Some Rules of Probability. More formulae: P(B|A) = = Thus, P(B|A) is not the same as P(A|B). P(A B)...
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Transcript of Some Rules of Probability. More formulae: P(B|A) = = Thus, P(B|A) is not the same as P(A|B). P(A B)...
Some Rules of Probability 10 AP 1SP
BPAPBAPBA then , If
APAP c 1
CPBPAPCBAP
CBCABA
then
,, If
BAPBPAPBAP
BA
then
, If
BPAPBAP
BA
then
t independen are , If
t?independennot
are , ifWhat BA
More formulae:
• P(B|A) = =
Thus, P(B|A) is not the same as P(A|B).
• P(AB) = P(A|B)·P(B)
• P(AB) = P(B|A)·P(A)
CONDITIONAL PROBABILITYCONDITIONAL PROBABILITY
P(A)A)P(B
P(A)B)P(A
)()()|(
BPBAPBAP
AIDS Testing Example
ELISA test: + : HIV positive
- : HIV negative
Correctness: 99% on HIV positive person (1% false negative)
95% on HIV negative person
(5% false alarm)
Mandatory ELISA testing for people applying for marriage licenses in MA.
“low risk” population: 1 in 500 HIV positive
Suppose a person got ELISA = +.Q: HIV positive?
Bayes’ Theorem
cc APABPAPABP
APABPBAP
||
||
…some peoplemake a living outof this formula
Try Michael Birnbaum’s (former UIUC psych faculty) Bayesian calculatorhttp://psych.fullerton.edu/mbirnbaum/bayes/BayesCalc.htm
Bayes’ Theorem
cc APABPAPABP
APABPBAP
||
||
notHIVPnotHIVPHIVPHIVP
HIVPHIVPHIVP
||
||
500
4995001
5001
05.99.
99.|
HIVP
038.| HIVP 99.| HIVP
The Binomial DistributionThe Binomial Distribution
Today: Today:
We have already talked about the most famous continuous random variable, which,
because it is so heavily used, even has a name:
The Normal Random Variable (and, associated with it, the Normal Distribution)
Today we will talk about a famous discrete random variable, which, because it is so heavily used, also has a name:
The Binomial Random Variable (and, associated with it, the Binomial Distribution)
FAIR COIN“POPULATION”
THEORETICAL PROBABILITY OF HEADS IS ½
If you toss a fair coin10 times,
what is the probability of
x many heads?
Binomial Distribution, p=.5, n=10
0
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10
# of successes
pro
ba
bil
ity
Binomial Random Variable
Potential Examples:Repeat a simple dichotomous experiment n times and count…
Coin Tossing: # headsMarketing: # purchasesMedical procedure: # patients curedFinance: # days stock Politics: # voters favoring a given billTesting: # number of test items of a given difficulty
that you solve correctlySampling: # of females in a random sample of peopleEducation: # of high school students who drink alcohol
Population
Repeat simple experiment n many times independently.
)success" ofy probabilit" (e.g.,
parameter a is
10
1
:RV sdichotomou a is
p
pYP
pYP
Y
n
in
yyy
YYYYY
,...,,
:DataObtain
. ofcopy a is Each ,...,,
21
21
RV) (a statistic a is
...
successes) # (i.e., s1' # Let
21
X
YYYX
X
n
, ..., n, X 10 valuescan take
onDistributi Binomial
, ..., n, kkXP
a called is
10for
ondistributiy probabilit The
XonDistributi Sampling of theis This
X: number of successes in n many independent repetitions of an experiment,each repetition having a probability p of success
Binomial Random Variable
),(~ pnBX
The Binomial DistributionBinomial Distribution, p=.25, n=10
0
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10
# of successes
pro
ba
bili
ty
Trial 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0
Trial 2 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0
Trial 3 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
Trial 4 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
Probab.
(letting
q=1-p)
pppp
pppq
ppqp
ppqq
pqpp
pqpq
pqqp
pqqq
qppp
qppq
qpqp
qpqq
qqpp
qqpq
qqqp
qqqq
X 4 3 3 2 3 2 2 1 3 2 2 1 2 1 1 0
Probab.
(letting
q=1-p)
pppp
pppq
ppqp
ppqq
pqpp
pqpq
pqqp
pqqq
qppp
qppq
qpqp
qpqq
qqpp
qqpq
qqqp
qqqq
X 4 3 3 2 3 2 2 1 3 2 2 1 2 1 1 0
44 pppppXP )1(4)1(43 3 ppppppXP
22 161162 ppppppXP 31411141 ppppppXP
4111110 pppppXP
In general, for n many trials
knk ppkXP 1successesk get to waysof #
How do we figure that out
in general?
Factorial
!
!3
!5
n
12...1
123
12345
nn
Binomial Distribution Formula:
1!0
)1)(2)...(2)(1)((! and
called )!(!
! where
)1()(
bygiven is ...,,1,0each for
successes)many exactly ( )(
nnnn
tCoefficien Binomialknk
n
k
n
ppk
nkXP
nk
kPkXP
knk
TABLE CPages T-6 to T-10
in the book
Probab.
(letting
q=1-p)
pppp
pppq
ppqp
ppqq
pqpp
pqpq
pqqp
pqqq
qppp
qppq
qpqp
qpqq
qqpp
qqpq
qqqp
qqqq
X 4 3 3 2 3 2 2 1 3 2 2 1 2 1 1 0
44 pXP
)1(43 3 ppXP
22 162 ppXP
3141 ppXP
410 pXP
knk ppk
nkXP
)1()(
44 pXP
)1(43 3 ppXP
22 162 ppXP
3141 ppXP
410 pXP
1!0!4
!4:4 k
4
1123
1234
!34!3
!4:3
k
6
1212
1234
!24!2
!4:2
k
4
1231
1234
!14!1
!4:1
k
1!4!0
!4:0 k
knk ppk
nkXP
)1()( BINOMIAL COEFFICIENTS:
Binomial Distribution, p=.5, n=5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0 1 2 3 4 5
# of successes
pro
ba
bil
ity
Binomial Distribution, p=.5, n=10
0
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10
# of successes
pro
ba
bil
ity
Binomial Distribution, p=.5, n=20
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
# of successes
pro
ba
bil
ity
Binomial Distribution, p=.5, n=50
0
0.02
0.04
0.06
0.08
0.1
0.12
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50
# of successes
pro
ba
bil
ity
Binomial Distribution, p=.5, n=100
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.090 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
# of successes
pro
bab
ility
)1(
for Variance and
Value Expected
2 pnp
np
X~B(n,p)
X
X
Example:
# heads in 5 tosses of a coin: X~B(5,1/2)
Expectation Variance# heads in 5 tosses of a coin: 2.5 1.25
)1(
for Variance and
Value Expected
2 pnp
np
X~B(n,p)
X
X
PROOF:
YYYYYX in ofcopy t independenan is each where...21
pYPpYP
Y
10,1
:RV sdichotomou a is
pppppp
ppp
Y
Y
1101
101222
pnpn
npn
YYYYYYYX
YYYYYYYX
nn
nn
1...
...22222
...2
...
2121
2121
Another Statistic: The Sample Proportion
knk ppk
nkXP
)1()(Remember that X is a
random variable
The sample proportion is a linear transformation of X and thus a random variable too
kXn
kp
n
XXp n en exactly wh ˆ thusand ˆ 1
knk ppk
n
n
kpP
)1(ˆ
Sampling Distributionof the Sample Proportion:
The Sample Proportion
knk ppk
nkXP
)1()(
knk ppk
n
n
kpP
)1(ˆ
)1(
for Variance and
Value Expected
2 pnp
np
X~B(n,p)
X
X
n
pp
n
pp
n
pnp
n
pn
X~B(n,p)n
Xp
pX
p
Xp
)1(,
)1()1(
whereˆfor Variance and
Value Expected
ˆ22
22ˆ
ˆ
Unbiased Estimator
Let’s take another look at some Binomial Distributions
especially what happens as n gets bigger and bigger
Binomial Distribution, p=.1, n=5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0 1 2 3 4 5
# of successes
pro
ba
bil
ity
Binomial Distribution, p=.1, n=10
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0 1 2 3 4 5 6 7 8 9 10
# of successes
pro
ba
bil
ity
Binomial Distribution, p=.1, n=20
0
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
# of successes
pro
ba
bil
ity
Binomial Distribution, p=.1, n=50
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50
# of successes
pro
ba
bil
ity
Binomial Distribution, p=.1, n=100
0
0.02
0.04
0.06
0.08
0.1
0.12
0.140 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99
# of successes
pro
bab
ility
Normal Approximation of/to the Binomial
n
pppNp
pnpnpNpnBX
n
1,ely approximat is ˆ ly,consequent and
1,ely approximat is ,~
thenlarge, is If
0
0.05
0.1
0.15
0.2
0.25
0.3p
rob
ab
ilit
y
Normal Approximation
1100
1 ppp
1100
1 ppp
p
1100
1,~ˆ
1100 :Example
pppNp
n
Normal Approximation: Let’s check it out!
789.1,420,20~
,20 :Example
54
51
51
51
NNX
pn
kk
kkXP
20
54
51
20
TABLE CPages T-6 to T-10
in the book
0020.10
2010
0222.8
208
0576.1
201
:exampleFor
10
5410
51
12
548
51
19
541
51
XP
XP
XP
Normal Approximation: Let’s check it out!
789.1,420,20~
,20 :Example
54
51
51
51
NNX
pn
?8 isWhat XP
8 :ionApproximat Normal theUsing XP
X
X
X
XXP
8
Standardizing:
23.2789.1
48
ZPZP
0129.23.21 ZP
Normal Approximation: Let’s check it out!
789.1,420,20~
,20 :Example
54
51
51
51
NNX
pn
?8 isWhat XP
0222.8 XP
8 :Table Binomial theUsing XP
0074.9 XP
0020.10 XP
0005.11 XP
0001.12 XP
0322.
...0001.
0005.0020.
0074.0222.8
XP
Normal Approximation: Let’s check it out!
789.1,420,20~
,20 :Example
54
51
51
51
NNX
pn
?8 isWhat XP
0322.8 :Table Binomial theUsing XP
0129.8 :ionApproximat Normal theUsing XP
Are westuck
with a badapproximation??
Binomial Distribution, n=20
0
0.05
0.1
0.15
0.2
0.25
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
# of successes
pro
ba
bil
ity
0.2
0322.8
:onDistributi Binomial
for C Table Using
XP
Binomial Distribution, n=20
0
0.05
0.1
0.15
0.2
0.25
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
# of successes
pro
bab
ility
0.2
0129.8
:ionApproximat Normal
theUsing
XP
Binomial Distribution, n=20
0
0.05
0.1
0.15
0.2
0.25
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
# of successes
pro
bab
ility
0.2
0250.9750.1
789.1
45.75.7
:Correction Continuitywith
ionApproximat Normal theUsing
ZPXP
For now, that’s it …
We will revisit the Binomial:
• Based on the sample proportion as an estimate of the population proportion, we will develop “confidence intervals” for the population proportion.
• We will carry out hypothesis tests about the true population proportion, using the information gained from the sample proportion.