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Unformatted text preview: 74 STA 2023 c D.Wackerly  Lecture 6 STA 2023 c D.Wackerly  Lecture 6 75 The trouble with jogging is that, by the time you realize
you’re not in shape for it, it’s too far to walk back. Assignments Conditional Probability For Today : p. 121–124, 127–135 7 5§ £
86%1#¤¢
is # (Quiz # 2), similar graphs (different data set)–similar . The conditional probability of § #£
%A'¢
§# 4 £ ©§ 9 £
%$"¦¥'¢ %@#¤¦¥'¢ PROJECT 1 : NOT COLLECTED!!! BUT–on next quiz be two events, where questions will be asked about the INTERPRETATION of given B For Wednesday : p. 127–135, Read pages 163 – 170 and ¥ Def. (p. 122) Let ¥ 3.47, 3.49, 3.50, 3.52, 3.53, 3.55–3.57, 3.59, 3.60, 3.61 # For tomorrow: Probs 3.35, 3.37–3.39, 3.41, 3.43, 3.46 the graphs. # LAST TIME : ¥
(What proportion of ¡
¡ Additive Rule of Probability p. 117 # ¦¥¤¨¦¥¤¢
§ £ ¢ © § £ Complement p. 115 : is also in ?) ¥ ¡ Mutually exclusive (disjoint) p. 118 § # 4 £ ¢ § £ ¢ )§ £ ¢ ©§ # ! £
%$"(¥'321#'"0(¥'&%$"¦¥¤¢ 76 STA 2023 c D.Wackerly  Lecture 6 STA 2023 c D.Wackerly  Lecture 6 77 EXAMPLE : Pol. party – job type. ¡ If I know that a person is a Republican, what is the
hY probability that (s)he is a white collar worker? ¥ RW DW IW RB DB IB ¡
¡
¡ # ¡
¡ ¡ . Thus, §£
%1#¤
© %$"¦¥¢ ¤¢ %#'¦¥¤¢
§# 4 £ ©§ 9 £
Q © § # y£
hSvi%A'¢
Hi%1t¤¢
Q ©§# £
H ©§ u £
xvi1y'¢ xwviv¨1t'¢
H ©§ u £ ¡ get a number greater than 3 S ©§#£
7v%P¦¤¢
W ©§ u£
7viR¦¤¢ ¡ get an even number person is a Republican . B §£
21#'
©
© §2ps¦¥¢ '¢ %@#¤¦¥'¢
# 4 £ ©§ 9 £
©§# 4 £
%$"¦¥'¢
Y
C©#4
rqp"¥
©§ £
%1#'¢
hY
C
D©
Y
C©
8#
©§ £
i¨¦¥'¢
CD©gY
C©
f&¥
© §c £ ¢
ed(b¤a`
Y WF F SF QF HF C
EEXPV UTRIPIGED©
B
. person is a white collar worker . Y C8# ¡
©
C©
f&¥ ¡ Balanced Y EXAMPLE : Toss a balanced die. If a person is NOT a Democrat, what is the
probability that (s)he is a blue collar worker? ¡ 78 STA 2023 c D.Wackerly  Lecture 6 STA 2023 c D.Wackerly  Lecture 6 EXAMPLE 3.13: (p. 123)
develops cancer ©
q# Simple Events Probabilities #
$4 ¥
#$4 ¥
ps¥
#4
#4
$"¥ Q
G7
7
H
V
7 Why? Recall (Defn of Cond‘l Prob.): §£
(¥'
F %$"¦¥¢ ¤¢ ¨¥ p1#¤¢
§# 4 £ ©§ 9 £ H so §£
¨¦¥¤
© %$"¦¥¢ '¢ ¦¥¤&¥ p1#¤¢ ¦¥¤¢
§# 4 £ § £¢ ©§ 9 £ § £ ©¥ Given: Multiplicative Law of Prob. (p. 128) §%@#¤¦¥¤¢ %1#¤&©
9 £ § £¢
§ 9 £ § £¢ ©§# 4 £
¥ $1#'¢ ¨¦¥¤&%ps¦¥¤¢ individual smokes 79 ©i§ ¦¥¤¢
£
©§ £
i¨¦¥'¢ § £
¦¥¤¢ © Find the probability that a person develops cancer Useful for computing the probability of the intersection given that the person smokes. when : ©
©
§£
¨¦¥'¢
§£
¦¥¤¢
§#4 £
2$"¦¥'¢ © &1#'¢ ¥ p1#¤¢
§¥4 £ ©§ 9 £
¡ Conditional Probability is given ¡ 80 STA 2023 c D.Wackerly  Lecture 6 Conditional Probability easy to get without deﬁnition. STA 2023 c D.Wackerly  Lecture 6 81 EXAMPLE : (Screening for quality) Shipment of 30
cameras, 6 defective. Choose 2 at random. If both EXAMPLE : (#3.61, p. 137) Microchip: good, accept shipment. Find the prob. accept shipment. ©
R7 v§ camera good
© ¥ ¨¦
©§ . Find the probability that a microchip does not fail : If ﬁrst camera known to be good, what is (cond.) prob. during the ﬁrst year.
NOTE: If chip does not fail during ﬁrst year–It cannot fail second is good? on its ﬁrst use!!! H©
&#
Y when we select the second, . cameras, Note that good, shipment accepted ©§ 9 £ § £¢ © §# 4 £
i¨¥ p1#¤¢ ¨¦¥'&&2ps¦¥¤¢ Y ©
§¨¥¦'¢ ¥ p1#¤&%$"¦¥'¢ ¡
£ § 9 £¢ ©§# 4 £
©
¡
“Does not fail during ﬁrst year” there are bad. Y Chip lasts from ﬁrst use through ﬁrst year , CD©q#$"¥
4
©§ 9 £
¥ p1#'¢ ¡ . : Start with 30 cameras, 6 bad. If ﬁrst chosen is good, ¡ C©
8# ¡
©i¨¦¥'¢
§£
C©
f&¥ ¡ Microchip lasts through ﬁrst use , camera good ©§ 9 £
¥ $1#'¢ ¡¡
£¢ is £
'¢ If it lasts through its ﬁrst use, prob. that it lasts a year ©§ £
¦¥¤¢ fail on ﬁrst use . ¤
¥ 82 STA 2023 c D.Wackerly  Lecture 6 83 §£
%1#¤
© %$"¦¥¢ ¤¢ i2@#'¦¥¤¢
§# 4 £ ©§ 9 £ Independence
Does knowing that one event has occured alway have
an impact on the probability of another event? © 9£
i§ ¢ '¦¥¤¢ DEFINITION: (Def. 3.9, p. 131) Events
said to be independent if ¡ § c (b'¢ © £ © 4 £¢
§ ¢ "(¥'aY
C
D© ¢ "¥ ¡
4
©i2$"¦¥£'aY
§#4 ¢
C©#4
rqp"¥ ¡
©§ £
¤ ©§ £
¡¨©© § ¢ £'¢ ¤¡§i%#1¤¢ ¦ © ¡¥i¨¦¥'¢ ¡
Y W F SF C
EEXFPV UTRQ D©Y
C¢
8© £¡
YW V C
hEIFPIF S D© Y
C©
8# ¡
hEIUTRH CDgY
Y WF SF ©
C©
f&¥ ¡
Y WF F SF QF HF C
EXPV UTRIPIGED© ¡ “Not independent” = “DEPENDENT.”
Ex. Die Toss: ¥ and and , ©
§¥ 9p1#£'T§¦¥¤&%$"¦¥¤¢
¢ £¢ ©§# 4 £
#
¥ Note that if STA 2023 c D.Wackerly  Lecture 6 84 and are independent then
[by Mult. Law]
[by independence] , and ¥ # greater than 2 # # greater than 3 ¢ B get an even number or § £¢ ©§ 9 £
¦¥¤&i2@#'¦¥¤¢ and are 21#'&¨¥ p1#¤¢
§ £¢ ©§ 9 £ STA 2023 c D.Wackerly  Lecture 6 85 Ex. (LAST TIME!!!) Odds are 2:1 that when H and T This is another way to deﬁne independence: (p. 133)
21#'¢ ¨¦¥¤&%# 4 ¦¥'¢
§ £ § £¢ ©§
£ and winner of each recorded.
Odds are 2:1 that H wins means H wins with prob. WARNING Q H i§ d1HUH
© ) £ Independent play racquetball, H will win. H and T play 3 matches, . T wins with prob. Q
¡ EXAMPLE : Toss a balanced die. and ¥ ? # § £¢ ©§ 9 £
¨¦¥'&%@#¤¦¥¤¢ Is STA 2023 c D.Wackerly  Lecture 6 . Probs. given Do NOT confuse INDEPENDENT with MUTUALLY in the example for 8 possible outcomes were calculated EXCLUSIVE. assuming matches independent : HH § £ § §
&© hQ 1H hQ T£ hQ $ £ ¤ ¤ ¤
¤ TTT ¤ TTH ¤ THT ¥ 7 %1#¤¢ wvi¨¦¥'¢
© § £ © § £ #
¥
©§ 9 £
2@#'(¥'¢
§21#'
£
© §%$"¦¥¢ £¤¢ ©§%#9'¥¦¤¢
£
#4
#
are THH ¤ and HTT ¤ are mutually exclusive HTH H#S © hQ 1H hQ A£ hQ $ £
§ £ § H §
H&© hQ hQ T£ hQ 1H£
H § £ § §
H § £ § §
#S © hQ 1H hQ T£ hQ 1H£ and that and , Suppose that ¤ HHT ¡
¡ Independent – need to check PROBABILITIES. ¤ HHH Probability H ! § £ § H §
"© hQ 1H hQ A£ hQ 1H£ common. Outcome sets have no points in Mutually Exclusive ¡ 86 STA 2023 c D.Wackerly  Lecture 6 STA 2023 c D.Wackerly  Lecture 6 87 EXAMPLE : Space shuttle Challenger disaster. The
all six ﬁeld joints (sealed by Orings) work.
ﬁrst ﬁeld joint works C
D© ¡ u sixth ﬁeld joint works .
.
. . ¤
¥ (independence) Assuming joint failures indep.: Y ©
©
§ #£ § 9 £ ¢ © § # 4 £
%A'¢ %#'¦¥¤&%$"¦¥¤¢
§#! £
%$"¦¥¤¢ %@#¤¦¥¤IF 2p"(¥'¢
§9 £¢§#4 £
Q
H ©§ £
©§ £
%1#¤¢
¦¥¤¢
and Rogers commission concluded disaster results unless C
D© u Find are Y ¥ independent events: and # EXAMPLE : Like #3.52(a), p. 136
UQ ¦ ¤¢
§ W¡ £ ©
U7v© h! f§
£
! vi§ ¡ Au¤&P¥¤£§ Au¤¢ ¡ GW
W © £¡ ¢ © ¢ ¢ ¢ © £ 7
©
§
£
¤¢ ©
©
§
£
¤¢
© u£ © ¢¢¢ © u£
v§ ¡ e'¢ P¥¤£§ A¤¢
¡ UQ
§ ¡ A¤¢ § A'T§ ¨ A'¢ § ¤ A¤¢ § ¦ A'¢ § A'&©
u£
u£ ¢ u£
u£
u£
u£ ¢
§ ¡ u 4 u 4 ¨ u 4 ¤ $4 ¦ $4 e'&§
u u u£ ¢ ©
£
¤¢
all six work Under Challenger ﬂight conditions ( ©§ 9 £
%@#¤¦¥¤¢ (independence) F), estimate £ , so ¡
all six work at least 1 fails , © at least 1 fails 88 F. STA 2023 c D.Wackerly  Lecture 7 ¡¡
£¢ th of what it was at about , only ¡ Note: At ¡¡
£¢ ©§# ! £
%$"¦¥¤¢ (Add. Law) STA 2023 c D.Wackerly  Lecture 7 89 Thought: Whether you think you can or you think you EXAMPLE : Space shuttle Challenger disaster. The can’t, you’re right! Rogers commission concluded disaster results unless
all six ﬁeld joints (sealed by Orings) work. C
D© u ﬁrst ﬁeld joint works C
D© ¡ u sixth ﬁeld joint works .
.
. For tomorrow: Exercises 3.99, 3.100, 3.105109,
3.114, 4.3–4.5, 4.7, 4.11, 4.14, 4.16
For Monday: Read pages 172–176 Assuming joint failures indep.: QUIZ 2 – Chapter 3 – Project 1
UQ ¦ ¤¢
§ W¡ £ ©
U7v© h! f§
£
! vi§ ¡ Au¤&P¥¤£§ Au¤¢ ¡ GW
W © £¡ ¢ © ¢ ¢ ¢ © £ 7
©
§
£
¤¢ ©
! v©
£
©
£
¤¢
¡ § q© v§
© u£ © ¢¢¢ © u£
v§ ¡ e'¢ P¥¤£§ A¤¢
¡ UQ
§ ¡ A¤¢ § A'T§ ¨ A'¢ § ¤ A¤¢ § ¦ A'¢ § A'&©
u£
u£ ¢ u£
u£
u£
u£ ¢
§ ¡ u 4 u 4 ¨ u 4 ¤ $4 ¦ $4 e'&§
u u u£ ¢ ©
£
¤¢ For Tuesday:Exercises 4.22, 4.2729, all six work Under Challenger ﬂight conditions ( £ , so ¡
¢ ¡
¢ £ ¡
¢ ¤ £ ¡
¢ ¤ ¡
¢ all six work at least 1 fails , about th of what it was at ¡¡
£¢ at least 1 fails ¡ Note: At ¡¡
£¢ %A'¢ ¦¥¤&%$"¦¥¤¢ §
§ #£ § £ ¢ © § # 4 £
§ £¢ ©§ 9 £
%1#¤&¥ p1#¤¢ §
§ £¢ ©§ 9 £
¦¥¤&%@#¤¦¥¤¢ §
#
¥¡
¨¦¥¤¢ ¥ $1#'&%1#¤¢ %@#¤¦¥¤&i%i¨¦¥'¢
§ £ § 9 £¢ ©§ £ § 9 £¢ ©§#4 £
¤ ¤ Independent (p. 131, 133): F), estimate £ ¡ Multiplicative Rule (p. 128): F. , only Last Time : ¤ and Y Today : p. 163–169 Y Assignments : 90 STA 2023 c D.Wackerly  Lecture 7 STA 2023 c D.Wackerly  Lecture 7 EXAMPLE : (Like # 3.109, p. 158) : Telephone test for
Altzheimer’s disease. 91 A person age 6569 is randomly selected and tests : person tests positive positive. What is the (cond.) prob. that the person : person tests negative actually has Altzheimers ? : person HAS Altzheimers : person does not have Altzheimers § £ £ §£
@y9 '¢ § 1y¤¢ 1y'¢ ¡
¡
¡
¡ Since y
y ¡¢H! v§ 9 ¤¢
© y £
7 ©§ £
G! vy9 '¢ § 6569, 7 in 1000 have Altzheimers, then for person age : © 4£
i§ v1y¤¢ ¡
B § 1y¤a
© £¢ © ©§ £
i1y¤¢ ©
y 92 y
STA 2023 c D.Wackerly  Lecture 7 STA 2023 c D.Wackerly  Lecture 7 93 4 y § 4 1y£ !0§ 41y£q© § ¤¢
£
? are mutually exclusive § § 9 1y¤¢
£ , found . “Good” test? according to rules of chance ( the “random” part) Two Types of Random Variables
DISCRETE (Def. 4.2, p. 166)– the number of
different values of the variable is countable ( look for
variables where the values change in “jumps”).
– Number of accidents in Gville next week.
– Number of phone calls to ﬁre dept. in a day.
– last digit in social security number.
– Toss a coin, bet $1 on “heads”,
Payoff %)
4
y QG7 £
©
§ '¢
£
y£
§ v1y¤¢ § 9 A'¢
4 £ © § 9 1y¤¢
£
SQS
! v©
S ! wf) 7v©
WV7
) 7v©
WV7
)
©
§ 4 1y¤") § v1y¤&i§ ¤¢
£¢
4 £¢© £
? takes on different values (the “variable” part) ¡ ¡
¡ How about Intuitively : a random variable is a quantity that ¡ § Thus and Chapter 4 : Discrete Random Variables ¡ § Note that §£
@y9 '¢ KNEW . © £
§ @y9 ¤¢
H ©
! v§ @y9 ¤¢
£
7 ©§ £
G! i@y9 '¢
§ '¢
£
£
§ v1y¤¢ i§ 9 1y¤¢
4 £ © “speciﬁcity of test” According to the Framingham study, for those age How about , and “sensitivity of test” 6569 selected at random , Know Gville Sun, 11/12/01, “good test” because Want , §
@y9 ¤¢
£ Know or . ¡ 94 STA 2023 c D.Wackerly  Lecture 7 STA 2023 c D.Wackerly  Lecture 7 CONTINUOUS (Def. 4.3, p. 166) – the variable can 95 take on all values on a line interval (look for ¡ variables where the possible values change
smoothly, without jumps)
– Time to assemble automobile DEFINITION : (Def. 4.4, p. 169) The probability – Height, Weight , Blood Pressure distribution of a discrete r.v. is a formula, table, or – Amount of rainfall (in inches) in a month graph giving : ¡ Each possible value for the variable ¡ that assigns one and only one numerical value to each
sample point.
Ex. Roll two dice. 36 sample points. Let value of the r.v. for all poss. values ; ©
§ £ ¡ §¡
¦
¢ ¡¢
D¥§ £ ¤£7 ¡ © §HF
IT£
§ T£
F
.
.
. , associated with each possible Note: Must have (grey shaded box, p. 169): sum of up faces.
Sample Point the prob., § 3¡
£ Def. A RANDOM VARIABLE (Def. 4.1, p. 164) is a rule (sum is over all values of ). .
.
. § WF
hXR1W£
96 STA 2023 c D.Wackerly  Lecture 7 STA 2023 c D.Wackerly  Lecture 7 97 Ex. Shipment of 30 cameras, 6 defective. Select 2 at
# good selected (0, 1, or 2).
: 1st camera selected is good : 2nd camera selected is good § £ ¡ ©
§ 9¥ 1#'¢ § ¦¥¤&§ $4 ¦¥¤&&© ¤&h(73¡
£
£¢© #
£ ¢ © §H £ ¢ © § £ #
¥
© random. Table of the dist. of : 0
1
2 ©
§
¨ 9¥ 1#'¢ ¨¦¥¤¢ ) § ¥ p1#¤¢ § ¦¥¤&©
£§£
9£
£¢
§ ps¦¥'"02p4 ¦¥¤&©
# 4 £ ¢ )§ #
£¢
Y # 4 !§ #
X§ $s¦¥£ 0%$4 ¦¥PC &§ © ¤&§ 3¡
£ ¢© £¢© £
©
§ 9 £ § £ ¢ © § # 4 £ ¢ © §H £ ¢ © § £
¥ p1#¤¢ ¦¥¤&%$"¦¥¤&&© ¤&1H3¡ 98 STA 2023 c D.Wackerly  Lecture 7 STA 2023 c D.Wackerly  Lecture 7 99 Ex. Sum of up faces on two dice.
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1) (1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2) (1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3) (1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4) (1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5) (1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6) Distribution of the Sum of Two Fair Dice
p(x) 2 (1,1) § £ ¡ 3 (1,2), (2,1) 2/36 4/36 4 (1,3), (2,2), (3,1) 3/36 3/36 5 ... 4/36 2/36 6 ... 5/36 1/36 7 ... 6/36 0 8 ... 5/36 9 ... 4/36 10 ... 3/36 11 (5,6), (6,5) 2/36 12 (6,6) 1/36 Sample Points STA 2023 c D.Wackerly  Lecture 7 © # defective. yF¦ F ﬁrst, then second. ¦ y y¦ y ¦ means select y y ¦ § £ ¡ ¡ ”Label” the bulbs 5/36 x 1 100 EXAMPLE : Have 3 light bulbs, 2 good, 1 defective.
Randomly select 2, 6/36 1/36 2 3 4 5 6 7 8 9 10 11 12 © B y ¡ ¡ ¡
¡ ied(b'¢ ¡ © §c £ ...
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 Spring '08
 Ripol
 Statistics, Conditional Probability, Probability

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