5.1 Patterns of probable inference (Page 3/6)

Applied probability Page 3 / 6

Independent evidence for a symptom

In the previous cases, we consider only a single item of evidence for a symptom. But it may be desirable to have a “second opinion.” We suppose the tests are for thesymptom and are not directly related to the hypothetical condition. If the tests are operationally independent, we could reasonably assume

\begin{matrix} P (E_{1} | S E_{2}) = P (E_{1} | S E_{2}^{c}) & {E_{1}, E_{2}} ci | S \\ P (E_{1} | S H) = P (E_{1} | S H^{c}) & {E_{1}, H} ci | S \\ P (E_{2} | S H) = P (E_{2} | S H^{c}) & {E_{2}, H} ci | S \\ P (E_{1} E_{2} | S H) = P (E_{1} E_{2} | S H^{c}) & {E_{1} E_{2}, H} ci | S \end{matrix}

This implies ${E_{1}, E_{2}, H} ci | S$ . A similar condition holds for S ^c . As for a single test, there are four cases,depending on the tie between S and H . We consider a "case a" example.

A market survey problem

A food company is planning to market nationally a new breakfast cereal. Its executives feel confident that the odds are at least 3 to 1 the productwould be successful. Before launching the new product, the company decidesto investigate a test market. Previous experience indicates that the reliability of the test market is such that if the national marketis favorable, there is probability 0.9 that the test market is also. On the other hand, if the national market is unfavorable, there is a probabilityof only 0.2 that the test market will be favorable. These facts lead to the following analysis. Let

H be the event the national market is favorable (hypothesis)

S be the event the test market is favorable (symptom)

The initial data are the following probabilities, based on past experience:

(a) Prior odds: $P (H) / P (H^{c}) = 3$
(b) Reliability of the test market: $P (S | H) = 0.9 P (S | H^{c}) = 0.2$

If it were known that the test market is favorable, we should have

\frac{P (H | S)}{P (H^{c} | S)} = \frac{P (S | H) P (H)}{P (S | H^{c}) P (H^{c})} = \frac{0.9}{0.2} \cdot 3 = 13.5

Unfortunately, it is not feasible to know with certainty the state of the test market. The company decision makers engage two market survey companiesto make independent surveys of the test market. The reliability of the companies may be expressed as follows. Let

E ₁ be the event the first company reports a favorable test market.
E ₂ be the event the second company reports a favorable test market.

On the basis of previous experience, the reliability of the evidence about the test market (the symptom) is expressed in the followingconditional probabilities.

P (E_{1} | S) = 0.9 P (E_{1} | S^{c}) = 0.3 P (E_{2} | S) = 0.8 B (E_{2} | S^{c}) = 0.2

Both survey companies report that the test market is favorable. What is the probability the national market is favorable, given this result?

Soution

The two survey firms work in an “operationally independent” manner. The report of either company is unaffected by the work of the other. Also, eachreport is affected only by the condition of the test market— regardless of what the national market may be. According to the discussion above, we should beable to assume

{E_{1}, E_{2}, H} ci | S and {E_{1}, E_{2}, H} ci | S^{c}

We may use a pattern similar to that in Example 2, as follows:

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

= 3 \cdot \frac{0.9 \cdot 0.9 \cdot 0.8 + 0.1 \cdot 0.3 \cdot 0.2}{0.2 \cdot 0.9 \cdot 0.8 + 0.8 \cdot 0.3 \cdot 0.2} = \frac{327}{32} \approx 10.22

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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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