Probability
An Experiment is any action that leads to an uncertain outcome.
A sample point is a simple outcome of an experiment.
The sample space , denoted S is the set of all possible outcomes.
Events are sets of outcomes or subsets of S and are denoted by
capital letters such as A, B, …
An event is simple if it consists of exactly 1 outcome.
An event is compound if it consists of more than 1 outcome.
Events = Sets
Outcomes = Elements
The order of elements in the set is not important.
A = {1, 2, 3} = {3, 2, 1} {1, 3, 2} ,…
Ex. Flip a fair coin twice.
S = {HH, HT, TH, TT}
Let B = event that exactly 1 H is flipped
Let C = {at least 1 H is flipped}
B = {HT, TH} = {TH, HT}
C = {HH, HT, TH}
Three basic set/event operations, Let E and F be two sets of S.
1. E compliment = E´ is the set of outcomes in S that are not in E.
E´ is the event that E does not occur.
2. E intersection F = E ∩ F is the set of sample points that are in
BOTH E and F or the set of sample points that are common to E
and F. E ∩ F is the event that both event E and event F occur.
3. E union F = E U F is the set of sample points in E or in F (or in
both). Any sample point that is in E is in E U F and any sample
point that is in F is in E U F. the union is the set of outcomes that
are in at least one of the events.
E U F is the event that E occurs or F occurs.
Note that E U F = F U E and E ∩ F = F ∩ E.
and (E´)´= E
The empty set or null set is the event that contains no outcomes.
Denoted {} = Ø
1
When E ∩ F = {}, E and F are said to be disjoint or mutually
exclusive.
Example: Roll a fair die. So S = {1,2,3,4,5,6}
Let A = {1, 3, 5} and B = {1, 2, 3, 4}
Find A´, B´, A U B, A ∩ B.
Draw Venn Diagram
A´ ={2,4,6}
A U B ={1,2,3,4,5}
A´∩B ={2,4}
(A ∩ B) ´ = {2,4,5,6}
B´={5,6}
A ∩ B ={1,3}
A´UB ={1,2,3,4,6}
(A U B)´= {6}
Extension: let C = {2, 4}
C´ = {1, 3, 5, 6}
A ∩ C = {}
A U C = {1, 2, 3, 4, 5}
B ∩ C = {2, 4}
B U C = {1, 2, 3, 4}
B ∩ C´ = {1, 3}
B U C´ = {1, 2, 3, 4, 5, 6}
A ∩ C´ = {1, 3, 5}
A U C´ = {1, 3, 5, 6}
In general 2 sets (call them A and B) partition the sample space S
into 4 regions.
2
I = A ∩B’
II = A ∩ B
III = A’ ∩ B
IV = A’ ∩ B’
Section 2.1 #2
3 vehicles exiting a freeway ramp
each vehicle has 3 options, R, L, S
Use T for sample space
Before answering the questions, one should determine how big
the sample space is and its elements (if feasible)
Each car has 3 options and there cars, so the sample space will
have 3 * 3 * 3 = 27 outcomes.
3
T={
RRR LLL
SSS
RRL
LLR
SSR
RLR LRL
SRS
LRR RLL
RSS
RRS LLS
SSL
RSR LSL
SLS
SRR SLL
LSS
RLS
LRS
SRL
RSL
LSR
SLR }
a. A = {RRR, LLL, SSS}
b. B = {RLS, LRS, SRL, RSL, LSR, SLR}
c. C = {RRL, RLR, LRR, RRS, RSR, SRR}
d. D = { RRL, RLR, LRR, RRS, RSR, SRR, LLR, LRL, RLL, LLS, LSL, SLL,
SSR, SRS, RSS, SSL, SLS, LSS}
e. D´ = {RRR, LLL, SSS, RLS, LRS, SRL, RSL, LSR, SLR}
CUD=D
C∩D=C
De Morgan’s Laws.
verify using Venn diagrams
(A U B)´ = A´ ∩ B´
(A ∩ B)´ = A´ U B´
Section 2.2 Definitions of Probability.
Simple: the probability of a sample point is the measure of the
likelihood that outcome will occur when the experiment is run.
The probability of an event, E, is sum of the probability of the
sample contains contained in E. It is denoted P(E).
You can think of P(E) in two ways:
1. If the experiment was run many many times, what fraction of
the time would E occur.
2. If all the outcomes are equally likely to occur, then
P(E) =
number of ways that E occurs
Total number outcomes of the experiment
P(E)
=
number of outcomes in E
number of outcomes in S
4
Axioms of Probability:
1.
For all events E: 0 ≤ P(E)
2.
P(S) = 1
3.a.
If A1, A2, …, Ak are a finite set of Mutually exclusive
events, then
P(A1 U A2 U … Ak) = P(A1) + P(A2) + … + P(Ak) = ΣP(Ai) as i = 1, 2, …
k
3.b. If A1, A2, … are an infinite set of Mutually exclusive events,
then
P(A1 U A2 U …) = P(A1) + P(A2) + … = ΣP(Ai) as i = 1, 2, … ∞
Note that Axioms 1 and 2 together mean the 0 ≤ P(E) ≤ 1
Back to the roll a fair die example:
S = {1,2,3,4,5,6} A = {1, 3, 5} B = {1, 2, 3, 4}
Find P(A), P(B), P(A U B), P(A ∩ B), P(A´), P(B´)
A fair die means that the P(1) = P(2) =…=P(6)
and since P(1) + P(2) + … + P(6) = 1, therefore if we let x = P(1),
we have 6x = 1 so x = 1/6
Meaning that P(1) = 1/6, P(2) = 1/6, …, P(6) = 1/6
so since all the outcomes are mutually exclusive events:
P(A) = P(1) + P(3) + P(5) = 3/6 = ½ = .5
P(B) = 4/6 = 2/3 = .667
P(A U B) = 5/6 = .833
P(A ∩ B) = 2/6 = 1/3 = .333
P(A´) = 3/6 = ½ = .5
P(B´) = 2/6 = 1/3 = .333
Looking at this example we can guess at 2 very important
formulas for probability.
1.
P(E´) = 1 – P(E)
2.
P(E U F) = P(E) + P(F) – P(E ∩ F) (The additive rule of
Probability)
Note that in the above example
P(A U B) = 5/6 = P(A) + P(B) – P(A ∩ B) = 3/6 + 4/6 – 2/6
P(B´) = 1/3 = 1 – 2/3 = 1 – P(B)
5
Proof that P(E´) = 1 – P(E) in general.
E U E´ = S
P(E U E´) = 1
P(E) + P(E´) – 0 = 1
P(E´) = 1 – P(E)
In order to prove 2, note first that E U F = E U (F ∩ E´).
Pf. (Show that E U F is a subset of E U (F ∩ E´) and then show
E U (F ∩ E´) is a subset of E U F.
a. Show that E U F is a subset of E U (F ∩ E´)
Let x be an element of E U F, then x is in E or x is not in E.
If x is in E then x is in E U (F ∩ E´). If x is not in E then x has to be in
F, because x is in E U F, therefore x is in F ∩ E´, so x is in
E U (F ∩ E´). Therefore E U F is a subset of E U (F ∩ E´).
b. Show E U (F ∩ E´) is a subset of E U F.
Let y be in E U (F ∩ E´), then y is in E or y is not in E.
If y is in E, then y is in E U F.
If y is not in E then y is in F ∩ E´, because y is in E U (F ∩ E´).
Therefore y is in F so y is in E U F.
Therefore E U (F ∩ E´) is a subset of E U F and
E U (F ∩ E´) = E U F. Call this result *
Also note that E and F ∩ E´ are mutually exclusive events.
Finally note that for any 2 sets G and H, G = (G ∩ H) U (G ∩ H´)
Proof: for you to do. Hint: draw a picture. Use argument similar
to above. Call this result **.
This means that P(G) = P(G∩H) + P(G ∩ H´) and
P(G ∩ H´) = P(G) - P(G∩H). Call this result ***
Prove P(E U F) = P(E) + P(F) – P(E ∩ F)
Proof.
P(E U F) = P(E) + P(F ∩ E´) by * and axiom 3.
P(E U F) = P(E) + P(F) – P(F ∩ E) by ***. QED.
An extension of this result to 3 sets is
P(EUFUG) = P(E) + P(F) + P(G) – P(E∩F) – P(E∩G) – P(F∩G)
+P(E∩F∩G).
Draw Venn diagram
6
Labels:
I = A ∩ B’ ∩ C’
III = A’ ∩ B ∩ C
V=A∩B∩C
VII = A’ ∩ B’∩ C
II = A ∩ B ∩ C’
IV = A ∩ B’ ∩ C
VI = A’ ∩ B ∩ C
VIII = A’ ∩ B’∩ C’
Another way to calculate the P(E) for an event E:
Let E = {a1, a2, …, ak}
It may be easy to determine P(ai) for i = 1, 2, … , k
P(E) = P(a1) + P(a2) + … + P(ak)
Examples from Section 2.2: 12, 15, 18, 27
12. A = {student has a visa} B = {student has a master card}
P(A) = .5, P(B) = .4 and P(A ∩ B) = .25
a. P(A U B) = .5 + .4 - .25 = .65
b. P(A´ ∩ B´) = P((A U B)´) = 1 - .65 = .35
c. P(A ∩ B´) = .5 - .25 = .25
15. Let A = {at most 1 purchase an electric dryer}
This means 0 or 1 customers purchase electric.
Since there are 5 people, the compliment of A is:
A´ = { at least 2 people purchase an electric dryer}
P(A´) = 1 - .428 = .572
b. P(all gas ) = .116 and P(all electric) = .005
P(at least one of each) = 1 - .116 - .005 = .879
18. 4 40W, 5 60W and 6 75W bulbs.
P(at least 2 bulbs are selected to get one 75W)
P(1st selected is not 75W) = 9/15
**** Added questions ****
7
1. Find the probability that the first 2 bulbs selected are 60W.
2. Find the probability that the first 2 bulbs selected are 60W
and a 75W.
Draw Tree diagram!
27. A = Anderson, B = Box, C = Cox, D = Cramer, F = Fisher
Hint: A tree diagram may help determine S. Then list the
outcomes, there are 10:
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Section 2.3 Counting is essential to probability.
Ex. How many ways can I order the letters A B C?
ABC, ACB, BAC, BCA, CAB, CBA: which is 6.
You can think of this as you have 3 choices for the first letter and
2 choices for the second letter and 1 choice for the 3 letter. 3*2*1
=6
when you say “and” you multiply.
This is known as the multiplicative rule of counting.
If the first part of an experiment can be done in n1 ways and the
second part can be done in n2 ways, then the experiment can be
done in n1 * n2 ways.
This can be generalized to an experiment with any number of
parts.
If an experiment has k parts and the ith part can be done in ni ways
(ni choices) as i = 1, 2, … , k then the entire experiment can be
done in n1 * n2 * … nk ways.
Menu Example:
You own a restaurant and want to offer combination dinners. The
patrons get a choice of 1 of 3 appetizers, 1 of 5 entrees and 1 of 2
desserts. How many combination meals are there?
3 * 5 * 2 = 30.
Two Dice Example:
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How many ways can I order the letters A B C D?
24 = 4 * 3 * 2 * 1
How many ways can I order n distinct letters (objects)?
n! = n factorial.
0! = 1
n! = n * (n – 1)! for any positive integer n
n! = n * (n-1) * (n-2) * …(2) * (1)
1! = 1
2! = 2
3! = 6
4! = 24
to find 5! On TI83/84, hit [5], [MATH], go over to PRB and the
fourth option is 4:!(hit [4]).
On screen you should see 5!
Hit [ENTER]
Answer is 120, which is 5 * 4! = 5 * 24
n! gets big really fast.
n! = the number of ways to order n distinct items.
Ex. How many ways can I select and order 2 letters from
{A B C D E}?
AB AC AD AE BC BD BE CD CE DE
BA CA DA EA CB DB EB DC EC ED
20 ways = 5 * 4
This can also be looked at as a permutation. When you have n
distinct objects and you select r of them and order is important,
meaning AB ≠ BA, then you have n P r or P(r, n).
n!
n Pr 
(n  r )!
Notes: n will always ≥ r and n P r will reduce to
n * (n – 1) * … * (n – r + 1)
For the above example n = 5 and r = 2 so
n P r = 5 P 2 = 5! / 3! = 120 / 6 = 20
Ex. How many ways can I select 2 letters from {A B C D E}?
(here it assumed that order is not important)
AB AC AD AE BC BD BE CD CE DE = 10 ways.
10
This can also be looked at as a combination. When you have n
distinct objects and you select r of them and order is NOT
important, meaning AB = BA, then you have n C r or C(r, n).
Other Notation:
nCr 
n!
r!(n  r )!
In the above example:
n = 5, r = 2 so
n C r = 5 C 2 = 5! / (2! * 3!) = 120 / (2 * 6) = 10
nCn = 1
nC1 = n
nC0 = 1
nC2 = n*(n-1)/2
Note:
6C0 = 1
6C1 = 6
6C2 = 15
6C3 = 20
6C4 =15
6C5 = 6
6C6 =1
Take the sum: 1+6+15+20+15+6+1 = 64 = 26.
In Class exercises section 2.3
29, 32, 43, 44
43. 5 card poker.
First find the ways to deal 5 cards: 52 C 5 = 2598960
How many types of straights are there?
(A, 2, 3, 4, 5), (2, 3, 4, 5, 6), … (10, J, Q, K, A)
So there are 10. How many ways can one get each of these?
There are 4 of each type of card so 4 * 4 * 4 * 4 *4 = 45.
Therefore P(straight) = 10 * 45 / 52 C 5 = 10240/2598960 = .004
Technically this is not 100% correct, why? A straight flush is better
than a straight.
How many way can one get a straight flush?
10 different types of straight flushes, but only 4 ways to get each.
P(straight flush) = 40 / 52 C 5 = 40 / 2598960 = .00002
P(Straight) = 10200/2598960 = .004
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Extra question: What is the probability of getting a flush?
P(Flush) = 4 * 13 C 5 / 52 C 5 = 5148 / 2598960 = .002
Again you need to subtract P(Straight Flush)
P(Flush) = 5108/2598960 = .002
Note that a flush beats a straight, because the probability of
getting a flush is less than that of a straight.
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The Partitions Rule:
There are N distinct items. You wish to partition them into k sets.
The ith set will contain ni items as i =1,2, …, k. For example the 1st
set will contain n1 items. The number of distinct partitions is:
N!
n1!n2!...nk !
Note that n1 + n2 + … nk = N
Ex.
I have 15 students. How many ways can I break them up into 3
groups of 4, 5, and 6 students each? Unknown to me 2 of the
students are sick, what is the probability that both sick students
are in group of 6, assuming the students are randomly selected to
the groups..
Ways: 15! /(4! 5! 6!) = 630630
P(Both in group of 6) = [13!/(4! 5! 4!)/630630] * 2!/2! =
90090/630630 = .143
Ex. How many ways can arrange the letters of the word LOOK?
What makes this problem different is that the items are not
distinct.
Can you list them:
However you can think of this as a partitions rule problem.
You are partitioning the N (= 4) letters into L’s O’s and K’s.
The answer is 4! /( 1! 2! 1!) = 12
Ex. How many ways can arrange the letters of the word
MISSISSIPPI?
11!/(1! 4! 4! 2!) = 34650
Ex. My fan club has 10 members.
1. How many ways can I pick a President, Vice-President and
Treasurer?
10 * 9 * 8 = 10 P 3 = 720
Order is important
2. How many ways can I pick 3 officers?
10 C 3 = 120
Order is not important
3. There are 6 males and 4 females in the club. If I pick 3 officers,
what is the probability that they are all male?
13
6 C 3/ 10 C 3 = 20/120
4. For the same scenario as in 3, what is the probability that there
is at least 1 female officer?
1 – 20/120 = 100/120
P(NONE) = 1 – P(AT LEAST 1) because {At least 1}’ = {None}
5. There are 6 males and 4 females in the club. If I pick 3 officers,
what is the probability that exactly 2 are males and 1 is a female?
(6 C 2) * (4 C 1) / (10 C 3)
44. Prove that n C k = n C (n – k)
nCk 
n!
(n  k )!k!
n!
n!

(n  (n  k ))!(n  k )! k!(n  k )!
because n – (n – k) = k
nC (n  k ) 
14
Section 2.4 Conditional Probability and Section 2.5
Independence
Example
One thousand people were asked to give their primary source of
news. The results are given in the following table. A random
subject is selected.
Male
Female
Radio
275
200
TV
135
175
Paper
35
70
Other
60
50
Find the probability that
a. the subject gets news from the Radio.
P(Radio) = (275 + 200)/ 1000 = 475 / 1000 = .475
b. the subject gets news from Paper and is Female.
P(Paper ∩ Female) = 70 / 1000 = .070
c. the subject gets news from TV or is Male
P(TV U Male) = P(TV) + P(Male) – P(TV ∩ Male) =
310/1000 + 505/1000 – 135/1000 = 680/1000 = 0.680
d. the subject gets news from Paper GIVEN she is Female.
This is a new type of probability called Conditional Probability.
We are given additional information here, that the subject is
Female. So we are limiting our Sample Space.
P(Paper | Female) = 70 / 495 = .141
In general P(A | B) = P(A ∩ B)
P(B)
Provided that the P(B) ≠0
for the last example P(Paper ∩ Female) = 70/1000 = .07 and the
P(Female) = 495/1000 = .495
so the P(Paper | Female) = .07 / .495 = .141
e.
Find the probability that the subject is Female given she
gets news from Paper.
P(Female | Paper) = .07 / .105 = .667 = 70 / 105
Notes In general P(A | B ) ≠ P(B | A)
whereas
P(A ∩ B) = P(B ∩ A) and P( A U B ) = P( B U A)
15
P(A | B) is a probability so it is between 0 and 1, if you get a
number bigger than 1 it means you made a mistake! Fix it!
You may give answers as fractions or decimals but not both.
Ex. Let P(A) = .32 and P(B) = .38 and P(A U B) = .55.
Find the P(A | B) and P(B | A).
First you need to find P(A ∩ B).
P(A ∩ B) = .15 Why?
P(A | B) = .15 / .38 = .395
P(B | A) = .15 / .32 = .469
Multiplicative Rule of Probability
P(A ∩ B) = P(A | B) P(B)
If you knew P(A | B) = .42 and P(B) = .27 then
P(A ∩ B) = .42 * .27 = .1134
Two events (A and B) are independent when the fact that A
occurs does not affect the probability that B will occur and vice
versa.
Equivalent definitions of Independent events A and B:
P( A ∩ B ) = P(A) P(B)
P(A | B) = P(A)
P(B | A ) = P(B)
So P(A ∩ B) = P(A) P(B) only if A and B are independent.
Ex. the experiment is to draw a single card from a deck of 52. Let
A be the event that the card is a Jack and let B be the event that the
card is Black. Find the probability of A, B and A ∩ B, A|B, B|A ?
Are A and B independent?
P(A) = 4/52 = 1/13
P(B) = 26/52 = ½
P(A ∩ B) = 2/52 = 1/26
P(A | B) = 2/26 = 1/13
P(B | A) = 2/4 = ½
A and B are independent because:
P(A ∩ B) = 1/26 = 1/13 * ½ = P(A) P(B)
P(A | B) = 1/13 = P(A)
P(B |A) = ½ = P(B)
16
Ex. Roll a fair die. Let A = {1, 3, 5} C = {1, 2, 3}
Are A and C independent?
P(A) = ½
P(C) = ½
P(A ∩ C) = 2/6 ≠ ½ * ½ = ¼ so A and C are not independent.
P(A | C ) = 2/3 ≠ P(A)
P(C | A) = 2/3 ≠ P(C)
Ex. 2 fair dice are rolled.
Let A = {Sum is odd}
B = {sum > 9}
C = {at least one die shows a 4}
Recall that:
P(A) = 18/36 = ½
P(B) = 6/36 = 1/6
P(C) = 11/36
P(A ∩ B) = 2/36 = 1/18
P(A ∩ C) = 6/36
P(B∩C) = 2/36 = 1/18
Now find P(A | B), P(B | A), P(A | C), P(C | A), P(B | C), P(C | B),
P(A|B) = (2/36) / (6/36) = 2/6 = 1/3
P(B|A) = (2/36) / (18/36) = 2/18 = 1/9
P(A|C) = (6/36) / (11/36) = 6/11
P(C|A) = (6/36) / (18/36) = 6/18 = 1/3
P(C|B) = (2/36) / (6/36) = 2/6 = 1/3
P(B|C) = (2/36) / (11/36) = 2/11
** DO not confuse Independent with Mutually Exclusive Events!
if A and B are independent then P(A ∩ B) = P(A) P(B)
if A and B are mutually exclusive then P(A ∩ B) = 0
Diagnostic testing.
Let S = {condition or disease is present}
Then S´ = {condition or disease is NOT present}
Let T = { test for S is positive}
Then T´ = { test for S is negative}
Sensitivity = P(T | S )
Sensitivity = probability that the test is + given we know the
subject has the condition
Specificity = P(T´ | S´)
Specificity = probability that the test is - given we know the
subject does not have the condition
P(False Positive) = P(T | S´)
P(False Negative) = P(T´ | S)
17
Note that
P(False Positive) = P(T | S´) = 1 – P(T´ | S´)
and
P(False Negative) = P(T´ | S) = 1 – P(T | S)
Prevalence of a condition or disease is the percent or probability of
the population that has it.
Prevalence = P(S)
What is usually of interest is the probability the subject actually
has the condition given the test is + or the probability the subject
actually does not have the condition given the test -.
Tree Diagram
---P(T |S)
---------P(S)---|
|
---P(T´|S)
|
|
|
---P(T| S´)
---------P(S´)---|
---P(T´| S´)
From the tree diagram we can calculate the intersection
probabilities.
18
---P(T | S)
---------P(S)---|
|
---P(T´|S)
|
|
|
---P(T|S´)
---------P(S´)---|
---P(T´|S´)
P(T ∩ S ) = P(T|S) * P(S)
P(T´ ∩ S ) = P(T´|S) * P(S)
P(T ∩ S´ ) = P(T|S´) * P(S´)
P(T´ ∩ S´) = P(T´|S´) * P(S´)
Note that P(S) = P(S ∩ T) + P(S ∩ T´) and
P(T) = P(T ∩ S) + P(T ∩ S´)
Another methodology is to use Bayes’s rule
Let A be any set. Then A = (A ∩ B) U (A ∩ B´).
P(A) = P[(A ∩ B) U (A ∩ B´)] = P(A ∩ B) + P (A ∩ B´) – 0
P(A) = P(A | B) P(B) + P(A | B´) P(B´) (from Multiplicative Rule)
P(B | A) = P(A ∩ B)/ P(A)
and P(A ∩ B) = P(A |B) P(B)
so
P( A | B) P( B)
P( B | A) 
P( A | B) P( B)  P( A | B' ) P( B' )
It gives you a way to reverse the order of the conditional.
Ex. It is known that about 1% of women who get a mammogram
have breast cancer. The sensitivity of a mammogram is 0.86 and
the specificity is 0.88.
Real Questions:
Given that the mammogram is +, what is the probability the
woman has breast cancer?
Given that the mammogram is -, what is the probability the woman
does not have breast cancer?
To get these answers we will make a tree diagram.
We will then find P(T), P(T´), P(S ∩ T) and P(S´ ∩ T´)
These will then be used to find P(S | T) and P(S´ | T´).
Note what we are given in the problem:
P(S) = .01, P(T|S) = .86 and P(T´|S´) = .88
which means: P(S´) = .99, P(T´|S) = .14 and P(T| S´) = .12
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P(T|S)=.86
P(T ∩ S ) = .01 * .86 = .0086
---P(S) = .01 |
|
P(T´|S)=.14 P(T´ ∩ S ) = .01 * .14 = .0014
|
|
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P(T|S´)=.12 P(T ∩ S´ ) = .99 * .12 = .1188
--P(S´) = .99 |
P(T´|S´)=.88 P(T´ ∩ S´) = .99 * .88 = .8712
P(T) = P(T ∩ S) + P(T ∩ S´)
P(T) = .0086 + .1188 = .1274
P(T´) = P(T´ ∩ S) + P(T´ ∩ S´)
P(T´) = .0014 + .8712 = .8726 = 1 – P(T)
To answer the questions:
P(S | T) = P(T ∩ S) / P(T) = .0086 / .1274 = .0675 = 6.75%
P(S´ | T´) = P(T´ ∩ S´) / P(T´) = .8712 / .8726 = .9984 = 99.84%
Using Bayes’s Rule:
P(S) = .01
P(T | S) = .86 P(T´ | S´) = .88
we want P(S | T) and P(S´ |T´)
P( S | T ) 
P(T | S ) P( S )
.86 * .01
.0086


P(T | S ) P( S )  P(T | S ' ) P( S ' ) .86 * .01  .12 * .99 .1274
P(S | T) = .0675
by the same reasoning
P(S´|T´) = .9984
The generalized version for Bayes’s Rule is given in your book:
Given k mutually exclusive and exhaustive events: A1, A2, …, Ak
such that P(A1) + P(A2) + … +P(Ak) = 1 (which means that A1,
…, Ak form a partition of the sample space) and an observed event
B, then
P( Ai  B)
P( Ai) P( B | Ai)
P( Ai | B) 

P( B)
P( A1) P( B | A1)  ...  P( Ak ) P( B | Ak )
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Ex. A construction company employs 3 engineers, E1, E2 and E3.
They use E1 for 30% of their jobs, E2 for 20% and E3 for 50%.
The following probabilities are given:
P(Error | E1) = .01, P(Error | E2) = .03, P(Error | E3) = .02,.
If an error was made find the probability that it was made by E1.
If an error was made find the probability that it was made by E2.
If an error was made find the probability that it was made by E3.
If an error was made who was most likely to make it?
The denominator will be the same for all 3 probabilities.
P(E|E1)P(E1) + P(E|E2)P(E2) + P(E|E3)P(E3)
.01*.3 + .03*.2 + .02*.5 = .019
P(E1 | E) = P(E|E1)P(E1) / .019 = .003/.019 = .158
P(E2 | E) = P(E|E2)P(E2) / .019 = .006/.019 = .316
P(E3 | E) = P(E|E3)P(E3) / .019 = .010/.019 = .526
Events A1, …, An are mutually independent if for every k
(k = 2, 3, …, n) and every subset i1, i2, …, ik,
P(Ai1 ∩ Ai2 ∩ … ∩ Aik) = P(Ai1) * P(Ai2) * …* P(Aik)
Note that this is a much stronger condition than pairwise
independent.
Ex. Roll 2 fair dice. one is red and one is green.
A={red = 3} B={green = 4}
C = {sum = 7}
Are these events pairwise independent?
Are these events mutually independent?
P(A) = 1/6
P(B) = 1/6
P(C) = 1/6
Note 1/6 * 1/6 = 1/36
P(A ∩B) = 1/36
P(A ∩C) = 1/36
P(B∩C) = 1/36
Therefore the 3 events are pairwise independent
But,
P(A∩B∩ C) = 1/36 ≠ 1/6* 1/6 * 1/6 = 1/216 so they are not
mutually independent.
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