5.2 Problems on conditional independence (Page 4/4)

Applied probability Page 4 / 4

A software company has developed a new computer game designed to appeal to teenagers and young adults.It is felt that there is good probability it will appeal to college students, and that if it appeals to college students it will appeal to a general youthmarket. To check the likelihood of appeal to college students, it is decided to test first by a sales campaign at Rice and University of Texas, Austin.The following analysis of the situation is made.

$H =$ the event the sales to the general market will be good
$S =$ the event the game appeals to college students
$E_{1} =$ the event the sales are good at Rice
$E_{2} =$ the event the sales are good at UT, Austin

Since the tests are for the reception are at two separate universities and are operationally independent, it seems reasonable to assume ${H, E_{1}, E_{2}} ci | S$ and ${H, E_{1}, E_{2}} ci | S^{c}$ . Because of its previous experience in game sales, the managers think $P (S) = 0.80$ . Also, experience suggests

$P (H \| S) = 0.95$	$P (E_{1} \| S) = 0.90$	$P (E_{2} \| S) = 0.95$
$P (H \| S^{c}) = 0.30$	$P (E_{1} \| S^{c}) = 0.20$	$P (E_{2} \| S^{c}) = 0.25$

Determine the posterior odds favoring H if sales results are satisfactory at both schools.

\frac{P (H | E_{1} E_{2})}{P (H^{c} | E_{1} E_{2})} = \frac{P (H E_{1} E_{2} S) + P (H E_{1} E_{2} S^{c})}{P (H^{c} E_{1} E_{2} S) + P (H^{c} E_{1} E_{2} S^{c})}

= \frac{P (S) P (H | S) P (E_{1} | S) P (E_{2} | S) + P (S^{c}) P (H | S^{c}) P (E_{1} | S^{c}) P (E_{2} | S^{c})}{P (S) P (H^{c} | S) P (E_{1} | S) P (E_{2} | S) + P (S^{c}) P (H^{c} | S^{c}) P (E_{1} | S^{c}) P (E_{2} | S^{c})}

= \frac{0.80 • 0.95 • 0.90 • 0.95 + 0.20 • 0.30 • 0.20 • 0.25}{0.80 • 0.05 • 0.90 • 0.95 + 0.20 • 0.70 • 0.20 • 0.25} = 15.8447

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In a region in the Gulf Coast area, oil deposits are highly likely to be associated with underground salt domes. If H is the event that an oil deposit is present in an area, and S is the event of a salt dome in the area, experience indicates $P (S | H) = 0.9$ and $P (S | H^{c}) = 0.1$ . Company executives believe the odds favoring oil in the area is at least 1 in 10. It decides to conduct two independentgeophysical surveys for the presence of a salt dome. Let $E_{1}, E_{2}$ be the events the surveys indicate a salt dome. Because the surveys are testsfor the geological structure, not the presence of oil, and the tests are carried out in an operationally independent manner, it seems reasonable to assume ${H, E_{1}, E_{2}} ci | S$ and $ci | S^{c}$ . Data on the reliability of the surveys yield the following probabilities

P (E_{1} | S) = 0.95 P (E_{1} | S^{c}) = 0.05 P (E_{2} | S) = 0.90 P (E_{2} | S^{c}) = 0.10

Determine the posterior odds $\frac{P (H | E_{1} E_{2})}{P (H^{c} | E_{1} E_{2})}$ . Should the well be drilled?

\frac{P (H | E_{1} E_{2})}{P (H^{c} | E_{1} E_{2})} = \frac{P (H E_{1} E_{2} S) + P (H E_{1} E_{2} S^{c})}{P (H^{c} E_{1} E_{2} S) + P (H^{c} E_{1} E_{2} S^{c})}

P (H E_{1} E_{2} S) = P (H) P (S | H) P (E_{1} | S H) P (E_{2} | S H E_{1}) = P (H) P (S | H) P (E_{1} | S) P (E_{2} | S)

with similar expressions for the other terms.

\frac{P (H | E_{1} E_{2})}{P (H^{c} | E_{1} E_{2})} = \frac{1}{10} • \frac{0.9 • 0.95 • 0.90 + 0.10 • 0.05 • 0.10}{0.1 • 0.95 • 0.90 + 0.90 • 0.05 • 0.10} = 0.8556

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A sample of 150 subjects is taken from a population which has two subgroups. The subgroup membership of each subject in the sample is known. Each individual is asked a batteryof ten questions designed to be independent, in the sense that the answer to any one is not affected by the answer to any other. The subjects answer independently. Data on the resultsare summarized in the following table:

GROUP 1 (84 members)	GROUP 2 (66 members)
Q	Yes	No	Unc	Yes	No	Unc
1	51	26	7	27	34	5
2	42	32	10	19	43	4
3	19	54	11	39	22	5
4	24	53	7	38	19	9
5	27	52	5	28	33	5
6	49	19	16	19	41	6
7	16	59	9	37	21	8
8	47	32	5	19	42	5
9	55	17	12	27	33	6
10	24	53	7	39	21	6

Assume the data represent the general population consisting of these two groups, so that the data may be used to calculate probabilities and conditionalprobabilities.

Several persons are interviewed. The result of each interview is a “profile” of answers to the questions. The goal is to classify the person in one of the two subgroups

For the following profiles, classify each individual in one of the subgroups

y, n, y, n, y, u, n, u, y. u
n, n, u, n, y, y, u, n, n, y
y, y, n, y, u, u, n, n, y, y

% file
npr05_16.m % Data for
[link] A = [51 26  7; 42 32 10; 19 54 11; 24 53  7; 27 52  5;
     49 19 16; 16 59  9; 47 32  5; 55 17 12; 24 53  7];
B = [27 34  5; 19 43  4; 39 22  5; 38 19  9; 28 33  5;19 41  6; 37 21  8; 19 42  5; 27 33  6; 39 21  6];disp('Call for oddsdf')
npr05_16 Call for oddsdf
oddsdfEnter matrix A of frequencies for calibration group 1  A
Enter matrix B of frequencies for calibration group 2  BNumber of questions = 10
Answers per question = 3Enter code for answers and call for procedure "odds"
y = 1;n = 2;
u = 3;odds
Enter profile matrix E  [y n y n y u n u y u]Odds favoring Group 1:   3.743
Classify in Group 1odds
Enter profile matrix E  [n n u n y y u n n y]Odds favoring Group 1:   0.2693
Classify in Group 2odds
Enter profile matrix E  [y y n y u u n n y y]Odds favoring Group 1:   5.286
Classify in Group 1

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The data of [link] , above, are converted to conditional probabilities and probabilities, as follows (probabilities are rounded to two decimal places).

GROUP 1 $P (G_{1}) = 0.56$	GROUP 2 $P (G_{2}) = 0.44$
Q	Yes	No	Unc	Yes	No	Unc
1	0.61	0.31	0.08	0.41	0.51	0.08
2	0.50	0.38	0.12	0.29	0.65	0.06
3	0.23	0.64	0.13	0.59	0.33	0.08
4	0.29	0.63	0.08	0.57	0.29	0.14
5	0.32	0.62	0.06	0.42	0.50	0.08
6	0.58	0.23	0.19	0.29	0.62	0.09
7	0.19	0.70	0.11	0.56	0.32	0.12
8	0.56	0.38	0.06	0.29	0.63	0.08
9	0.65	0.20	0.15	0.41	0.50	0.09
10	0.29	0.63	0.08	0.59	0.32	0.09

For the following profiles classify each individual in one of the subgroups.

y, n, y, n, y, u, n, u, y, u
n, n, u, n, y, y, u, n, n, y
y, y, n, y, u, u, n, n, y, y


npr05_17 % file
npr05_17.m % Data for
[link] PG1 = 84/150;
PG2 = 66/125;A = [0.61 0.31 0.08
     0.50 0.38 0.120.23 0.64 0.13
     0.29 0.63 0.080.32 0.62 0.06
     0.58 0.23 0.190.19 0.70 0.11
     0.56 0.38 0.060.65 0.20 0.15
     0.29 0.63 0.08];B = [0.41 0.51 0.08
     0.29 0.65 0.060.59 0.33 0.08
     0.57 0.29 0.140.42 0.50 0.08
     0.29 0.62 0.090.56 0.32 0.12
     0.29 0.64 0.080.41 0.50 0.09
     0.59 0.32 0.09];
disp('Call for oddsdp')Call for oddsdp
oddsdpEnter matrix A of conditional probabilities for Group 1  A
Enter matrix B of conditional probabilities for Group 2  BProbability p1 an individual is from Group 1  PG1
Number of questions = 10Answers per question = 3
 Enter code for answers and call for procedure "odds"y = 1;
n = 2;u = 3;
oddsEnter profile matrix E  [y n y n y u n u y u]
Odds favoring Group 1:   3.486Classify in Group 1
oddsEnter profile matrix E  [n n u n y y u n n y]
Odds favoring Group 1:   0.2603Classify in Group 2
oddsEnter profile matrix E  [y y n y u u n n y y]
Odds favoring Group 1:   5.162Classify in Group 1

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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$P (H \| S) = 0.95$	$P (E_{1} \| S) = 0.90$	$P (E_{2} \| S) = 0.95$
$P (H \| S^{c}) = 0.30$	$P (E_{1} \| S^{c}) = 0.20$	$P (E_{2} \| S^{c}) = 0.25$