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Chapter 2: Q88E (page 91)

The probability that an individual randomly selected from a particular population has a certain disease is \({\rm{.05}}\). A diagnostic test correctly detects the presence of the disease \({\rm{98\% }}\)of the time and correctly detects the absence of the disease \({\rm{99\% }}\)of the time. If the test is applied twice, the two test results are independent, and both are positive, what is the (posterior) probability that the selected individual has the disease? (Hint: Tree diagram with first-generation branches corresponding to Disease and No Disease, and second- and third-generation branches corresponding to results of the two tests.)

Short Answer

Expert verified

The selected individual has the disease

\(\begin{array}{c}{\rm{P( disease }}\mid {\rm{ positive twice )}} \rm &=& 0{\rm{.7016}}\\ \rm &=& 70{\rm{.16\% }}\end{array}\)

Step by step solution

01

Definition of Independent Probability

If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

02

Finding the (posterior) probability that the selected individual has the disease?

Given:

\({\rm{P( disease ) = 0}}{\rm{.05}}\)

In 98 percent of cases, the disease is correctly recognised, and the test is thus positive:

\(\begin{array}{c}{\rm{P( positive}}\mid \rm disease ) &=& 98\% \\ \rm &=& 0{\rm{.98}}\end{array}\)

In 99 percent of cases, the absence of the disease is correctly recognised, and the test is consequently negative:

\(\begin{array}{c}{\rm{P( not positive}}\mid \rm no disease ) &=& 99\% \\ \rm &=& 0{\rm{.99}}\end{array}\)

03

Apply complement rule

Complement rule:

\({\rm{P(notA) = 1 - P(A)}}\)

General multiplication rule:

\(\begin{array}{c}\rm P(A and B) &=& P(A)*P(B\mid {\rm{A)}}\\ \rm &=& P(B)*P(A \mid {\rm{B)}}\end{array}\)

For discontinuous or mutually exclusive occurrences, use the following addition rule:

\({\rm{P(A or B) = P(A) + P(B)}}\)

Multiplication rule for independent events:

\({\rm{P(A and B) = P(A)*P(B)}}\)

Definition Conditional probability:

\({\rm{P(B}}\mid {\rm{A) = }}\frac{{{\rm{P(A and B)}}}}{{{\rm{P(A)}}}}\)

Use the complement rule:

\(\begin{array}{c} \rm P( No disease ) &=& 1 - P( disease ) \\ \rm &=& 1 - 0 \rm .05 &=& 0{\rm{.95}}\\{\rm{P( detect }}\mid \rm no disease ) &=& 1 - P( not positive \mid \rm no disease )\\ \rm &=& 1 - 0{\rm{.99}}\\ \rm &=& 0{\rm{.01}}\end{array}\)

04

 Step 4: Explanation of the Solution

Use the general multiplication rule:

\(\begin{array}{c}\rm P( positive and disease ) &=& P( disease )*P( positive \mid {\rm{ disease )}}\\ \rm &=& 0{\rm{.05*0}}{\rm{.98}}\\ \rm &=& 0{\rm{.049}}\end{array}\)

\(\begin{array}{c}{\rm{P}}\left( {{\rm{ positive and no disease}}} \right)\rm &=& P\left( {{\rm{no disease}}} \right){\rm{*P(positive}}\mid {\rm{no disease )}}\\ \rm &=& 0{\rm{.950}}{\rm{.01}}\\ \rm &=& 0{\rm{.0095}}\end{array}\)

Add the corresponding probabilities:

\(\begin{array}{c} \rm P( positive ) &=& P( positive and disease ) + P( positive and no disease )\\ \rm &=& 0{\rm{.049 + 0}}{\rm{.0095}}\\ \rm &=& 0{\rm{.0585}}\end{array}\)

05

Explain the probability

Because the two tests are independent, we can apply the multiplication rule to both. We're going to assume

\(\begin{array}{c}\rm P( positive twice ) &=& P( positive )*P( positive )\\ \rm &=& 0{\rm{.0585*0}}{\rm{.0585}}\\ \rm &=& 0{\rm{.00342225}}\end{array}\)

P(disease and positive twice )=P( positive and disease ) \(*\)P(positive and disease )

\(\begin{array}{c}\rm &=& 0{\rm{.049*0}}{\rm{.049}}\\\rm &=& 0{\rm{.002401}}\end{array}\)

Use the following conditional probability definition:

\(\begin{array}{c}{\rm{P( disease }}\mid \rm positive twice ) &=& \frac{{{\rm{P( disease and positive twice )}}}}{{{\rm{P( positive twice )}}}}\\ &=& \frac{{0.002401}}{{0.00342225}} \approx 0.7016\\ \rm &=& 70{\rm{.16\% }}\end{array}\)

Therefore , the selected individual has the disease

\(\begin{array}{c}{\rm{P( disease }}\mid {\rm{ positive twice )}} \rm &=& 0{\rm{.7016}}\\ \rm &=& 70{\rm{.16\% }}\end{array}\)

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Most popular questions from this chapter

The population of a particular country consists of three ethnic groups. Each individual belongs to one of the four major blood groups. The accompanying joint probability table gives the proportions of individuals in the various ethnic group-blood group combinations.

Suppose that an individual is randomly selected from the population, and define events by \({\rm{A = }}\)\{type A selected), \({\rm{B = }}\) (type B selected), and \({\rm{C = }}\) (ethnic group \({\rm{3}}\) selected).

a. Calculate\({\rm{P(A),P(C)}}\), and \({\rm{P(A{C}C)}}\).

b. Calculate both \({\rm{P(A}}\mid {\rm{C)}}\)and \({\rm{P(C}}\mid {\rm{A)}}\), and explain in context what each of these probabilities represents.

c. If the selected individual does not have type B blood, what is the probability that he or she is from ethnic group\({\rm{1}}\)?

A certain shop repairs both audio and video components. Let A denote the event that the next component brought in for repair is an audio component, and let B be the event that the next component is a compact disc player (so the event B is contained in A). Suppose that \(P\left( A \right) = 0.6\)and\(P\left( B \right) = 0.05\). What is\(P\left( {B|A} \right)\)?

An engineering construction firm is currently working on power plants at three different sites. Let Aidenote the event that the plant at site i is completed by the contract date. Use the operations of union, intersection, and complementation to describe each of the following events in terms of \({A_1}\), \({A_2}\), and \({A_3}\), draw a Venn diagram, and shade the region corresponding to each one.

a. At least one plant is completed by the contract date.

b. All plants are completed by the contract date.

c. Only the plant at site 1 is completed by the contract date.

d. Exactly one plant is completed by the contract date.

e. Either the plant at site 1 or both of the other two plants are completed by the contract date.

The three most popular options on a certain type of newcar are a built-in GPS (A), a sunroof (B), and an automatictransmission (C). If 40% of all purchasers request A, 55% request B, 70% request C, 63% request Aor B,77% request Aor C, 80% request Bor C, and 85% request Aor Bor C, determine the probabilities of the following events. (Hint:鈥淎or B鈥� is the event that at leastone of the two options is requested; try drawing a Venn

diagram and labeling all regions.)

a. The next purchaser will request at least one of thethree options.

b. The next purchaser will select none of the three options.

c. The next purchaser will request only an automatictransmission and not either of the other two options.

d. The next purchaser will select exactly one of thesethree options.

Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The first inspector detects \({\rm{90\% }}\)of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on \({\rm{20\% }}\)of all defective components. What is the probability that the following occur?

a. A defective component will be detected only by the first inspector? By exactly one of the two inspectors?

b. All three defective components in a batch escape detection by both inspectors (assuming inspections of different components are independent of one another)?
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