Free Math Worksheets

- Treat all award recipients as the new total group. - Find what fraction of that group is in the music program.

1. The total number of award recipients is \(144+72=216\). 2. Of these recipients, \(144\) are in the music program. 3. Therefore, the probability is \(\frac{144}{216}=\frac{2}{3}\).

\(\frac{2}{3}\approx66.7\%\)

- Draw a tree beginning with the scanner used. - Add all complete paths that end in an incorrect reading. - In the fraction in part c, identify the joint event in the numerator and the conditioned event in the denominator. - Correct and incorrect readings are complementary events.

1. Add the two error paths: \(P(F)=0.70\cdot 0.02+0.30\cdot 0.05=0.014+0.015=0.029\). 2. The joint probability of Scanner A and an error is \(0.70\cdot 0.02=0.014\). Therefore, \(P(A\mid F)=\frac{0.014}{0.029}=\frac{14}{29}\approx 0.4828\). 3. In the given expression, \(0.30\cdot 0.95=P(B\cap\overline{F})\), and the denominator is \(P(\overline{F})\). Thus, the expression is \(P(B\mid\overline{F})\), the probability that a correctly read package was processed by Scanner B.

a) \(P(F)=0.029\), or \(2.9\%\). b) \(P(A\mid F)=\frac{14}{29}\approx 0.4828\), or about \(48.28\%\). c) The expression is the probability that a package was processed by Scanner B, given that it was read correctly.

- An alarm can occur through a true-positive path or a false-positive path. - Reverse conditional probabilities by comparing the relevant joint path with all paths in the conditioned group. - For part c, use the paths that end in no alarm. - In part d, first find the undamaged proportion among all alarmed packages.

1. Add the alarm paths: \(P(S)=0.02\cdot 0.94+0.98\cdot 0.04=0.0188+0.0392=0.058\). 2. \(P(D\mid S)=\frac{P(D\cap S)}{P(S)}=\frac{0.0188}{0.058}=\frac{47}{145}\approx 0.3241\). 3. The no-alarm probability is \(P(\overline{S})=1-0.058=0.942\). Also, \(P(D\cap\overline{S})=0.02\cdot 0.06=0.0012\). Thus, \(P(D\mid\overline{S})=\frac{0.0012}{0.942}=\frac{1}{785}\approx 0.001274\). 4. The undamaged proportion among alarmed packages is \(P(\overline{D}\mid S)=\frac{0.0392}{0.058}=\frac{98}{145}\approx 0.6759\). The expected count is \(200\cdot\frac{98}{145}\approx 135.17\).

a) \(P(S)=0.058\), or \(5.8\%\). b) \(P(D\mid S)=\frac{47}{145}\approx 0.3241\), or about \(32.41\%\). c) \(P(D\mid\overline{S})\approx 0.001274\), or about \(0.1274\%\). d) About \(135\) packages.

- Enter all given counts in the table before using row and column totals. - For a tree branch, divide the branch count by the total at the preceding node. - In part 3, the conditioned group consists only of defective components. - Distinguish \(P(D\mid M_1)\) from \(P(M_1\mid D)\).

1. The machine totals are \(n(M_1)=2500\) and \(n(M_2)=4000-2500=1500\). The quality totals are \(n(D)=110\) and \(n(A)=4000-110=3890\). 2. The inner counts are \(n(M_1\cap D)=50\), \(n(M_1\cap A)=2500-50=2450\), \(n(M_2\cap D)=110-50=60\), and \(n(M_2\cap A)=1500-60=1440\). 3. The first-stage probabilities are \(P(M_1)=\frac{2500}{4000}=0.625\) and \(P(M_2)=\frac{1500}{4000}=0.375\). 4. The second-stage probabilities are \(P(D\mid M_1)=\frac{50}{2500}=0.02\), \(P(A\mid M_1)=0.98\), \(P(D\mid M_2)=\frac{60}{1500}=0.04\), and \(P(A\mid M_2)=0.96\). 5. Reversing the condition, \(P(M_1\mid D)=\frac{50}{110}=\frac{5}{11}\approx 0.4545\).

1. <table> <tr><td></td><td>\(D\)</td><td>\(A\)</td><td>Total</td></tr> <tr><td>\(M_1\)</td><td>\(50\)</td><td>\(2450\)</td><td>\(2500\)</td></tr> <tr><td>\(M_2\)</td><td>\(60\)</td><td>\(1440\)</td><td>\(1500\)</td></tr> <tr><td>Total</td><td>\(110\)</td><td>\(3890\)</td><td>\(4000\)</td></tr> </table> 2. \(P(M_1)=0.625\), \(P(M_2)=0.375\), \(P(D\mid M_1)=0.02\), \(P(A\mid M_1)=0.98\), \(P(D\mid M_2)=0.04\), and \(P(A\mid M_2)=0.96\). 3. \(P(M_1\mid D)=\frac{5}{11}\approx 0.4545\), or about \(45.45\%\).

- Branch first by disease status and then by test result. - A positive result can occur on two different paths. - For part b, compare the true-positive path with all positive-result paths. - Consider whether disease status and test results are symmetric, equally likely outcomes.

1. Let \(D\) represent having the disease and \(T\) a positive result. Then \(P(D)=0.02\), \(P(\overline{D})=0.98\), \(P(T\mid D)=0.98\), and \(P(T\mid\overline{D})=1-0.95=0.05\). 2. Add the positive-result paths: \(P(T)=0.02\cdot 0.98+0.98\cdot 0.05=0.0196+0.049=0.0686\). 3. Reverse the condition: \(P(D\mid T)=\frac{0.0196}{0.0686}=\frac{2}{7}\approx 0.2857\). 4. The prevalence, sensitivity, and specificity come from observed long-run relative frequencies. The outcomes are not assumed to be equally likely, so an equally likely outcomes model does not apply.

a) \(P(T)=0.0686\), or \(6.86\%\). b) \(P(D\mid T)=\frac{2}{7}\approx 0.2857\), or about \(28.57\%\). c) The probabilities are empirical because they are estimated from observed population and test data, not from a set of equally likely outcomes.

- Identify the marginal and conditional probabilities stated in Report A. - Use the two possible brands to find the total probability of being younger than \(30\). - Reverse each condition with the definition of conditional probability or Bayes' rule. - A tree diagram can organize the path probabilities.

1. Let \(T\) be the event that a respondent uses TechX, and let \(Y\) be the event that the respondent is younger than \(30\). Report A gives \(P(T)=0.35\), \(P(Y\mid T)=0.70\), and \(P(Y\mid\overline{T})=0.40\). 2. By the law of total probability, \(P(Y)=0.70\cdot 0.35+0.40\cdot 0.65=0.505\), which matches \(50.5\%\). 3. The younger-group market share is \(P(T\mid Y)=\frac{P(Y\mid T)P(T)}{P(Y)}=\frac{0.245}{0.505}\approx 0.4851\), which rounds to \(48.5\%\). 4. For respondents age \(30\) or older, \(P(T\mid\overline{Y})=\frac{P(\overline{Y}\mid T)P(T)}{P(\overline{Y})}=\frac{0.30\cdot 0.35}{0.495}\approx 0.2121\), which rounds to \(21.2\%\). 5. All three values in Report B agree with Report A after rounding.

Yes. Report A implies that \(50.5\%\) of respondents are younger than \(30\), TechX has about \(48.5\%\) market share among younger respondents, and about \(21.2\%\) market share among respondents age \(30\) or older.

- Write each given percentage as a marginal or conditional probability. - First find the total probability that a vehicle owner lives in a large city. - Use path probabilities to find the intersections involving electric vehicles. - Compare the two conditional probabilities using a ratio.

1. Let \(E\) mean owning an electric vehicle and \(C\) mean living in a large city. The data give \(P(E)=0.15\), \(P(C\mid E)=0.80\), and \(P(C\mid\overline{E})=0.55\). 2. The total probability of living in a large city is \(P(C)=0.80\cdot 0.15+0.55\cdot 0.85=0.5875\). 3. Therefore, \(P(E\mid C)=\frac{0.80\cdot 0.15}{0.5875}\approx 0.2043\). 4. Also, \(P(E\mid\overline{C})=\frac{0.20\cdot 0.15}{1-0.5875}=\frac{0.03}{0.4125}\approx 0.0727\). 5. The ratio is \(\frac{0.2043}{0.0727}\approx 2.81\), so the first probability is more than twice the second.

The claim is true. \(P(E\mid C)\approx 0.2043\), or \(20.43\%\), and \(P(E\mid\overline{C})\approx 0.0727\), or \(7.27\%\). The first is about \(2.81\) times the second.

- Use the given conditional percentage to find the business-and-electric intersection. - Complete the table by subtracting from row and column totals. - In part 3, identify the event after “of electric vehicles” as the condition. - A second-stage branch probability is conditioned on the first-stage event.

1. The business-and-electric cell is \(P(G\cap E)=P(G)P(E\mid G)=0.60\cdot 0.25=0.15\). 2. The remaining cells are \(P(G\cap\overline{E})=0.60-0.15=0.45\), \(P(\overline{G}\cap E)=0.20-0.15=0.05\), and \(P(\overline{G}\cap\overline{E})=0.40-0.05=0.35\). 3. \(P(E\mid\overline{G})=\frac{P(E\cap\overline{G})}{P(\overline{G})}=\frac{0.05}{0.40}=0.125\). 4. The requested proportion in part 3 is \(P(G\mid E)\). A tree that branches first by \(E\) displays this conditional probability on a second-stage branch from the \(E\) node.

1. <table> <tr><td></td><td>\(E\)</td><td>\(\overline{E}\)</td><td>Total</td></tr> <tr><td>\(G\)</td><td>\(0.15\)</td><td>\(0.45\)</td><td>\(0.60\)</td></tr> <tr><td>\(\overline{G}\)</td><td>\(0.05\)</td><td>\(0.35\)</td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.20\)</td><td>\(0.80\)</td><td>\(1\)</td></tr> </table> 2. \(P(E\mid\overline{G})=0.125\), or \(12.5\%\). 3. Branch first by \(E\), because \(P(G\mid E)\) is then a second-stage branch probability from the \(E\) node.

- Convert each conditional percentage into a joint relative frequency. - Complete each row by subtracting from its marginal probability. - In part 3, use students with poor sleep quality as the conditioned group.

1. The joint relative frequencies involving poor sleep are \(P(S\cap Q)=0.40\cdot 0.70=0.28\) and \(P(\overline{S}\cap Q)=0.60\cdot 0.20=0.12\). 2. The other cells are \(P(S\cap\overline{Q})=0.40-0.28=0.12\) and \(P(\overline{S}\cap\overline{Q})=0.60-0.12=0.48\). 3. The poor-sleep marginal probability is \(P(Q)=0.28+0.12=0.40\). 4. Reversing the condition, \(P(S\mid Q)=\frac{P(S\cap Q)}{P(Q)}=\frac{0.28}{0.40}=0.70\).

1. <table> <tr><td></td><td>\(Q\)</td><td>\(\overline{Q}\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(0.28\)</td><td>\(0.12\)</td><td>\(0.40\)</td></tr> <tr><td>\(\overline{S}\)</td><td>\(0.12\)</td><td>\(0.48\)</td><td>\(0.60\)</td></tr> <tr><td>Total</td><td>\(0.40\)</td><td>\(0.60\)</td><td>\(1\)</td></tr> </table> 2. \(P(Q)=0.40\), or \(40\%\). 3. \(P(S\mid Q)=0.70\), or \(70\%\).

- Identify which percentages describe subgroups. - Use complements to find the probabilities of not using the library daily. - Part 1 reverses the direction of the given condition. - In each part, identify the group named after “given that.”

1. Let \(M\) represent a master’s student, \(U\) an undergraduate, and \(L\) daily library use. Then \(P(M)=0.42\), \(P(U)=0.58\), \(P(L\mid M)=0.15\), and \(P(L\mid U)=0.25\). 2. The complementary conditional probabilities are \(P(\overline{L}\mid M)=0.85\) and \(P(\overline{L}\mid U)=0.75\). 3. The overall probability of not using the library daily is \(P(\overline{L})=0.42\cdot 0.85+0.58\cdot 0.75=0.357+0.435=0.792\). 4. Therefore, \(P(U\mid\overline{L})=\frac{0.58\cdot 0.75}{0.792}=\frac{145}{264}\approx 0.5492\). 5. The second probability is the direct branch value \(P(\overline{L}\mid U)=0.75\).

1. \(P(U\mid\overline{L})=\frac{145}{264}\approx 0.5492\), or about \(54.92\%\). 2. \(P(\overline{L}\mid U)=0.75\), or \(75\%\).

- Translate each stated percentage into conditional probability notation. - Distinguish “field sales and no company car” from “field sales given no company car.” - A tree diagram or two-way table can organize the proportions. - Part 2 asks for the complement of a percentage given directly in the problem.

1. Let \(F\) mean field sales and \(C\) mean commuting in a company car. Then \(P(F)=0.35\), \(P(\overline{F})=0.65\), \(P(C\mid F)=0.80\), and \(P(C\mid\overline{F})=0.10\). 2. The complementary conditional probabilities are \(P(\overline{C}\mid F)=0.20\) and \(P(\overline{C}\mid\overline{F})=0.90\). 3. The total probability of not using a company car is \(P(\overline{C})=0.20\cdot 0.35+0.90\cdot 0.65=0.655\). 4. Therefore, \(P(F\mid\overline{C})=\frac{0.20\cdot 0.35}{0.655}=\frac{0.07}{0.655}\approx 0.1069\). 5. For part 2, \(P(\overline{C}\mid F)=0.20\).

1. \(P(F\mid\overline{C})\approx 0.1069\), or about \(10.69\%\). 2. \(P(\overline{C}\mid F)=0.20\), or \(20\%\).

- Complete complementary branches so that probabilities leaving each node sum to \(1\). - Multiply along a complete path. - A positive result can occur on two paths. - For part b, compare the trait-and-positive path with all positive paths.

1. The missing branch probabilities are \(P(\overline{T})=0.95\), \(P(-\mid T)=0.05\), and \(P(-\mid\overline{T})=0.90\). 2. The four path probabilities are \(P(T\cap +)=0.05\cdot 0.95=0.0475\), \(P(T\cap -)=0.05\cdot 0.05=0.0025\), \(P(\overline{T}\cap +)=0.95\cdot 0.10=0.095\), and \(P(\overline{T}\cap -)=0.95\cdot 0.90=0.855\). 3. The total positive probability is \(P(+)=0.0475+0.095=0.1425\). 4. Therefore, \(P(T\mid +)=\frac{0.0475}{0.1425}=\frac{1}{3}\).

a) The missing branches are \(0.95\), \(0.05\), and \(0.90\). The path probabilities are \(P(T\cap +)=0.0475\), \(P(T\cap -)=0.0025\), \(P(\overline{T}\cap +)=0.095\), and \(P(\overline{T}\cap -)=0.855\). b) \(P(T\mid +)=\frac{1}{3}\approx 0.3333\), or about \(33.33\%\).

- Each inner table cell is a complete path probability. - Add down each column to find the total probability for that satisfaction level. - In each conditional probability, use the conditioned satisfaction group as the denominator.

1. Multiply along each path: \(P(T\cap S)=0.60\cdot 0.50=0.30\), \(P(T\cap N)=0.60\cdot 0.30=0.18\), \(P(T\cap D)=0.60\cdot 0.20=0.12\), \(P(C\cap S)=0.40\cdot 0.30=0.12\), \(P(C\cap N)=0.40\cdot 0.40=0.16\), and \(P(C\cap D)=0.40\cdot 0.30=0.12\). 2. The column totals are \(P(S)=0.42\), \(P(N)=0.34\), and \(P(D)=0.24\). 3. \(P(T\mid S)=\frac{0.30}{0.42}=\frac{5}{7}\approx 0.7143\). Thus, about \(71.43\%\) of satisfied commuters use the train. 4. \(P(C\mid D)=\frac{0.12}{0.24}=0.50\).

a) <table> <tr><td></td><td>\(S\)</td><td>\(N\)</td><td>\(D\)</td><td>Total</td></tr> <tr><td>\(T\)</td><td>\(0.30\)</td><td>\(0.18\)</td><td>\(0.12\)</td><td>\(0.60\)</td></tr> <tr><td>\(C\)</td><td>\(0.12\)</td><td>\(0.16\)</td><td>\(0.12\)</td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.42\)</td><td>\(0.34\)</td><td>\(0.24\)</td><td>\(1\)</td></tr> </table> b) \(P(T\mid S)=\frac{5}{7}\approx 0.7143\). About \(71.43\%\) of satisfied commuters use the train. c) \(P(C\mid D)=0.50\), so \(50\%\) of dissatisfied commuters use a car.

- Add all complete paths that end in \(B\). - The same joint probabilities appear in either ordering of the tree. - Divide a joint path probability by the new first-stage probability to find a reversed branch.

1. Add the two paths ending in \(B\): \(P(B)=0.30\cdot 0.60+0.70\cdot 0.10=0.18+0.07=0.25\). 2. Therefore, \(P(\overline{B})=0.75\). 3. Reverse the branches from \(B\): \(P(A\mid B)=\frac{0.18}{0.25}=0.72\) and \(P(\overline{A}\mid B)=0.28\). 4. The joint probability \(P(A\cap\overline{B})=0.30\cdot 0.40=0.12\). Thus, \(P(A\mid\overline{B})=\frac{0.12}{0.75}=0.16\), and \(P(\overline{A}\mid\overline{B})=0.84\).

a) \(P(B)=0.25\). b) First stage: \(P(B)=0.25\), \(P(\overline{B})=0.75\). After \(B\): \(P(A\mid B)=0.72\), \(P(\overline{A}\mid B)=0.28\). After \(\overline{B}\): \(P(A\mid\overline{B})=0.16\), \(P(\overline{A}\mid\overline{B})=0.84\).

- Multiply along each complete path. - Add the paths ending in the same test result. - In the reversed tree, divide each joint probability by its new first-stage probability.

1. Add the positive-test paths: \(P(T)=0.20\cdot 0.80+0.80\cdot 0.30=0.16+0.24=0.40\). 2. Therefore, \(P(\overline{T})=0.60\). 3. After \(T\), \(P(C\mid T)=\frac{0.16}{0.40}=0.40\), so \(P(\overline{C}\mid T)=0.60\). 4. The joint probability \(P(C\cap\overline{T})=0.20\cdot 0.20=0.04\). Thus, \(P(C\mid\overline{T})=\frac{0.04}{0.60}=\frac{1}{15}\), and \(P(\overline{C}\mid\overline{T})=\frac{14}{15}\).

a) \(P(T)=0.40\). b) First stage: \(P(T)=0.40\), \(P(\overline{T})=0.60\). After \(T\): \(P(C\mid T)=0.40\), \(P(\overline{C}\mid T)=0.60\). After \(\overline{T}\): \(P(C\mid\overline{T})=\frac{1}{15}\), \(P(\overline{C}\mid\overline{T})=\frac{14}{15}\).

- Complete each pair of complementary branches. - Add both paths that lead to a flagged result. - Compare true flagged defects with all flagged chips. - In part c, consider both the sensitivity and the false-discard proportion.

1. The missing branches are \(P(\overline{D})=0.98\), \(P(\overline{T}\mid D)=0.05\), and \(P(\overline{T}\mid\overline{D})=0.98\). 2. The overall flagged probability is \(P(T)=0.02\cdot 0.95+0.98\cdot 0.02=0.019+0.0196=0.0386\). 3. Therefore, \(P(D\mid T)=\frac{0.019}{0.0386}=\frac{95}{193}\approx 0.4922\). 4. The test detects \(95\%\) of defective chips, but \(P(\overline{D}\mid T)=\frac{98}{193}\approx 0.5078\), so slightly more than half of flagged chips are actually working. Whether that false-discard rate is acceptable depends on the company’s cost and quality thresholds; a confirmatory test could reduce unnecessary disposal.

a) \(P(\overline{D})=0.98\), \(P(\overline{T}\mid D)=0.05\), and \(P(\overline{T}\mid\overline{D})=0.98\). b) \(P(D\mid T)=\frac{95}{193}\approx 0.4922\), or about \(49.22\%\). c) The test catches most defective chips, but about \(50.78\%\) of flagged chips are working. Suitability therefore depends on the acceptable false-discard cost or threshold; confirmation before disposal would reduce waste.

- Probabilities leaving the same node must sum to \(1\). - Multiply branch probabilities to obtain a complete path probability. - For a reversed conditional, first find the total probability of the new condition.

1. Since \(P(A)=0.40\), \(P(\overline{A})=0.60\). 2. The diagram gives \(P(A\cap B)=0.10\). Because \(P(A\cap B)=P(A)P(B\mid A)\), \(P(B\mid A)=\frac{0.10}{0.40}=0.25\). Therefore, \(P(\overline{B}\mid A)=0.75\). 3. Since \(P(\overline{B}\mid\overline{A})=0.30\), \(P(B\mid\overline{A})=0.70\). 4. The remaining joint probabilities are \(P(A\cap\overline{B})=0.40\cdot 0.75=0.30\), \(P(\overline{A}\cap B)=0.60\cdot 0.70=0.42\), and \(P(\overline{A}\cap\overline{B})=0.60\cdot 0.30=0.18\). 5. Thus, \(P(B)=0.10+0.42=0.52\) and \(P(\overline{B})=0.30+0.18=0.48\). 6. Therefore, \(P(A\mid B)=\frac{0.10}{0.52}=\frac{5}{26}\approx 0.1923\), and \(P(A\mid\overline{B})=\frac{0.30}{0.48}=0.625\).

1. The missing values are \(P(\overline{A})=0.60\), \(P(B\mid A)=0.25\), \(P(\overline{B}\mid A)=0.75\), and \(P(B\mid\overline{A})=0.70\). 2. \(P(A\mid B)=\frac{5}{26}\approx 0.1923\), and \(P(A\mid\overline{B})=0.625\).

- Probabilities leaving the same node sum to \(1\). - Multiply along a path to find a joint probability. - Add all paths that produce the same test result. - Distinguish a joint probability from a conditional probability. - For part f, divide the true-positive path probability by the probability of all positive results.

1. Complete each pair of complementary branches: \(P(\overline{A})=0.98\), \(P(-\mid A)=0.05\), and \(P(-\mid\overline{A})=0.96\). 2. Add the two negative paths: \(P(-)=0.02\cdot 0.05+0.98\cdot 0.96=0.001+0.9408=0.9418\). 3. The requested conditional probability is the branch value \(P(-\mid A)=0.05\). 4. \(P(A\cap +)=0.02\cdot 0.95=0.019\). 5. \(P(+\mid A)\) is the probability of a positive result among people who have the allergy. 6. The overall positive probability is \(P(+)=1-0.9418=0.0582\). Thus, \(P(A\mid +)=\frac{0.019}{0.0582}=\frac{95}{291}\approx 0.3265\).

a) \(P(\overline{A})=0.98\), \(P(-\mid A)=0.05\), and \(P(-\mid\overline{A})=0.96\). b) \(P(-)=0.9418\). c) \(P(-\mid A)=0.05\), or \(5\%\). d) \(P(A\cap +)=0.019\). e) Among people who have the allergy, \(95\%\) receive a positive result. f) \(P(A\mid +)=\frac{95}{291}\approx 0.3265\), or about \(32.65\%\).

- Complete complementary branches first. - Add all paths that lead to a positive test. - Reverse a condition by dividing the joint probability by the new conditioned event. - The same four joint path probabilities appear in both tree orderings.

1. The missing probabilities are \(P(\overline{D})=0.97\), \(P(\overline{T}\mid D)=0.04\), and \(P(\overline{T}\mid\overline{D})=0.98\). 2. \(P(T)=0.03\cdot 0.96+0.97\cdot 0.02=0.0288+0.0194=0.0482\). 3. \(P(D\mid T)=\frac{0.0288}{0.0482}=\frac{144}{241}\approx 0.5975\). 4. The first-stage probabilities in the reversed tree are \(P(T)=0.0482\) and \(P(\overline{T})=0.9518\). 5. After \(T\), \(P(D\mid T)=\frac{144}{241}\approx 0.5975\) and \(P(\overline{D}\mid T)=\frac{97}{241}\approx 0.4025\). 6. After \(\overline{T}\), \(P(D\mid\overline{T})=\frac{0.0012}{0.9518}=\frac{6}{4759}\approx 0.001261\), and \(P(\overline{D}\mid\overline{T})=\frac{0.9506}{0.9518}=\frac{4753}{4759}\approx 0.9987\).

a) \(P(\overline{D})=0.97\), \(P(\overline{T}\mid D)=0.04\), and \(P(\overline{T}\mid\overline{D})=0.98\). b) \(P(T)=0.0482\), or \(4.82\%\). c) \(P(D\mid T)=\frac{144}{241}\approx 0.5975\), or about \(59.75\%\). d) First stage: \(0.0482\), \(0.9518\). After \(T\): \(\frac{144}{241}\), \(\frac{97}{241}\). After \(\overline{T}\): \(\frac{6}{4759}\), \(\frac{4753}{4759}\).

- Identify every complete path that ends at a defective chip. - Add the probabilities of those paths for part a. - For part b, divide the defective path from Plant 1 by the probability of all defective chips.

1. The branch probabilities are \(P(P_1)=0.60\), \(P(P_2)=0.40\), \(P(D\mid P_1)=0.04\), and \(P(D\mid P_2)=0.07\). 2. Add the defective paths: \(P(D)=0.60\cdot 0.04+0.40\cdot 0.07=0.024+0.028=0.052\). 3. The Plant 1 defective path has probability \(P(P_1\cap D)=0.024\). Therefore, \(P(P_1\mid D)=\frac{0.024}{0.052}=\frac{6}{13}\approx 0.4615\).

a) \(P(D)=0.052\), or \(5.2\%\). b) \(P(P_1\mid D)=\frac{6}{13}\approx 0.4615\), or about \(46.15\%\).

- Which event forms the reference group in the first conditional probability? - Find the probabilities of both paths that end at \(A\).

1. The statement \(P(A\mid B)=0.80\) means that among outcomes in event \(B\), \(80\%\) are also in event \(A\). 2. The two paths ending in \(A\) have probabilities \(P(A\cap B)=0.30\cdot 0.80=0.24\) and \(P(A\cap\overline{B})=0.70\cdot 0.20=0.14\). 3. Therefore, \(P(A)=0.24+0.14=0.38\). 4. Reverse the condition: \(P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{0.24}{0.38}=\frac{12}{19}\approx 0.6316\).

\(P(A\mid B)=0.80\) means that \(80\%\) of the outcomes in \(B\) are also in \(A\). Also, \(P(B\mid A)=\frac{12}{19}\approx 0.6316\), or about \(63.16\%\).

- Find the four complete path probabilities first. - Then regroup the paths by the second-stage event.

1. The four joint probabilities are \(P(A\cap B)=0.40\cdot 0.75=0.30\), \(P(A\cap\overline{B})=0.40\cdot 0.25=0.10\), \(P(\overline{A}\cap B)=0.60\cdot 0.25=0.15\), and \(P(\overline{A}\cap\overline{B})=0.60\cdot 0.75=0.45\). 2. Since \(P(B)=0.30+0.15=0.45\), \(P(A\mid B)=\frac{0.30}{0.45}=\frac{2}{3}\). 3. Since \(P(\overline{B})=0.10+0.45=0.55\), \(P(A\mid\overline{B})=\frac{0.10}{0.55}=\frac{2}{11}\).

\(P(A\mid B)=\frac{2}{3}\), and \(P(A\mid\overline{B})=\frac{2}{11}\).

- Which event becomes the restricted sample space after the condition is stated? - Add all complete paths that end in that event.

1. The two paths ending in \(B\) have probabilities \(P(A\cap B)=0.60\cdot 0.30=0.18\) and \(P(\overline{A}\cap B)=0.40\cdot 0.20=0.08\). 2. Therefore, \(P(B)=0.18+0.08=0.26\). 3. The correct conditional probability is \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.18}{0.26}=\frac{9}{13}\approx 0.6923\). 4. Jordan used the wrong denominator. The denominator must represent the condition \(B\), not event \(A\).

Jordan used \(P(A)\) as the denominator instead of \(P(B)\). The correct value is \(P(A\mid B)=\frac{9}{13}\approx 0.6923\), or about \(69.23\%\).

- Find each supplier's contribution to the defective group. - Compare Supplier C's defective contribution with all defective components.

1. The defective contributions are \(0.50\cdot 0.01=0.005\) from A, \(0.30\cdot 0.03=0.009\) from B, and \(0.20\cdot 0.05=0.010\) from C. 2. The overall defective probability is \(0.005+0.009+0.010=0.024\). 3. Therefore, \(P(C\mid\text{defective})=\frac{0.010}{0.024}=\frac{5}{12}\).

\(P(C\mid\text{defective})=\frac{5}{12}\approx 0.4167\), or about \(41.67\%\).

- Complete each branch so that its leaf counts add to the group count. - Then combine all respondents who usually wear a helmet into the conditioned group.

1. The number who do not regularly ride a bicycle is \(1200-450=750\). 2. The remaining missing counts are \(450-360=90\) bicycle riders without helmets and \(750-150=600\) nonriders without helmets. 3. The total number who usually wear a helmet is \(360+150=510\). 4. Therefore, \(P(\text{bicycle rider}\mid\text{helmet})=\frac{360}{510}=\frac{12}{17}\approx 0.7059\).

The missing counts are \(750\), \(90\), and \(600\). \(P(\text{bicycle rider}\mid\text{helmet})=\frac{12}{17}\approx 0.7059\), or about \(70.59\%\).

- Find the probability of each path that ends with passing. - Add those path probabilities to find the overall pass probability. - Divide the intensive-use-and-pass probability by the overall pass probability. - Consider variables that could affect both platform use and exam performance.

1. The probability of intensive use and passing is \(0.40\cdot 0.85=0.34\). 2. The probability of rare use and passing is \(0.60\cdot 0.55=0.33\). 3. Therefore, the overall probability of passing is \(0.34+0.33=0.67\). 4. The reverse conditional probability is \(P(\text{intensive use}\mid\text{passed})=\frac{0.34}{0.67}=\frac{34}{67}\approx 0.5075\). 5. The different pass rates show an association, but the data do not establish causation. Motivation, prior achievement, study time, or other variables could influence both platform use and exam success.

\(P(\text{intensive use}\mid\text{passed})=\frac{34}{67}\approx 0.5075\), or about \(50.75\%\). The observed association does not prove that the platform caused students to pass.

- Add all complete paths that lead to a positive result. - What fraction of the positive-result probability comes from the condition branch?

1. Let \(D\) be the event that the person has the condition. The true-positive path has probability \(P(D\cap +)=0.05\cdot 0.90=0.045\). 2. The false-positive path has probability \(P(\overline{D}\cap +)=0.95\cdot 0.10=0.095\). 3. Thus, \(P(+)=0.045+0.095=0.14\). 4. Therefore, \(P(D\mid +)=\frac{0.045}{0.14}=\frac{9}{28}\approx 0.3214\).

\(P(D\mid +)=\frac{9}{28}\approx 0.3214\), or about \(32.14\%\).

- Find the total probabilities of positive and negative results separately. - For each conditional probability, restrict attention to the matching test-result group.

1. The total positive probability is \(P(+)=0.10\cdot 0.80+0.90\cdot 0.10=0.17\). Therefore, \(P(I\mid +)=\frac{0.08}{0.17}=\frac{8}{17}\approx 0.4706\). 2. The total negative probability is \(P(-)=0.10\cdot 0.20+0.90\cdot 0.90=0.83\). Therefore, \(P(\overline{I}\mid -)=\frac{0.81}{0.83}=\frac{81}{83}\approx 0.9759\).

\(P(I\mid +)=\frac{8}{17}\approx 0.4706\), or about \(47.06\%\); \(P(\overline{I}\mid -)=\frac{81}{83}\approx 0.9759\), or about \(97.59\%\).

- A negative result can occur for a person who has the condition. - Compare that path with all paths that end in a negative result.

1. The probability of having the condition and testing negative is \(0.20\cdot 0.05=0.01\). 2. The probability of not having the condition and testing negative is \(0.80\cdot 0.90=0.72\). 3. The total negative-result probability is \(0.01+0.72=0.73\). 4. Therefore, \(P(D\mid -)=\frac{0.01}{0.73}=\frac{1}{73}\approx 0.0137\). The condition is unlikely but not ruled out.

The claim is false. \(P(D\mid -)=\frac{1}{73}\approx 0.0137\), or about \(1.37\%\).

- Use the definition of conditional probability to find the probability that both events occur. - How can the intersection and the other conditional probability be used to find \(P(A)\)? - Express all category probabilities with a common denominator, then check whether the resulting counts can be reduced.

1. Find the intersection: \(P(A\cap B)=P(A\mid B)P(B)=\frac{3}{4}\cdot\frac{2}{5}=\frac{3}{10}\). 2. Use \(P(B\mid A)=\frac{P(A\cap B)}{P(A)}\): \(P(A)=\frac{3/10}{1/2}=\frac{3}{5}\). 3. Write the relevant probabilities with denominator \(10\): \(P(A)=\frac{6}{10}\), \(P(B)=\frac{4}{10}\), and \(P(A\cap B)=\frac{3}{10}\). Thus, \(10\) tokens can represent the probabilities. 4. Of the \(10\) tokens, \(3\) have both properties. Then \(6-3=3\) have only property \(A\), \(4-3=1\) has only property \(B\), and \(10-(3+3+1)=3\) have neither property. 5. Because the four category counts \(3,3,1,3\) have no common factor greater than \(1\), \(10\) is the smallest possible total.

The smallest model has \(10\) tokens: \(3\) have both properties, \(3\) have only property \(A\), \(1\) has only property \(B\), and \(3\) have neither property.

- Draw a tree beginning with the box selection. - For each observed result, identify every path that can produce it. - When selecting without replacement, update the number of bulbs after the first draw. - For part c, compare how the prior probabilities of the two boxes change.

1. Each box initially has probability \(0.50\). The red probabilities are \(P(R\mid B_1)=\frac{5}{20}=0.25\) and \(P(R\mid B_2)=\frac{10}{20}=0.50\). 2. The overall red probability is \(P(R)=0.50\cdot 0.25+0.50\cdot 0.50=0.375\). Thus, \(P(B_2\mid R)=\frac{0.50\cdot 0.50}{0.375}=\frac{2}{3}\). 3. The probabilities of selecting two blue bulbs are \(P(\text{two blue}\mid B_1)=\frac{5}{20}\cdot\frac{4}{19}=\frac{1}{19}\) and \(P(\text{two blue}\mid B_2)=\frac{2}{20}\cdot\frac{1}{19}=\frac{1}{190}\). 4. Therefore, \(P(B_1\mid\text{two blue})=\frac{0.50\cdot\frac{1}{19}}{0.50\cdot\frac{1}{19}+0.50\cdot\frac{1}{190}}=\frac{10}{11}\). 5. If Box 1 becomes more likely to be selected, Box 2 becomes less likely. A red bulb still favors Box 2 because its red rate is higher, but the lower prior probability for Box 2 reduces \(P(B_2\mid R)\).

a) \(P(B_2\mid R)=\frac{2}{3}\approx 0.6667\), or about \(66.67\%\). b) \(P(B_1\mid\text{two blue})=\frac{10}{11}\approx 0.9091\), or about \(90.91\%\). c) The probability would decrease because Box 2 would be selected less often before the red bulb is observed.

- A tree diagram can show the two paths that lead to a positive result. - Add the probabilities of the true-positive and false-positive paths. - For independence, compare a conditional probability with the corresponding marginal probability. - Compare the numbers of true positives and false positives among \(20{,}000\) employees.

1. Use the law of total probability: \(P(T)=P(A)P(T\mid A)+P(\overline{A})P(T\mid\overline{A})=0.01\cdot 0.98+0.99\cdot 0.03=0.0395\). 2. The joint probability is \(P(A\cap T)=0.01\cdot 0.98=0.0098\). Therefore, \(P(A\mid T)=\frac{0.0098}{0.0395}=\frac{196}{790}\approx 0.2481\). About \(24.8\%\) of employees with a positive result actually have the reaction. 3. The events are not independent because \(P(T\mid A)=0.98\neq P(T)=0.0395\). 4. Among \(20{,}000\) employees, about \(200\) have the reaction, producing \(196\) true positive results. About \(19{,}800\) do not have the reaction, producing \(594\) false positive results. The group without the reaction is so large that its small false-positive rate creates more positive results than the much smaller group with the reaction.

a) \(P(T)=0.0395\), or \(3.95\%\). b) \(P(A\mid T)\approx 0.2481\). About \(24.8\%\) of employees with a positive result actually have the reaction. c) \(A\) and \(T\) are not independent. d) The reaction has a very low base rate. The \(3\%\) false-positive rate is applied to the much larger group of employees without the reaction.

- Compare how likely each observed result is under the two coin hypotheses. - Update the relative weights of the hypotheses after each flip. - A result that is less likely under one hypothesis should reduce that hypothesis's posterior probability.

1. Let \(F\) be the fair-coin hypothesis and \(G\) the biased-coin hypothesis. Initially, \(P(F)=P(G)=0.50\). 2. After one head, \(P(G\mid H)=\frac{0.50\cdot 0.80}{0.50\cdot 0.80+0.50\cdot 0.50}=\frac{8}{13}\approx 0.6154\). 3. After two heads, \(P(G\mid HH)=\frac{0.50(0.80)^2}{0.50(0.80)^2+0.50(0.50)^2}=\frac{64}{89}\approx 0.7191\). 4. After heads, heads, tails, \(P(G\mid HHT)=\frac{0.50(0.80)^2(0.20)}{0.50(0.80)^2(0.20)+0.50(0.50)^2(0.50)}=\frac{128}{253}\approx 0.5059\). 5. The tail is more likely under the fair-coin hypothesis than under the biased-coin hypothesis, so it reduces the posterior probability of the biased coin from about \(71.91\%\) to about \(50.59\%\).

After the first flip: \(\frac{8}{13}\approx 0.6154\), or about \(61.54\%\). After the second flip: \(\frac{64}{89}\approx 0.7191\), or about \(71.91\%\). After the third flip: \(\frac{128}{253}\approx 0.5059\), or about \(50.59\%\). The third flip sharply lowers the probability of the biased-coin hypothesis because tails is less likely for that coin.

- Decide which values are branch probabilities and which require multiplying or adding paths. - For a reversed conditional, divide the relevant joint probability by the new conditioned event's probability. - In part b, identify which outcomes are already included in the event \(M_1\).

1. \(P(M_2\cap R)=0.25\cdot 0.10=0.025\). 2. Add the two rejected paths: \(P(R)=0.75\cdot 0.05+0.25\cdot 0.10=0.0375+0.025=0.0625\). 3. The branch probability is \(P(S\mid M_2)=0.30\). 4. Reverse the condition: \(P(M_1\mid R)=\frac{P(M_1\cap R)}{P(R)}=\frac{0.0375}{0.0625}=0.60\). 5. The event \(M_1\cup E\) contains every component from \(M_1\). The only additional outcomes are excellent components from \(M_2\), so the two disjoint pieces are \(M_1\) and \(M_2\cap E\). 6. Therefore, \(P(M_1\cup E)=0.75+0.25\cdot 0.60=0.90\).

a) (1) \(P(M_2\cap R)=0.025\). (2) \(P(R)=0.0625\). (3) \(P(S\mid M_2)=0.30\). (4) \(P(M_1\mid R)=0.60\). b) The union consists of the disjoint events \(M_1\) and \(M_2\cap E\), because the excellent components from \(M_1\) are already included in \(M_1\). Thus, \(P(M_1\cup E)=0.90\).

- Identify the probabilities stated directly in the problem. - Probabilities on branches leaving the same node must add to \(1\). - Multiply along each path that ends in a positive result. - Divide the disease-and-positive path probability by the total positive-result probability.

1. Since \(P(K)=0.1\), the complement is \(P(\overline{K})=0.9\). 2. Sensitivity gives \(P(+\mid K)=0.9\), so \(P(-\mid K)=0.1\). Specificity gives \(P(-\mid\overline{K})=0.8\), so \(P(+\mid\overline{K})=0.2\). 3. The positive path probabilities are \(P(K\cap +)=0.1\cdot0.9=0.09\) and \(P(\overline{K}\cap +)=0.9\cdot0.2=0.18\). 4. Therefore, \(P(+)=0.09+0.18=0.27\). 5. The reverse conditional probability is \(P(K\mid +)=\frac{0.09}{0.27}=\frac{1}{3}\approx33.3\%\).

a) \(P(\overline{K})=0.9\), \(P(-\mid K)=0.1\), and \(P(+\mid\overline{K})=0.2\) b) \(P(K\mid +)=\frac{1}{3}\approx33.3\%\)

- Use a separate tree diagram for each group. - Specificity gives the negative-test rate among people without the intolerance. Find the complementary false-positive rate. - The desired probability is the true-positive path divided by all positive-test paths. - Compare the sizes of the unaffected groups that can generate false positives.

1. For Group A, the positive-test probability is \(P(T)=0.01\cdot 0.98+0.99\cdot 0.04=0.0494\). Thus, \(P(I\mid T)=\frac{0.0098}{0.0494}\approx 0.1984\). 2. For Group B, the positive-test probability is \(P(T)=0.15\cdot 0.98+0.85\cdot 0.04=0.181\). Thus, \(P(I\mid T)=\frac{0.147}{0.181}\approx 0.8122\). 3. With the same sensitivity and specificity, a positive result is more likely to be a true positive in the higher-prevalence group. In the low-prevalence group, false positives from the much larger unaffected population make up a larger share of all positive results.

a) Group A: \(P(I\mid T)\approx 0.1984\), or \(19.84\%\). Group B: \(P(I\mid T)\approx 0.8122\), or \(81.22\%\). b) The positive result is much more reliable in Group B because the intolerance is more common in that group.

- A conditional probability such as \(P(T\mid E)\) restricts attention to people in event \(E\). - Use the multiplication rule to find \(P(E\cap T)\). - For part c, divide the probability of a true positive by the probability of any positive result. - Independent events satisfy \(P(T\mid E)=P(T)\).

1. The sensitivity is \(P(T\mid E)=0.98\). This means that \(98\%\) of people who have the disease receive a positive result. 2. The joint probability is \(P(E\cap T)=P(E)P(T\mid E)=0.005\cdot0.98=0.0049\), or \(0.49\%\). 3. The total probability of a positive result is \(P(T)=0.0049+0.995\cdot0.03=0.03475\). 4. Therefore, \(P(E\mid T)=\frac{P(E\cap T)}{P(T)}=\frac{0.0049}{0.03475}\approx0.1410\), or about \(14.1\%\). 5. If \(E\) and \(T\) were independent, \(P(T\mid E)\) would equal \(P(T)\). Since \(0.98\ne0.03475\), the events are dependent.

a) \(P(T\mid E)=0.98\); among people with the disease, \(98\%\) test positive. b) \(P(E\cap T)=0.0049=0.49\%\) c) \(P(E\mid T)\approx0.141=14.1\%\) d) The events are dependent because \(P(T\mid E)=0.98\ne P(T)=0.03475\).

- Add the two complete paths that end in a defective adapter. - For part b, determine what fraction of all defective adapters come from Manufacturer B. - Independence means that knowing the manufacturer does not change the defect probability.

1. Add the defective paths: \(P(D)=0.70\cdot 0.01+0.30\cdot 0.04=0.007+0.012=0.019\). 2. The joint probability for Manufacturer B and defective is \(P(B\cap D)=0.30\cdot 0.04=0.012\). Thus, \(P(B\mid D)=\frac{0.012}{0.019}=\frac{12}{19}\approx 0.6316\). 3. For defect status to be independent of manufacturer, the conditional defect rates for both manufacturers must equal the same overall defect rate. Since Manufacturer A's rate is fixed at \(1\%\), Manufacturer B's rate must also be \(1\%\).

a) \(P(D)=0.019\), or \(1.9\%\). b) \(P(B\mid D)=\frac{12}{19}\approx 0.6316\), or about \(63.16\%\). c) Manufacturer B must reduce its defect rate to \(1\%\). Equal conditional defect rates make defect status independent of manufacturer.

- Multiply along each path to find the four joint probabilities. - Add all paths ending in \(S\). - For the reversed tree, divide each joint probability by its new first-stage probability. - Pay attention to which group is the condition in each comparison.

1. The joint probabilities are \(P(P\cap S)=0.40\cdot 0.70=0.28\), \(P(P\cap\overline{S})=0.40\cdot 0.30=0.12\), \(P(\overline{P}\cap S)=0.60\cdot 0.30=0.18\), and \(P(\overline{P}\cap\overline{S})=0.60\cdot 0.70=0.42\). 2. Thus, \(P(S)=0.28+0.18=0.46\), and \(P(\overline{S})=0.54\). 3. In the reversed tree, \(P(P\mid S)=\frac{0.28}{0.46}=\frac{14}{23}\), \(P(\overline{P}\mid S)=\frac{9}{23}\), \(P(P\mid\overline{S})=\frac{0.12}{0.54}=\frac{2}{9}\), and \(P(\overline{P}\mid\overline{S})=\frac{7}{9}\). 4. \(P(S\mid P)=0.70\) is the proportion of podcast listeners who use the paid streaming service. \(P(P\mid S)=\frac{14}{23}\approx 0.6087\) is the proportion of paid streaming users who listen to podcasts.

a) \(P(S)=0.46\), or \(46\%\). b) First stage: \(P(S)=0.46\), \(P(\overline{S})=0.54\). After \(S\): \(P(P\mid S)=\frac{14}{23}\), \(P(\overline{P}\mid S)=\frac{9}{23}\). After \(\overline{S}\): \(P(P\mid\overline{S})=\frac{2}{9}\), \(P(\overline{P}\mid\overline{S})=\frac{7}{9}\). c) \(P(S\mid P)=0.70\) describes streaming use among podcast listeners. \(P(P\mid S)=\frac{14}{23}\approx 0.6087\) describes podcast listening among streaming users.

- Follow the same positive-then-negative path from each initial disease-status branch. - Divide the probability of the disease path by the total probability of both matching paths.

1. The probability of having the disease and receiving the sequence positive, then negative is \(0.02\cdot0.95\cdot0.05=0.00095\). 2. The probability of not having the disease and receiving the same sequence is \(0.98\cdot0.02\cdot0.98=0.019208\). 3. The total probability of the observed sequence is \(0.00095+0.019208=0.020158\). 4. Therefore, \(P(\text{disease}\mid+,-)=\frac{0.00095}{0.020158}\approx0.0471\).

Approximately \(4.71\%\).

- First calculate the four terminal probabilities in the given tree. - Regroup the terminal probabilities by the event that will appear first in the new tree.

1. The terminal probabilities from tree a) are \(P(D\cap +)=0.2\cdot0.6=0.12\) and \(P(D\cap -)=0.8\cdot0.1=0.08\). 2. Therefore, \(P(D)=0.12+0.08=0.2\) and \(P(\overline{D})=0.8\). 3. After \(D\), \(P(+\mid D)=\frac{0.12}{0.2}=0.6\), so \(P(-\mid D)=0.4\). 4. Also, \(P(\overline{D}\cap +)=0.2-0.12=0.08\). Thus, \(P(+\mid\overline{D})=\frac{0.08}{0.8}=0.1\), so \(P(-\mid\overline{D})=0.9\).

\(P(D)=0.2\), \(P(\overline{D})=0.8\); after \(D\): \(P(+\mid D)=0.6\), \(P(-\mid D)=0.4\); after \(\overline{D}\): \(P(+\mid\overline{D})=0.1\), \(P(-\mid\overline{D})=0.9\).

- Split the joint event by sunny and cloudy weather. - Look for a third variable that affects both observed events.

1. The probability of buying ice cream and getting sunburned is \(0.6\cdot0.8\cdot0.5+0.4\cdot0.2\cdot0.05=0.244\). 2. The total probability of buying ice cream is \(0.6\cdot0.8+0.4\cdot0.2=0.56\). 3. Therefore, \(P(\text{sunburn}\mid\text{ice cream})=\frac{0.244}{0.56}\approx0.4357\), or about \(43.6\%\). 4. Weather affects both ice-cream purchases and sunburn risk. This common factor can create an association without ice cream causing sunburn.

\(P(\text{sunburn}\mid\text{ice cream})\approx43.6\%\). Weather is a common influencing variable, so the association does not establish causation.

- Use a population size that turns the small percentages into whole-number counts. - The flagged group includes both true flags and false flags.

1. Of \(100{,}000\) people, \(100{,}000\cdot0.003=300\) have the trait. Of these, \(300\cdot0.9=270\) are flagged. 2. The remaining \(99{,}700\) people do not have the trait. Of these, \(99{,}700\cdot0.02=1994\) are flagged. 3. The total number flagged is \(270+1994=2264\). 4. The proportion of flagged people with the trait is \(\frac{270}{2264}=\frac{135}{1132}\approx0.1193\), or about \(11.9\%\).

\(270\) true flags and \(1994\) false flags; approximately \(11.9\%\) of flagged people have the trait.

- Represent the unknown sensitivity with a variable. - Write the true-positive probability and the false-positive probability. - Express the proportion of true positives among all positive results as an equation.

1. Let \(s=P(+\mid D)\) be the sensitivity. The true-positive probability is \(0.10s\). 2. The false-positive probability is \(0.90\cdot 0.05=0.045\). 3. The given positive predictive value produces the equation \(\frac{0.10s}{0.10s+0.045}=0.50\). 4. Multiply both sides by the denominator: \(0.10s=0.50(0.10s+0.045)=0.05s+0.0225\). Thus, \(0.05s=0.0225\), so \(s=0.45\).

The sensitivity is \(0.45\), or \(45\%\).

- Represent the unknown prevalence with a variable. - Write the true-positive and false-positive path probabilities in terms of that variable. - Express the proportion of true positives among all positive results as an equation.

1. Let \(D\) be the event that a person has the condition. The true-positive probability is \(0.90p\), and the false-positive probability is \(0.05(1-p)\). 2. The given posterior probability gives \(\frac{0.90p}{0.90p+0.05(1-p)}=0.75\). 3. Solving, \(0.90p=0.675p+0.0375(1-p)\), so \(0.2625p=0.0375\). 4. Therefore, \(p=\frac{1}{7}\approx 0.1429\).

\(p=\frac{1}{7}\approx 0.1429\), or about \(14.29\%\).

- First find the proportion of the whole group that produces a true positive. - Represent the unknown positive rate among people without the condition with a variable. - Use the given proportion of true positives among all positive results.

1. The true-positive probability is \(0.20\cdot 0.85=0.17\). 2. Let \(f\) be the false-positive rate. The false-positive probability is \(0.80f\). 3. The given posterior probability produces \(\frac{0.17}{0.17+0.80f}=0.68\). 4. Solving, \(0.17=0.1156+0.544f\), so \(f=0.10\).

The false-positive rate is \(0.10\), or \(10\%\).

- Follow the two-positive path from each initial disease-status branch. - Divide the probability of the disease path by the total probability of both matching paths.

1. The probability of having the disease and testing positive twice is \(0.02\cdot0.95^2=0.01805\). 2. The false-positive rate is \(1-0.98=0.02\). The probability of not having the disease and testing positive twice is \(0.98\cdot0.02^2=0.000392\). 3. The total probability of two positive results is \(0.01805+0.000392=0.018442\). 4. Therefore, \(P(\text{disease}\mid+,+)=\frac{0.01805}{0.018442}\approx0.9787\).

Approximately \(97.9\%\).

- Follow the two-positive path from each initial disease-status branch. - The two test stages use different conditional probabilities.

1. The probability of having the disease and testing positive on both tests is \(0.1\cdot0.9\cdot0.8=0.072\). 2. The probability of not having the disease and testing positive on both tests is \(0.9\cdot0.05\cdot0.1=0.0045\). 3. The total probability of two positive results is \(0.072+0.0045=0.0765\). 4. Therefore, \(P(\text{disease}\mid A+,B+)=\frac{0.072}{0.0765}=\frac{16}{17}\approx0.9412\).

\(\frac{16}{17}\approx94.1\%\)

- Do not confuse the positive rate among users with the user rate among positive results. - Include false-positive results from the much larger group that does not use the substance.

1. Let \(D\) be the event that the participant used a banned substance. The true-positive probability is \(0.02\cdot 0.99=0.0198\). 2. The false-positive probability is \(0.98\cdot 0.03=0.0294\). 3. The total positive probability is \(0.0198+0.0294=0.0492\). 4. Therefore, \(P(D\mid +)=\frac{0.0198}{0.0492}=\frac{33}{82}\approx 0.4024\). The statement confuses sensitivity with the probability of substance use after a positive result.

The statement is false. The probability is \(P(D\mid +)=\frac{33}{82}\approx 0.4024\), or about \(40.24\%\).

- For each program, find the true-positive and false-positive probabilities. - Within all positive results, determine the proportion from the condition branch.

1. For Program A, the true-positive probability is \(0.01\cdot 0.95=0.0095\), and the false-positive probability is \(0.99\cdot 0.01=0.0099\). Thus, \(P(D\mid +)=\frac{0.0095}{0.0095+0.0099}\approx 0.4897\). 2. For Program B, the true-positive probability is \(0.01\cdot 0.99=0.0099\), and the false-positive probability is \(0.99\cdot 0.03=0.0297\). Thus, \(P(D\mid +)=\frac{0.0099}{0.0099+0.0297}=0.25\). 3. A positive result is more reliable under Program A. Program B detects slightly more affected people but produces substantially more false positives.

Program A. Its positive predictive probability is about \(48.97\%\), compared with \(25\%\) for Program B.