Nathan VanHoudnos
10/13/2014
to fill in
Dogs are inbred for such desirable characteristics as blue eye color; but an unfortunate byproduct of such inbreeding can be the emergence of characteristics such as deafness. A 1992 study of Dalmatians (by Strain and others, as reported in The Dalmatians Dilemma) found the following:
(i) 31% of all Dalmatians have blue eyes.
(ii) 38% of all Dalmatians are deaf.
(iii) 42% of blueeyed Dalmatians are deaf.
Based on the results of this study is “having blue eyes” independent of “being deaf”?
Write this out:
(i) 31% of all Dalmatians have blue eyes.
\[ P(B) = .31 \]
(ii) 38% of all Dalmatians are deaf.
\[ P(D) = .38 \]
(iii) 42% of blueeyed Dalmatians are deaf.
\[ P(DB) \text{ or } P(BD) \]
\[ P(DB) = .42 \]
\[ \begin{aligned} P(B) & = .31 & P(D) & = .38 \\ P(DB) & = .42 \ \end{aligned} \]
Based on the results of this study is “having blue eyes” independent of “being deaf”?
Write out the symbols…
\[ \begin{aligned} P(B) & = .31 & P(D) & = .38 \\ P(DB) & = .42 \ \end{aligned} \]
Based on the results of this study is “having blue eyes” independent of “being deaf”?
\[ \begin{aligned} P(B) & = .31 & P(D) & = .38 \\ P(DB) & = .42 \ \end{aligned} \]
Based on the results of this study is “having blue eyes” independent of “being deaf”?
Independence if and only if: \[ P(DB) = P(D) \] therefore b) is the correct choice.
Let \( S \) be the sample space, \( A \) any event, \( A^c \) its complement, and \( B \) another event.
\( 0 \le P(A) \le 1 \)
\( P(S) = 1 \)
\( P(A^c) = 1  P(A) \)
\( P(A \text{ or } B ) = P(A) + P(B)  P(A \text{ and } B) \)
If and only if \( A \) and \( B \) are independent, then
\[ P(A \text{ and } B ) = P(A) \times P(B) \]
Recall the definition of conditional probability:
\[ P(AB) = \frac{P(A \text{ and } B )}{P(B)} \]
Therefore, for all events \( A \) and \( B \):
\[ P(A \text{ and } B )= P(AB) \times P(B) \]
The general rule:
\[ P(A \text{ and } B ) = P(AB) \times P(B) \]
Recall that \( A \) is independent of \( B \) if and only if
\[ P(AB) = P(A) \]
Therefore, if \( A \) is independent of \( B \),
\[ \begin{aligned} P(A \text{ and } B ) & = P(AB) P(B) \\ & = P(A) P(B) \end{aligned} \]
which is rule #5.
Let \( S \) be the sample space, \( A \) any event, \( A^c \) its complement, and \( B \) another event.
\( 0 \le P(A) \le 1 \)
\( P(S) = 1 \)
\( P(A^c) = 1  P(A) \)
\( P(A \text{ or } B ) = P(A) + P(B)  P(A \text{ and } B) \)
\( P(A \text{ and } B ) = P(AB) P(B) \)
Independent: if and only if \( P(AB) = P(A) \).
Note that: \[ \begin{aligned} P(AB) & = \frac{P(A \text{ and } B )}{P(B)} \\ P(BA) & = \frac{P(A \text{ and } B )}{P(A)} \end{aligned} \]
Therefore:
\[ P(A \text{ and } B ) = P(AB) P(B) = P(BA) P(A) \]
Both are correct.
In later courses you will work with objects like this:
\[ \begin{aligned} P(X, & \mu, \text{ and } \sigma) \\ & = P(X, \big\{ \mu \text{ and } \sigma \big\} ) \\ & = P(X\big\{ \mu \text{ and } \sigma \big\}) P( \big\{ \mu \text{ and } \sigma \big\} ) \\ & = P(X \mu, \sigma ) P(\mu\sigma)P(\sigma) ) \end{aligned} \]
This chaining of the general multiplication rule is important for:
In a certain region, one in every thousand people (0.001) of all individuals are infected by the HIV virus that causes AIDS. Tests for presence of the virus are fairly accurate but not perfect. If someone actually has HIV, the probability of testing positive is 0.95.
Let \( H \) denote the event of having HIV, and \( T \) the event of testing positive.
\[ \begin{aligned} P(H) & = ? & P(T) & = ? \\ P(H \text{ and } T) & = ? & P(H \text{ or } T) & = ? \\ P(HT) & = ? & P(TH) & = ? \end{aligned} \]
In a certain region, one in every thousand people (0.001) of all individuals are infected by the HIV virus that causes AIDS. Tests for presence of the virus are fairly accurate but not perfect. If someone actually has HIV, the probability of testing positive is 0.95.
Let \( H \) denote the event of having HIV, and \( T \) the event of testing positive.
\[ \begin{aligned} P(H) & = 0.001 & P(T) & = ? \\ P(H \text{ and } T) & = ? & P(H \text{ or } T) & = ? \\ P(HT) & = ? & P(TH) & = 0.95 \end{aligned} \]
What is the probability that someone chosen at random tests has HIV and tests positive?
\[ \begin{aligned} P(H) & = 0.001 & P(T) & = ? \\ P(H \text{ and } T) & = ? & P(H \text{ or } T) & = ? \\ P(HT) & = ? & P(TH) & = 0.95 \end{aligned} \]
We need: \[ \begin{aligned} P(H \text{ and } T) & = P(TH)P(H) \\ & = (0.95) (0.001) = 0.00095 \end{aligned} \]
implying that approximately 1/10 of 1% of people will have HIV and will test positive for it.
A sales representative tells his friend that the probability of landing a major contract by the end of the week, resulting in a large commission, is .4. If the commission comes through, the probability that he will indulge in a weekend vacation in Bermuda is .9. Even if the commission doesn't come through, he may still go to Bermuda, but only with probability .3.
\[ \begin{aligned} P(C) & = ? & P(V) & = ? \\ P(VC) & = ? & P(V\text{not } C) & = ? \\ \end{aligned} \]
A sales representative tells his friend that the probability of landing a major contract by the end of the week, resulting in a large commission, is .4. If the commission comes through, the probability that he will indulge in a weekend vacation in Bermuda is .9. Even if the commission doesn't come through, he may still go to Bermuda, but only with probability .3.
\[ \begin{aligned} P(C) & = 0.40 & P(V) & = ? \\ P(VC) & = 0.90 & P(V\text{not } C) & = 0.3 \\ \end{aligned} \]
“the probability of landing a major contract … is .4”
+0.40
/
C
/
<
\
not C
\
+0.60
“If the commission comes through, the probability [of a] vacation … is .9.”
/0.90
V
/
+0.40<
/ \
C not V
/ \[
<
\ /[
not C V
\ /
+0.60<
\
not V
\[
“If the commission comes through, the probability [of a] vacation … is .9.”
/0.90
V
/
+0.40<
/ \
C not V
/ \0.10
<
\ /[
not C V
\ /
+0.60<
\
not V
\[
“Even if the commission doesn't come through, he may still go … with probability .3.”
/0.90
V
/
+0.40<
/ \
C not V
/ \0.10
<
\ /0.30
not C V
\ /
+0.60<
\
not V
\0.70
/P(VC) = 0.90
V
P(C) /
+0.40<
/ \
C not V
/ \P(not VC) = 0.10
<
\ /P(Vnot C) = 0.30
not C V
\ /
+0.60<
P(not C) \
not V
\P(not Vnot C) = 0.70
 V  not V  Total

C    0.40

not C    0.60

Total    1
From the tree we have
How do we get \( P(V \text{ and } C) \) etc. to put in the table?
There are two ways to take a vacation, with and without the commission:
\[ P(V) = P(V \text{ and } C) + P(V \text{ and not } C) \]
By the general multiplication rule we have:
\[ \begin{aligned} P(V \text{ and } C) & = P(VC) P(C) \\ P(V \text{ and not } C) & = P(V\text{not } C) P(\text{not } C) \end{aligned} \]
Therefore:
\[ P(V) = P(VC) P(C) + P(V\text{not } C) P(\text{not } C) \]
\( P(V) = P(VC) P(C) + P(V\text{not } C) P(\text{not } C) \)
+0.90 # P(VC)P(C) = 0.90 * 0.40
V = 0.36

+0.40+
 
C not V
 +0.10
<
 +0.30
not C V
 
+0.60+

not V
+0.70
\( P(V) = P(VC) P(C) + P(V\text{not } C) P(\text{not } C) \)
+0.90 # P(VC)P(C) = 0.90 * 0.40
V = 0.36

+0.40+
 
C not V
 +0.10
<
 +0.30 # P(Vnot C)P(not C)
not C V = 0.60 * 0.30
  = 0.18
+0.60+

not V
+0.70
\( P(V) = P(VC) P(C) + P(V\text{not } C) P(\text{not } C) \)
+0.90 # P(VC)P(C) = 0.90 * 0.40
V = 0.36

+0.40+
 
C not V
 +0.10
<
 +0.30 # P(Vnot C)P(not C)
not C V = 0.60 * 0.30
  = 0.18
+0.60+ ______________________
 { Therefore: }
not V { P(V) = 0.36 + 0.18 }
+0.70 { = 0.54 }
After multiplying out the tree:
 V  not V  Total

C  0.36   0.40

not C  0.18   0.60

Total  0.54   1
And finding the rest:
 V  not V  Total

C  0.36  0.04  0.40

not C  0.18  0.42  0.60

Total  0.54  0.46  1
Two way tables (or Venn Diagrams)
Probability trees
Can convert backandforth between them as needed.
Suppose the friend left for a week and came back to the office. When the friend returned, the salesman had left for Bermuda.
What is the probability that the salesman received a commission given that he is on vacation in Bermuda?
\[ P(CV) = ? \]
\[ P(CV) = ? \]
 V  not V  Total

C  0.36  0.04  0.40

not C  0.18  0.42  0.60

Total  0.54  0.46  1
\[ P(CV) = \frac{P(C \text{ and } V)}{P(V)} = \frac{.36}{.54} = .667 \]
implying that there is a 67% chance that he received the commission.
The probability of landing a major contract by the end of the week is .4. If the commission comes through, the probability that he will vacation in Bermuda is .9. Even if the commission doesn't come through, he may still go to Bermuda, but only with probability .3.
\[ \begin{aligned} P(C) & = 0.40 & P(V) & = ? \\ P(VC) & = 0.90 & P(V\text{not } C) & = 0.3 \\ \end{aligned} \]
What is the probability that the salesman received a commission given that he is on vacation in Bermuda? \[ P(CV) = ? \]
Recall: \[ \begin{aligned} P(AB) & = \frac{P(A \text{ and } B )}{P(B)} \\ P(BA) & = \frac{P(A \text{ and } B )}{P(A)} \\ P(A \text{ and } B ) & = P(AB) P(B) = P(BA) P(A) \end{aligned} \]
Therefore: \[ P(AB) = \frac{P(BA) P(A)}{P(B)} \]
This is Bayes' Rule.
Bayes' Rule:
\[ P(AB) = \frac{P(BA) P(A)}{P(B)} \]
Law of Total Probability
\[ \begin{aligned} P(B) & = P(B \text{ and } A) + P(B \text{ and not} A) \\ & = P(BA)P(A) + P(B\text{not }A)P(\text{not A}) \end{aligned} \]
\[ \begin{aligned} P(C) & = 0.40 & P(V) & = ? \\ P(VC) & = 0.90 & P(V\text{not } C) & = 0.3 \\ \end{aligned} \]
What is the probability that the salesman recieved a commission given that he is on vacation in Bermuda? \[ P(CV) = ? \]
\[ \begin{aligned} P(CV) & = \frac{P(VC) P(C)}{P(V)} \\ P(V) & = P(VC)P(C) + P(V\text{not }C)P(\text{not C}) \end{aligned} \]
\[ \begin{aligned} P(C) & = 0.40 & P(V) & = ? \\ P(VC) & = 0.90 & P(V\text{not } C) & = 0.3 \\ \end{aligned} \]
What is the probability that the salesman recieved a commission given that he is on vacation in Bermuda? \[ P(CV) = ? \]
\[ \begin{aligned} P(CV) & = \frac{ 0.90 * 0.40 }{P(V)} \\ P(V) & = .90 * 0.40 + 0.3 * (1  0.40) = 0.667 \end{aligned} \]
\[ \begin{aligned} P(C) & = 0.40 & P(V) & = ? \\ P(VC) & = 0.90 & P(V\text{not } C) & = 0.3 \\ \end{aligned} \]
What is the probability that the salesman recieved a commission given that he is on vacation in Bermuda?
\[ \begin{aligned} P(CV) & = \frac{ 0.90 * 0.40 }{0.54} \\ & = 0.667 \end{aligned} \]
implying that there is a 67% chance that he recieved the comission.
Recall that Stat 202 will let you solve probability problems your own way.
brute force: (i) make a probability tree, then (ii) make a table, then (iii) find the relevant conditional probability.
elegant: Bayes' Rule, the Law of Total Probability, and other probability rules.
Polygraph (liedetector) tests are often routinely administered to employees or prospective employees in sensitive positions. Lie detector results are “better than chance, but well below perfection.” Typically, the test may conclude someone is a spy 80% of the time when he or she actually is a spy, but 16% of the time the test will conclude someone is a spy when he or she is not.
Let us assume that 1 in 1,000, or .001, of the employees in a certain highly classified workplace are actual spies.
Test may conclude someone is a spy 80% of the time when he or she actually is a spy, but 16% of the time the test will conclude someone is a spy when he or she is not. Assume that 1 in 1,000, or .001, are actual spies.
\[ P(S) = ? \]
\[ P(S) = 0.001 \]
\[ P(DS) \text{ or } P(SD)? \]
\[ P(DS) = 0.80 \]
\[ P(D\text{not }S) = ? \]
\[ P(D\text{not }S) = .16 \]
\[ P(S) = 0.001 \]
\[ P(DS) = 0.80 \]
\[ P(D\text{not }S) = .16 \]
If the polygraph detects a spy, are you convinced that the person is actually a spy?
\[ P(SD) \text{ or } P(DS)? \]
\[ P(SD) = \frac{P(DS)P(S)}{P(D)} \]
\[ P(D) = P(DS)P(S) \\ + P(D\text{not }S)P(\text{not }S) \]
Law of total probability: \[ \begin{aligned} P(D) & = P(DS)P(S) + P(D\text{not }S)P(\text{not }S) \\ & = .80 * .001 + .16 * (1  .001) \\ & = .161 \end{aligned} \]
Bayes' Rule \[ \begin{aligned} P(SD) & = \frac{P(DS)P(S)}{P(D)} \\ & = \frac{ 0.80 * 0.001 }{ .161} \\ & = 0.005 \end{aligned} \]
If the polygraph detects a spy, are you convinced that the person is actually a spy?
\[ P(SD) = 0.005 \]
implying that about one half of one percent of “detections” are actual spies.
Are you convinced?
The order of conditioning matters:
\[ P(SD) = 0.005 \]
implying that about one half of one percent of “detections” are actual spies.
\[ P(DS) = 0.80 \]
implying that 80% of actual spies are dectected by the test.
Careful attention is required!