Probability
Notes Outline - Key
The outcome of a chance process cannot be predicted in the short term.
However, per the "Law of Large Numbers", if we repeat a chance process enough times, the
proportion of times an outcome occurs over the long-tun can be predicted.
The probability of an outcome is the proportion of times the outcome is expected to occur
if the number of repetitions is very large
The Probability of an outcome of a chance process is always a number between:
Outcome is….
Certain to occur (will definitely/always occur)
Very likely
Equally likely to occur or not occur (50/50 chance)
Very Unlikely
Impossible
.50
.25
0
less likely to
outcome is
0 and 1 (inclusive)
Probability
1
close to 1
0.5
close to 0
0
1.0
.75
very likely
50/50
always
If an event has already occurred, the probability of an outcome is exactly 0 or 1 .
(either the event happened or it didn't)
Simulation:
process of modeling chance behavior with numerous repetitions, to estimate
the answer (e.g., to a probability related question) based on the simulated results
Basic Probability Rules
Sample Space (S) Event -
the set of all possible chance outcomes
a particular combination of outcomes; e.g., Event A
Sum of probabilities of all possible outcomes =
For any event A, 0 < P(A) < 1
1
If all outcomes in the sample space are equally likely:
P(A) =
number of favorable outcomes corresponding to A
total number of possible outcomes in sample space
AC means the "Complement of A" (i.e., "Not A")
P(AC) + P(A) = 1
P(AC) =
1 - P(A)
P(AC) means P(Not A)
P(A) =
1 - P(AC)
A
AC
Examples:
(1) Toss 2 coins. Find P(2 Tails). Assume fair coin (outcomes are equally likely)
Possible outcomes:
HT, HH, TT, TH
No. of favorable outcomes:
1
P(TT) =
1/4
(2)
Family has 3 children. Find P(1 or more boys). Assume 50/50 chance of B vs G (equally likely outcomes)
Possible outcomes for 1st, 2nd, 3rd child:
1st 2nd 3rd All-3
B
BBB
Total # Possible Outcomes:
8
B
G
BBG
# of Favorable Outcomes:
7
B
B
BGB
P(1+ boys):
7/8
G
G
BGG
B
GBB
Event A = have 1 or more boys
B
G
GBG
Event AC =
have No boys
G
C
B
GGB
P(A ) =
1/8
G
C
G
GGG
P(A) = 1 - P(A ) =
1 - 1/8 = 7/8
(3)
Toss coin 3 times. Find P(exactly 2 tails in 3 tosses). Assume fair coin (equally likely outcomes)
Possible outcomes for 1st, 2nd, 3rd tosses:
1st 2nd 3rd All-3
H
HHH
Total # Possible Outcomes:
8
H
T
HHT
# of Favorable Outcomes:
3
H
H
HTH
P(2 tails):
3/8
T
T
HTT
H
THH
Events:
H
T
THT
Event A = have exactly 2 tails in 3 tosses
T
H
TTH
Event AC =
have 0, 1 or 3 tails
T
C
T
TTT
P(A ) = 1 - P(A) =
1 - 3/8 = 5/8
(4)
Toss coin 4 times. Find total number of possible outcomes:
24 = 16
(5)
Toss coin 5 times. Find total number of possible outcomes:
P(at least 1 tail in 5 tosses) =
1 - P(no tails)
1 - 1/32
= 31/32
25 = 32
(6)
Roll 2 standard dice. Find P(sum is at least 9). Assume fair dice (equally likely outcomes).
Sample Space Possible Outcomes on 2 dice:
11
12
13
14
15
16
Total # Possible Outcomes:
36
21
22
23
24
25
26
# of Favorable Outcomes:
10
31
32
33
34
35
36
P(sum > 9):
10/36
41
42
43
44
45
46
51
52
53
54
55
56
Event AC =
sum is 8 or less
C
61
62
63
64
65
66
P(A ) = 1 - P(A) =
1 - 10/36 = 26/36
(7)
Roll die 3 times. Find total number of possible outcomes:
63 = 216
(8)
Roll die 4 times. Find total number of possible outcomes:
P(Not all 5's) =
1 - P(all 5's)
1 - 1/1296 = 1295/1296
64 = 1296
(9)
Distribution of a sample is given. Randomly select 1 subject from sample.
P(male and age 65+) =
0-19 20-39 40-64 65+ Total
P(age 65+) =
18
Male
15
14
7
54
P(age < 65) =
20
Fem
17
16
8
61
P(male) =
Total
32
30
38
15
115
P(age 20-64) =
P(female and age 20-64)=
P(Not age 40-64) =
7 / 115
15 / 115
1 - 15/115 = 100 / 115
54 / 115
(30+38)/115 = 68 / 115
(16+20)/115 = 36 / 115
1 - (38/115) = 77 / 115
Addition Rule for P(A or B) i.e., P(A U B)
Let A and B represent 2 chance Events
A ∩ B denotes the intersection of events A and B (the set of all outcomes in both A and B)
A U B denotes the union of events A and B (the set of all outcomees in either event A or B or both.
Addition Rule:
P(A or B) = P(A) + P(B) - P(A & B)
Mutually Exclusive ("disjoint") Events - have no outcomes in common (can never occur together)
Iff A and B are mutually exclusive,
P(A & B) = 0
P(A or B) = P(A) + P(B) - P(A&B)=0
P(A or B) = P(A) + P(B)
for mutually exclusive events
Mutually Exclusive events A & B:
No Intersection; P(A & B)=0
For Events A & B below, label the following regions of the graph:
A ∩ BC
A∩B
B ∩ AC
AC ∩ BC
Not(A U B)
A ∩ BC
A∩B
B ∩ AC
AC ∩ BC
= Not(A U B)
(10) Given standard deck of cards. Randomly select 1 card. Equally likely outcomes:
Standard deck:
P(heart) =
#hearts/52 = 13/52
• 4 suits:
1 / 52
♥ ♠ ♦ ♣
P(K♠) = P(K ∩ ♠ ) =
• 13 cards per suit: Ace, 1-10, J,Q,K
P(7 or 3) =
8 / 52
• 3 face cards per suit: J,Q,K
P(face card) =
12/52
• 52 cards total
P(ace) =
4/52
P(face card ∩ club) =
3/52
P(Ace or club) =
=P(A) + P(C) - P(A ∩ C)
= 4/52 + 13/52 - 1/52
= 16/52
Ace
P(face card or diamond) =
=P(FC) + P(D) - P(FC ∩ D)
=12/52 + 13/52 - 3/52
= 22/52
Club
FC
9/52
3/52
1/52
12/52
P(Not (7 or heart)) =
= 1 - P(7 or heart)
= 1 - [4/52 + 13/52 - 1/52]
= 1 - 16/52 = 36/52
Diam
3/52
7
3/52
1/52
heart
12/52
10/52
36/52
(11) Survey college students. 56% live on campus; 62% have campus meal plan; 42% do both.
Sketch Venn Diagram
for reference below:
OC
MP
.14
.42
.20
Find:
a) P(on campus or have meal plan) =
P(OC) + P(MP) - P(OC ∩ MP) = .56 + .62 - .42 = .76
b) P(on campus and no meal plan) =
P(OC) - P(OC ∩ MP) = .56 - .42 = .14
c) P(meal plan and off campus) =
P(MP) - P(MP ∩ OC) = .62 - .42 = .20
d) P(off campus and no meal plan) =
1 - P(on campus or meal plan) = 1 - .76 = .24
e) P(meal plan or no meal plan) =
.62 + (1-.62) = 1
f) P(meal plan, among students who live on campus) =
.42 / .56 = 42/56 = 3/4
(12) Survey of customers found the following:
Walk-ins Appt Total
Not satisfied
21 9
30
Neutral
18 25
43
Satisfied
36 43
79
Very satisfied
28 31
59
Total
103 108
211
For a customer selected at random, find the probability that the customer is
that the customer is
a) Not satisfied
= 30 / 211
b) Not satisfied and walk-in
= 21 / 211
c) Very satisfied
= 59 / 211
d) Very satisfied and had an appt
= 31 / 211
e) Walk-in customer
= 103 / 211
f) Very satisfied, given that s/he had an appt
=P(VS ∩ Appt) / P(Appt) = 31 / 108
g) Not satisfied, given that s/he was a walk-in
=P(NS ∩ Walkin) / P(Walkin) = 21 / 103
h) Very satisfied or had an appt
= P(VS U Appt)
VS
Appt
= P(VS) + P(Appt) - P(VS ∩ Appt)
28
31
77
= 59/211 + 108/211 - 31/211
= 136 / 211
i) Walk-in or Neutral
= P(WI U Neutral)
W-I
Neut
= P(WI) + P(N) - P(WI ∩ N)
= 103/211 + 43/211 - 18/211
85
18
25
= 128 / 211
j) walk-in or Neutral or both
means the same as i) = 128/211
Conditional Probability
Conditional probability : The probability that one event happens given that another event is already
known to have happened. Suppose we know that event A has happened.
Then the probability that event B happens given that event A has happened is denoted by P(B | A).
P(B | A) = P(B, given A)
B
A
P(A ∩ B)
B ∩ AC
P(B | A) =
A∩B
P(A)
A ∩ BC
=> What fraction of A is also B
i.e., P(B | A) = P(Both) / P(A)
P(A) = P(A ∩ B) + P(A ∩ BC)
General Multiplication Rule for P(A ∩ B) :
P(A ∩ B) = P(A) ∙ P(B | A)
=
P(B) ∙ P(A | B)
i.e., P(A) x P(B occurs given that event A has already occurred)
Can use either formula, since these are equivalent:
P(A) ∙ P(B | A) = P(A) ∙ P(A & B)/P(A) = P(A & B)
These are equivalent:
P(B) ∙ P(A | B) = P(B) ∙ P(A & B)/P(B) = P(A & B)
Two-way tables and venn diagrams make it easy to think about conditional probability:
(13) Given: 78% of suspected drunk drivers get a breath test, 36% a blood test, and 22% both. Find:
a) P(getting a blood test (i.e., both), given that a breath test is administered)
b) P(no blood test, given that breath test is administered)
c) P(getting a breath test (i.e., both), given that a blood test is administered)
d) P(blood test only)
e) P(neither test is given)
Given data:
Complete 2-way table:
Blood
Breath test
Yes
No Total
Yes 0.22
0.36
No
Total 0.78
1.00
a) P(both | breath):
b) P(no blood | breath):
c) P(both | blood):
d) P(blood only):
e) P(neither):
Blood
Test
Blood
Test
0.14
Breath test
Yes
No Total
Yes 0.22 0.14 0.36
No 0.56 0.08 0.64
Total 0.78 0.22 1.00
= P(breath ∩ blood) / P(breath)
= P(no blood ∩ breath) / P(breath)
= P(breath ∩ blood) / P(blood)
= .14
= 1 - P(either) = 1 - (.78 + .36 - .22)
0.22
0.56
Breath
= .22 / .78
= .56 / .78
= .22 / .36
= 1 - .92 = .08
Independent Events
Two events are independent if the occurrence of one event has no effect on the chance that the other event
will happen. In other words, events A and B are independent iff:
P(A | B) = P(A)
and
P(B | A) = P(B)
In other words, for independent events, the conditional probability, P(B|A),
is the same as the marginal probability of the event, P(B)
Multiplication rule for independent events
If A and B are independent events, then the probability that A and B both occur is:
P(A ∩ B) = P(A) ∙ P(B)
P(A ∩ B) =
P(A) ∙ P(B | A)
=
P(A) ∙ P(B), if independent
In other words, when multiplying probabilities for independent events,
don't need to use conditional probabilities;
can just use marginal (independent) probabilities
Examples
(14) Select 3 cards with replacement, reshuffling each time. Find P(3 consecutive hearts).
Since selecting with replacement, probabilities for each draw are not affected by prior draws.
i.e., each card represents an independent event
P(3 hearts) = (13/52) ∙ (13/52) ∙ (13/52) = (1/4)3 = 1/64
(15) Select 3 cards without replacement. Find P(3 consecutive hearts)
Here, consecutive draws are not independent; the P(heart) for a draw is affected by prior draws.
1st 2 cards: P(2 hearts) = P(heart on 1st) ∙ P(heart on 2nd | heart on 1st)
= 13/52 ∙ 12/51
1st 3 cards: P(3 hearts) = P(heart on 1st & 2nd) ∙ P(heart on 3rd | hearts on 1st two)
= ( 13/52 ∙ 12/51) ∙ 11/50
Consecutive coin tosses.
Tosses are independent.
e.g., even if you just had 5 Tails in a row, the probability of a Tail on the next toss is still 0.5
Find P(7 tails in a row).
= (1/2)7 =
1 / 128
Find P(7 consecutive tosses being H, H, T, H, T, T, T).
= (1/2)7 =
1 / 128
6
Find P(at least 1 heads in 6 tosses).
= 1 - P(no heads) = 1 - (1/2)
= 1 - 1/64 = 63/64
The following table shows the results of survey of individuals watching the premier of a movie:
It shows the percentage of individuals in each category
Are the events "Enjoyed Movie" and "Gender" independent?
Given:
Complete Table:
Yes
Male 0.45
Fem 0.30
No
0.15
0.10
Enjoyed Movie
Gender
Enjoyed Movie
Gender
(16)
(17)
(18)
(19)
Yes
No Total
Male 0.45 0.15
0.60 << conditional dist'n of Enjoyment if Male
Fem 0.30 0.10
0.40 << conditional dist'n of Enjoyment if Fem
Total
0.75 0.25 1.00 << marginal dist'n of Enjoyment
These are independent only if the conditional distributions and marginal distributions are equal.
Test:
Independent only if:
P(enjoyed movie) = P(enjoyed movie | gender)
P(enjoyed movie) =
.75 / 1.00 = .75
P(enjoyed movie | male) =
.45 / .60 = .75
P(enjoyed movie | female) =
.30 / .40 = .75
Similarly, we could have tested whether:
Since these are
all equal, the events
are independent
P(gender) = P(gender | enjoyed movie)
Tree Diagrams for Probabilities
Tree diagram Used to display the sample space for a chance process that involves a sequence of outcomes.
Ex. (20) An airline reported that 60% of its customers buy advance-purchase tickets, instead of regular fare
tickets. The "no-show" rate among people purchasing advance-purchase tickets was 5%, compared
to 30% for customoers purchasing regular fare tickets.
a) what percent of all ticket holders are no-shows?
b) what's the probability that a customer who didn't "show" had an advance-purchase ticket.
c) is being a no-show independent of the type of ticket purchased. Explain
0.60
0.05
AP & no show
(0.6)(0.05)=
0.03
0.95
AP & shows up
(0.6)(0.95)=
0.57
0.30
RF & no show
(0.4)(0.3)=
0.12
0.70
RF & shows up
(0.4)(0.7)=
0.28
Note: should check that total is 100% :
1.00
Adv. Purch.
customer
0.40
a)
b)
c)
Reg. Fare
% of customers who are "No shows":
3%
+
12%
=
15%
P("advance purchase" given "no show") = P(AP ∩ NS) / P(NS)
= .03 / (.03 + .12) = .03 / .15 = 1/5
If independent, conditional probability = marginal probability
P(no show) =
(.03 + .12) / 1.00 = 15/100 = 3/20
Since these are
P(no show | AP) =
.03 / (.03 + .57) = 3/60 = 1/20
not equal, the events
P(no show | RF) =
.12 / (.12 + .28) = 12/40 = 6/20
are not independent