5.2 Problems on conditional independence (Page 2/4)

Applied probability Page 2 / 4

A manufacturer claims to have improved the reliability of his product. Formerly, the product had probability 0.65 of operating 1000 hours without failure. The manufacturerclaims this probability is now 0.80. A sample of size 20 is tested. Determine the odds favoring the new probability for various numbers of surviving units under theassumption the prior odds are 1 to 1. How many survivors would be required to make the claim creditable?

Let E ₁ be the event the probability is 0.80 and E ₂ be the event the probability is 0.65. Assume $P (E_{1}) / P (E_{2}) = 1$ .

\frac{P (E_{1} | S_{n} = k)}{P (E_{2} | S_{n} = k)} = \frac{P (E_{1})}{P (E_{2})} • \frac{P (S_{n} = k | E_{1})}{P (S_{n} = k | E_{2})}

k = 1:20;
odds = ibinom(20,0.80,k)./ibinom(20,0.65,k);disp([k;odds]')- - - - - - - - - - - -
   13.0000    0.295814.0000    0.637215.0000    1.3723   % Need at least 15 or 16 successes
   16.0000    2.955817.0000    6.3663
   18.0000   13.712119.0000   29.5337
   20.0000   63.6111

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A real estate agent in a neighborhood heavily populated by affluent professional persons is working with a customer. The agent is trying toassess the likelihood the customer will actually buy. His experience indicates the following: if H is the event the customer buys, S is the event the customer is a professional with good income, and E is the event the customer drives a prestigious car, then

P (S) = 0.7 P (S | H) = 0.90 P (S | H^{c}) = 0.2 P (E | S) = 0.95 P (E | S^{c}) = 0.25

Since buying a house and owning a prestigious car are not related for a given owner, it seems reasonable to suppose $P (E | H S) = P (E | H^{c} S)$ and $P (E | H S^{c}) = P (E | H^{c} S^{c})$ . The customer drives a Cadillac. What are the odds he will buy a house?

Assumptions amount to ${H, E} ci | S$ and $ci | S^{c}$ .

\frac{P (H | S)}{P (H^{c} | S)} = \frac{P (H) P (S | H)}{P (H^{c}) P (S | H^{c})}

P (S) = P (H) P (S | H) + [1 - P (H)] P (S | H^{c}) which implies

P (H) = \frac{P (S) - P (S | H^{c})}{P (S | H) - P (S | H^{c})} = 5 / 7 so that \frac{P (H | S)}{P (H^{c} | S)} = \frac{5}{2} • \frac{0.9}{0.2} = \frac{45}{4}

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In deciding whether or not to drill an oil well in a certain location, a company undertakes a geophysical survey. On the basis of past experience, thedecision makers feel the odds are about four to one favoring success. Various other probabilities can be assigned on the basis of past experience. Let

H be the event that a well would be successful
S be the event the geological conditions are favorable
E be the event the results of the geophysical survey are positive

The initial, or prior, odds are $P (H) / P (H^{c}) = 4$ . Previous experience indicates

P (S | H) = 0.9 P (S | H^{c}) = 0.20 P (E | S) = 0.95 P (E | S^{c}) = 0.10

Make reasonable assumptions based on the fact that the result of the geophysical survey depends upon the geological formations and not on the presence or absence of oil. Theresult of the survey is favorable. Determine the posterior odds $P (H | E) / P (H^{c} | E)$ .

\frac{P (H | E)}{P (H^{c} | E)} = \frac{P (H)}{P (H^{c})} • \frac{P (S | H) P (E | S) + P (S^{c} | H) P (E | S^{c})}{P (S | H^{c}) P (E | S) + P (S^{c} | H^{c}) P (E | S^{c})}

= 4 • \frac{0.90 • 0.95 + 0.10 • 0.10}{0.20 • 0.95 + 0.80 • 0.10} = 12.8148

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A software firm is planning to deliver a custom package. Past experience indicates the odds are at least four to one that it will pass customeracceptance tests. As a check, the program is subjected to two different benchmark runs. Both are successful. Given the following data, what are the odds favoring successfuloperation in practice? Let

H be the event the performance is satisfactory
S be the event the system satisfies customer acceptance tests
E ₁ be the event the first benchmark tests are satisfactory.
E ₂ be the event the second benchmark test is ok.

Under the usual conditions, we may assume ${H, E_{1}, E_{2}} ci | S$ and $ci | S^{c}$ . Reliability data show

P (H | S) = 0.95, P (H | S^{c}) = 0.45

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

Determine the posterior odds $P (H | E_{1} E_{2}) / P (H^{c} | E_{1} E_{2})$ .

\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 • 0.45 • 0.25 • 0.20}{0.80 • 0.05 • 0.90 • 0.95 + 0.20 • 0.55 • 0.25 • 0.20} = 16.64811

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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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