency of the event appears to settle down to a constant value, the probability of the event.
You have just flipped a fair coin 7 times getting all heads. Does the coin owe you a tail? Although the mythic
Averages predicts you are due for a tail, no, the probability of a head on the next toss hasn't changed.
2 Is a good hitter in baseball who has struck out the last six times due for a hit his next time up?
● Yes the hitter is due for a hit. This is an example of the Law of Averages
● This is an example of the so-called Law of Averages. No, the hitter is not due for a hit because the Law
● Yes, the hitter is due for a hit. This is an example of the Law of Large Numbers
● The Law of Large Numbers predicts the hitter is not due for a hit. 3 The running mean of the relative frequency of the event appears to settle down to a constant value.
● This is an example of the Law of Large Numbers.
● This is an example of the Law of Averages. Counting Outcomes
Outcomes with Replacement
Draw a card from a deck, replace it, and shuffle the deck. There are 52 equally likely outcomes on each draw
● P(draw ace of hearts) = 1/52 is the same from draw to draw.
● For each of the 52 ways in which the 1st draw occurs., there are 52 ways the 2nd draw occurs. So, there a
two draws can occur. P(acefollowed by ace) = 4/52 x 4/52
Outcomes without Replacement
Draw a card from a deck and not replace it. P(heart) = 13/52.
P(heart followed by spade) = 13/52 x 13/51 Outcomes with Replacement
Draw a card from a deck, replace it, and shuffle the deck. There are 52 equally likely outcomes on each draw
● P(draw ace of hearts) = 1/52 is the same from draw to draw.
● For each of the 52 ways in which the 1st draw occurs., there are 52 ways the 2nd draw occurs. So, there a
two draws can occur. P(acefollowed by ace) = 4/52 x 4/52
Outcomes without Replacement
Draw a card from a deck and not replace it. P(heart) = 13/52.
P(heart followed by spade) = 13/52 x 13/51 Permutations
A permutation is a sequence of outcomes where outcomes are drawn without replacement and order is imp
For a sequence of n things drawn from N things without replacement, there are N ways the 1st draw occurs.
For each way of the 1st trial, there are N-1 ways the 2nd trial can occur. For 2 draws, there are N x (N-1) poss
Continuing, there are N x (N-1) x (N-2) triples for 3 draws. In general, it's N!/(N - n)! , or =PERMUT( N, n) Combinations
A combination is a group of outcomes where outcomes are drawn without replacement and order is not imp
Example: How many groups of 4 can be formed from 7 people? We know there are 840 sequences possible.
4 things can arrange themselves 4 x 3 x 2 x 1 = 4! (read 4 factorial) ways. so, there are 840 / 4! groups. In ge
of combinations = the number of permutations/the number of sequences of n things =N! / (n! x (N - n)!),
or =COMBIN(N,n) Sequences of 4 of 7 things 840 Groups of 4 of 7 things 35 # of poker hands Click on the cells to
see the formulas 2,598,960 4 The number of sequences (order is important) formed by drawing without replacement of 5 The number of sequences (order is important) formed by drawing with replacement of 6 The number of committees of size 4 formed from 8 Mutually Exclusvie Events
If event A = Bill is a Democrat, and event B = Bill is a Republican, events A and B cannot occur jointly (simult
and B are called mutually exclusive. P(A and B) = 0. P(A or B) = P(A) + P(B), which is the Simple Addition Rul
mutually exclusive events = 1.
In the following, mark the correct answer.
7 P(Democrat and Republican) > 0
● Correct
● Incorrect 8 P(Democrat, Republican, or Independent) = 0, or impossible
● Correct
● Incorrect 9 P(Democrat, Republican, or Independent) = 1, or certain
● Correct
● Incorrect 10 P(Democrat and Republican) = 0, or impossible
● Correct
● Incorrect Example: Toss 2 dice, and let the event be the sum of the top face values. The possible events are
computed in the following table . Example: Toss 2 dice, and let the event be the sum of the top face values. The possible events are
computed in the following table . toss of
1st die 1
2
3
4
5
6 1
2
3
4
5
6
7 2
3
4
5
6
7
8 toss of 2nd die
3
4
5
6
7
8
9 4
5
6
7
8
9
10 5
6
7
8
9
10
11 6
7
8
9
10
11
12 The probabilities of the events are
Event Probability
2
1/36
3
2/36
4
3/36
5
4/36
6
5/36
7
6/36
8
5/36
9
4/36
10
3/36
11
2/36
12
1/36 Only one of the events 2 through 12 can occur in a toss of the dice. These are mutually exclusive event
Simple Addition Rule applies.
P(X not more than 7) or P(X at most 7) = P(X ≤ 7) = P(2) +P(3) +P(4) +P(5) +P(6) +P(7)
P(X not less than 7) or P(X at least 7) = P(X ≥ 7) = P(7) +P(8) +P(9) +P(10) +P(11) +P(12)
P(X less than 7) or P(X < 7) = P(2) +P(3) +P(4) +P(5) +P(6)
P(X greater than 7) or P(X > 7) = P(8) +P(9) +P(10) +P(11) +P(12)
P(X is not 9) = P(X≠9) = 1 – P(X=9)
P(no event) = 0
P(any or some event) = 1 11 Assume the mutually exclusive events {1,2,3,4,5} occur with the following probabilities:
X
P(X = x)
1
0.047
2
0.136
3
0.553
4
0.198
5
0.067
Total
1.000
Which applies?
● Simple Multiplication Rule: P(A and B) = P(A)*P(B)
● Simple Addition Rule: P(A or B) = P(A) + P(B)
● General Addition Rule: P(A or B) = P(A) + P(B) - P(A and B) 12
13 P(X not more than 4) =
P(X at least 3) = 0.933
0.817 Let Excel compute for you by first
typing = followed by the computatio
with addresses to values in the blue
cells. This will avoid typing errors. 14
15
16 P(X = 2) =
P(X > 1) =
P(X < 6) = 0.136
0.953
1.000 Let Excel compute for you by first
typing = followed by the computatio
with addresses to values in the blue
cells. This will avoid typing errors. Non-Mutually Exclusive Events
Non-mutually exclusive events can occur jointly. Example: In a deck of 52 cards, there are 4 suites (heart, di
spade), each suite has 4 face cards (ace, king, queen, jack) and 9 number cards (2 – 10). If event A is the car
heart and event B is ace, these events can occur jointly. Adding P(ace) + P(heart), we get = 4/52 + 13/52 = 1
know there are only 16 cards (13 hearts and 3 more aces). So that P(A and B) is not double counted, P(A or
P(B) – P(A and B), which is the General Addition Rule. So, P(A or B) = 13/52 + 4/52 – 1/52 = 16/32
∩ means "and". If you
add P(A) and P(B), an
extra P(A∩B) is added
17 P(A) =
P(B) =
P(C) =
P(A and B)= 0.418
0.242
0.524
0.184 Note: If P(A and B) = 0, A and B are mutually exclusive. Otherwise, A and B can 18 Which applies?
● Simple Multiplication Rule: P(A and B) = P(A)*P(B)
● Simple Addition Rule: P(A or B) = P(A) + P(B)
● General Addition Rule: P(A or B) = P(A) + P(B) - P(A and B) 19 P(A or B) = 20 P(A or B or C) = 0.559
1.000 Independent Sequential or Simultaneous Events
An event which is independent does not affect the probability of occurrence of the next event. Suppose that the probability of winning the lottery, event A, is 1 in 10 million, and that the probability of ex
airplane crash, event B, is 2.5 in 1 million (I looked this up). These events are independent. You have been g
trillion lifetimes to determine the probability of winning the lottery and crashing on the way to a resort to e
winnings. On average, of 10 trillion lifetimes, you won 10T x 1/10M = 1M lottery tickets. On the accompanyi
your plane crashes 2.5 times. So, the probability that you will win and crash is 2.5/10T. In general, P(A) of the time A occurs. P(B) of the time B occurs. When A and B share the same time, the eve
coincide P(A) x P(B) of the time, or P(A and B) = P(A) x P(B), which is the Simple Multiplication Rule. So, P(AB
2.5/M. You should fly, but you may consider buying something other than the lottery ticket. The assumption of independence lead to P(A and B) = P(A) x P(B). Conversely, if P(A and B) = P(A) x P(B), A a
independent. P(A) =
P(B) =
P(A and B) =
21 0.433
0.348
0.150 Are A and B independent events? ● Yes
● No
22 How do you test independence?
Testing to see if the occurance of one event doesn't make it less probable for the other event to occur.
Two events are independent if any of the following are true: 1.P(A|B)=P(A) 2. P(B|A)=P(B) 3. P(A and B) = P(A Two customers enter a store. Independently, they make decisions to purchase or not to purchase. The follow
shows how the events can occur and combine with red bolding of sequence Customer 1 does not purchase a
2 does not purchase The possible outcomes are shown on the right. If the event is the number of purchases
events: 0 purchases, 1 purchase, and 2 purchases 23 If each of the branches on the right of the figure are equally likely, the P(1 purchase) = 24 Which Rule applies to the sequence of independent events?
● Simple Multiplication Rule: P(A and B) = P(A)*P(B)
● Simple Addition Rule: P(A or B) = P(A) + P(B)
● General Addition Rule: P(A or B) = P(A) + P(B) - P(A and B)
Assume the following probabilities:
P(customer makes purchase) =
P(customer does not purchase) = 25
26
27 28 Event Purchases
0
2
1
(0,1,or 2) neither purchases
both purchase
one xor other purchases 0.887
0.113
Probability
0
1/3
2/3
1.000 Hint:
1=
P(Purchases = 0, 1, or 2) =
P(Purchases = 0 or 2) + P(Pu
P(Purchases = 1) = 1 - P(Purc Explain why P(Purchases = 0, 1, or 2) = 1
There are only three options, 0 1 and 2 and if purchases =0,1,or 2 then it’s a 100% chance one of the number Contingency Tables,
Joint, Marginal, and Conditional Probabilities
In a relative frequency contingency tables, joint probabilities are located inside the margins. In the margin of of a relative frequency continency table, you sum joint probabilites across a row or sum acr Contingency Tables,
Joint, Marginal, and Conditional Probabilities
In a relative frequency contingency tables, joint probabilities are located inside the margins. In the margin of of a relative frequency continency table, you sum joint probabilites across a row or sum acr
It is called a marginal probability. The sum of all marginal probabilities in the bottom row is 1. The sum of all
probabilities in the right column is 1. When the probability of event B is affected by the occurrence of event A, the events are not independent. Le
the probability of B given the condition that A has occurred. This is called a conditional probability. P(B | A) ≠
P(A | B) ≠ P(A). P(A and B) = P(A) x P(B | A). This is the General Multiplication Rule.
● Type | by holding down Shift and type \
29 Fill in the blanks in the following relative frequency contingency table for events A, B, C, D, E D
E A
0.25
0.14 B
0.13
0.33 C
0.17
0.11
0.28 0.62
0.80 0.11 means P(C and E) = 0.11
0.33 means P(B) = 0.33
Suppose we have the probabilities: P(Democrat and Yes) is called a joint probability because it involves joint events of Democrat and Yes. It can
directly from the table. P(Democrat and Yes) = 0.15. P(Democrat) is a marginal probability located in the bottom margin and column Democrat. It is computed by
joint probabilities P(Democrat and Yes) + P(Democrat and No) = 0.15 + 0.25 = 0.40. P(Yes) is a marginal probability located in the right margin and row Yes. It is computed by adding the joint pr
P(Democrat and Yes) + P(Republican and Yes) = 0.15 + 0.20 = 0.35. Conditional probability is simply a probability given a condition that some event has occurred. This reduces
the row or column of the event that occurred. P(Yes | Democrat) is the probability of event Yes given the co
the event Democrat has occurred. The table reduces to the Democrat column. In condition Democrat, Yes occurs at a rate of 0.15 in 0.40. So P(Yes | Democrat) = 0.15/0.40 = 0.37 30 Compute the marginal probabilities for following summary of poll results, Male
Female Republican
0.611
0.126 Democrat Independent
0.003
0.048
0.125
0.088 31 P(Male and Republican) is type
● Joint
● Marginal
● Conditional 32 P(Male and Republican)= 33 P(Male) is type
● Joint
● Marginal
● Conditional 34 P(Male) = 35 P(Male | Republican) is type
● Joint
● Marginal
● Conditional 36 P(Republican | Male)= (Compute using the values in the Male row) 37 P(Male | Republican) = (Compute using the values in the Republican column) Additional Reading ures help managers make educated decisions for ber of defects in 10,000 widgets was observed to n event subjectively y experience and conditions of the pesent, I feel e outcomes, counting the number of outcomes
of the outcomes of interest / the total number
stributed in a box, the probability of drawing an 100%, certainty). sn't occur. event becomes larger, the running mean of the
he probability of the event.
you a tail? Although the mythical Law of
next toss hasn't changed. s next time up?
due for a hit because the Law of Averages is non-existent and doesn't predict anything. to a constant value. y likely outcomes on each draw. he 2nd draw occurs. So, there are 52 x 52 ways y likely outcomes on each draw. he 2nd draw occurs. So, there are 52 x 52 ways t replacement and order is important.
are N ways the 1st draw occurs.
draws, there are N x (N-1) possible ways.
N - n)! , or =PERMUT( N, n) placement and order is not important.
re are 840 sequences possible. But
here are 840 / 4! groups. In general, the number
n things =N! / (n! x (N - n)!), people = 4 things from 8 things= 1680.000 4 things from 8 things= 70.000 d B cannot occur jointly (simultaneously). A
which is the Simple Addition Rule. P(All The possible events are 1/2 The possible events are se are mutually exclusive events. The ) +P(6) +P(7)
+P(11) +P(12) xcel compute for you by first
g = followed by the computation
addresses to values in the blue
This will avoid typing errors. xcel compute for you by first
g = followed by the computation
addresses to values in the blue
This will avoid typing errors. ds, there are 4 suites (heart, diamond, club,
ds (2 – 10). If event A is the card drawn is
eart), we get = 4/52 + 13/52 = 17/52, but we
) is not double counted, P(A or B) = P(A) +
4/52 – 1/52 = 16/32 clusive. Otherwise, A and B can occur jointly of the next event. n, and that the probability of experiencing an
independent. You have been granted 10
hing on the way to a resort to enjoy your
ery tickets. On the accompanying flights,
s 2.5/10T. B share the same time, the events A and B
ple Multiplication Rule. So, P(AB) = 1/10M x
e lottery ticket. y, if P(A and B) = P(A) x P(B), A and B are I don't understand what I am supposed to do here for #17. it says P(A and B
that would mean that .418x.242 is actually
0.1013077 he other event to occur.
(B|A)=P(B) 3. P(A and B) = P(A) * P(B) e or not to purchase. The following sketch
Customer 1 does not purchase and Customer
ent is the number of purchases, there are 3 Sketches make
probability intuitive 2/3 Hint:
1=
P(Purchases = 0, 1, or 2) =
P(Purchases = 0 or 2) + P(Purchases = 1). So,
P(Purchases = 1) = 1 - P(Purchases = 0 or 2) 00% chance one of the numbers are pulled de the margins. abilites across a row or sum across a column. de the margins. abilites across a row or sum across a column.
bottom row is 1. The sum of all marginal events are not independent. Let P(B | A) be
onditional probability. P(B | A) ≠ P(B). Also,
Rule. A, B, C, D, E and E) = 0.11 s: nts of Democrat and Yes. It can be read mn Democrat. It is computed by adding the
0.40. omputed by adding the joint probabilities ent has occurred. This reduces the table to
ability of event Yes given the condition that
n. | Democrat) = 0.15/0.40 = 0.375. publican column) s/boundless-algebra-textbook/combinatorics-and-probability-343/combinatorics-57/permutations-238-5523/ ks/boundless-statistics-textbook/probability-8/probability-rules-34/the-addition-rule-170-4444/ o here for #17. it says P(A and B)=.184 but P(Aand B) = P(A) x P(B)
??? ki100k/docs/Statistical_independence.html ks/boundless-statistics-textbook/probability-8/probability-rules-34/the-multiplication-rule-171-4445/ ?file=Algebra_ProbabilityMultiplicationRule.xml NKzEhttps://people.richland.edu/james/lecture/m170/ch05-rul.html ...

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- Spring '16
- Conditional Probability, Probability, Probability theory