- Identify the information that is already known in the question. - In conditional-probability notation, where is the known condition written? - Compare the two questions represented by the two possible orders.

1. The phrase “given that a package was damaged” makes \(D\) the condition. 2. The conditioned event is written after the vertical bar, so the required probability is \(P(A\mid D)\). 3. \(P(D\mid A)\) asks the different, forward question of how often Warehouse A packages are damaged.

\(P(A\mid D)\), because \(D\) is the given condition.

- Which event has already occurred for the people in the question? - Compare the groups named after the bar in the two expressions.

1. A backpack purchase is already known, so it goes after the bar. The requested probability is \(P(C\mid B)\). 2. The classmate wrote \(P(B\mid C)\), which means the chance of buying a backpack among people known to have used a coupon.

Incorrect. The requested expression is \(P(C\mid B)\). The classmate's \(P(B\mid C)\) conditions on coupon users instead.

- Which group does the phrase “of late buses” identify? - Which denominator represents that group?

1. The known group is the twelve buses that arrived late. Nine came from Route East, so the requested fraction is \(\frac{9}{12}=\frac{3}{4}\). 2. The other fraction, \(\frac{9}{20}\), is the share of Route East buses that arrived late.

\(\frac{9}{12}=\frac{3}{4}\). The fraction \(\frac{9}{20}\) describes late arrivals among Route East buses.

- Find the cell where Train and Delayed meet. - Which other arrivals satisfy the known condition “Delayed”?

1. The numerator is the \(4\) arrivals that were both trains and delayed. 2. The denominator is all delayed arrivals, \(5+4=9\). 3. Thus \(P(\text{Train}\mid\text{Delayed})=\frac{4}{9}\).

Numerator: \(4\). Denominator: \(9\). Probability: \(\frac{4}{9}\).

- Which posters can actually be selected? - Add only the counts in the relevant column. - Does the question allow an undisplayed poster to be chosen?

1. The displayed posters number \(7+4+9=20\). 2. Nine of those are green, so the probability is \(\frac{9}{20}\). 3. The full \(30\) posters include ten that were not displayed and therefore could not be chosen.

\(\frac{9}{20}\). Only the \(20\) displayed posters belong in the denominator.

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

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 \(P(D^c\cap +)=0.95\cdot 0.10=0.095\). 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 \(32.14\%\).

- Convert each route's forward delay rate into a delayed-trip count. - For the requested reverse conditional probability, what group must form the denominator? - Compare the event after the vertical bar in the requested probability with the event after the vertical bar in the given \(20\%\).

1. Route X has \(0.20\cdot40=8\) delayed trips. 2. Route Y has \(0.10\cdot60=6\) delayed trips, so there are \(8+6=14\) delayed trips altogether. 3. Conditioning on a delay means using those \(14\) delayed trips as the denominator. Thus, \(P(\text{Route X}\mid\text{delayed})=\frac{8}{14}=\frac{4}{7}\approx0.5714\). 4. The given \(20\%\) is \(P(\text{delayed}\mid\text{Route X})\), which conditions on Route X trips instead of delayed trips.

\(P(\text{Route X}\mid\text{delayed})=\frac{4}{7}\approx57.1\%\). The given \(20\%\) is the opposite conditional probability, \(P(\text{delayed}\mid\text{Route X})\).

- How many players actually made a free throw? - How many players started the game? - Which of those groups is given in each conditional probability?

1. Five players made a free throw, four of whom were starters, so \(P(S\mid F)=\frac{4}{5}\). 2. There were eight starters, four of whom made the shot, so \(P(F\mid S)=\frac{4}{8}=\frac{1}{2}\). 3. The shared count is four in both fractions, but the given group changes when the condition is reversed.

\(P(S\mid F)=\frac{4}{5}\); \(P(F\mid S)=\frac{1}{2}\). The denominators count different groups.

- For part a), which column contains everyone who could be selected? - For part b), which row contains everyone who could be selected? - Compare the denominators you used.

1. The guided group has \(8+22=30\) visitors, including \(8\) teens. So a) is \(\frac{8}{30}=\frac{4}{15}\). 2. The teen group has \(8+12=20\) visitors, including \(8\) on guided tours. So b) is \(\frac{8}{20}=\frac{2}{5}\). 3. Each probability uses the same shared cell but a different total as its denominator.

a) \(\frac{4}{15}\). b) \(\frac{2}{5}\). The given group changes when the condition is reversed.

- Which two paths end at Available? - Find the probability for both types of available room before changing the denominator. - Among available rooms, what fraction came from East?

1. Available rooms from East account for a path probability of \(0.4\cdot0.8=0.32\). 2. Available rooms from West account for \(0.6\cdot0.2=0.12\). 3. Given availability, the probability the room is from East is \(\frac{0.32}{0.32+0.12}=\frac{8}{11}\).

\(\frac{8}{11}\)

- Which outcomes remain possible if you know \(A\) happened? - What changes if you instead know \(B\) happened?

1. Condition on A: \(P(B\mid A)=\frac{0.09}{0.18}=0.50\). 2. Reverse the condition to B: \(P(A\mid B)=\frac{0.09}{0.30}=0.30\). 3. The numerators match, but the denominators do not, so the probabilities are unequal.

\(P(B\mid A)=0.50\); \(P(A\mid B)=0.30\). They are not equal.

- First find the probability that both events occur. - Which probability belongs in the denominator after reversing the condition?

1. The shared-event probability is \(P(R\cap S)=P(S\mid R)P(R)=0.60\cdot0.25=0.15\). 2. Reverse the denominator: \(P(R\mid S)=\frac{0.15}{0.50}=0.30\).

\(P(R\mid S)=0.30\)

- If you already know a package has a red sticker, which kinds of packages could it be? - What was already known when the \(\frac{9}{10}\) fraction was formed?

1. There are \(9+18=27\) packages with red stickers. 2. Nine of these are fragile, so the fraction is \(\frac{9}{27}=\frac{1}{3}\). 3. The fraction \(\frac{9}{10}\) uses fragile packages as the starting group. The fraction \(\frac{9}{27}\) uses all packages with red stickers.

\(\frac{1}{3}\). The fraction \(\frac{9}{10}\) refers to stickers among fragile packages, not fragile packages among stickered ones.

- Which group do you know a randomly selected winner belongs to? - Does the denominator count blue-shirted players or prize winners?

1. Eight of the ten blue-shirted players won, so \(P(\text{winner}\mid\text{blue})=\frac{8}{10}=80\%\). 2. Eight of the forty prize winners wore blue, so \(P(\text{blue}\mid\text{winner})=\frac{8}{40}=20\%\). 3. Liam reversed which group he was comparing within.

\(20\%\). The \(80\%\) refers to winners among blue-shirted players, not blue-shirted players among winners.

- 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, \(0.30\cdot 0.95=P(B\cap F^c)\), and the denominator is \(P(F^c)\). Thus, the expression is \(P(B\mid F^c)\), 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 \(P(S^c)=1-0.058=0.942\). Also, \(P(D\cap S^c)=0.02\cdot 0.06=0.0012\). Thus, \(P(D\mid S^c)=\frac{0.0012}{0.942}=\frac{1}{785}\approx 0.001274\). 4. The undamaged proportion among alarmed packages is \(P(D^c\mid S)=\frac{0.0392}{0.058}=\frac{98}{145}\approx 0.6759\). 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) \(P(D\mid S^c)\approx 0.001274\), 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 \(P(D)=0.02\), \(P(D^c)=0.98\), \(P(T\mid D)=0.98\), and \(P(T\mid D^c)=1-0.95=0.05\). 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 \(P(Y\mid T^c)=0.40\). 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, \(P(T\mid Y^c)=\frac{P(Y^c\mid T)P(T)}{P(Y^c)}=\frac{0.30\cdot 0.35}{0.495}\approx 0.2121\), 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.

- Distinguish the chance of living in a city given an electric vehicle from the reverse conditional probability. - Combine the city-resident portions of electric and non-electric owners to get the total city group. - For the outside-city comparison, consider both complements before finding the new denominator.

1. Let \(E\) mean owning an electric vehicle and \(C\) mean living in a large city. Then \(P(E)=0.15\), \(P(C\mid E)=0.80\), and \(P(C\mid E^c)=0.55\). 2. The total chance of living in a large city is \(P(C)=0.80\cdot0.15+0.55\cdot0.85=0.5875\). 3. Among city residents, \(P(E\mid C)=\frac{0.80\cdot0.15}{0.5875}\approx0.2043\). 4. Outside large cities, \(P(E\mid C^c)=\frac{0.20\cdot0.15}{1-0.5875}\approx0.0727\). 5. The ratio of the two unrounded probabilities is about \(2.81\), so the first chance is more than twice the second.

\(P(E\mid C)\approx0.2043\) (\(20.43\%\)) and \(P(E\mid C^c)\approx0.0727\) (\(7.27\%\)). Yes, 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 \(P(G\cap E^c)=0.60-0.15=0.45\), \(P(G^c\cap E)=0.20-0.15=0.05\), and \(P(G^c\cap E^c)=0.40-0.05=0.35\). 3. \(P(E\mid G^c)=\frac{P(E\cap G^c)}{P(G^c)}=\frac{0.05}{0.40}=0.125\). 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>\(E^c\)</td><td>Total</td></tr><tr><td>\(G\)</td><td>\(0.15\)</td><td>\(0.45\)</td><td>\(0.60\)</td></tr><tr><td>\(G^c\)</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. \(P(E\mid G^c)=0.125\), 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. - Compare the requested reverse conditional with the given poor-sleep rate among high-stress students; they need not be equal.

1. The joint relative frequencies involving poor sleep are \(P(S\cap Q)=0.40\cdot 0.75=0.30\) and \(P(S^c\cap Q)=0.60\cdot 0.20=0.12\). 2. The other cells are \(P(S\cap Q^c)=0.40-0.30=0.10\) and \(P(S^c\cap Q^c)=0.60-0.12=0.48\). 3. The poor-sleep marginal probability is \(P(Q)=0.30+0.12=0.42\). 4. Reversing the condition, \(P(S\mid Q)=\frac{P(S\cap Q)}{P(Q)}=\frac{0.30}{0.42}=\frac{5}{7}\approx 0.7143\).

1. <table><tr><td></td><td>\(Q\)</td><td>\(Q^c\)</td><td>Total</td></tr><tr><td>\(S\)</td><td>\(0.30\)</td><td>\(0.10\)</td><td>\(0.40\)</td></tr><tr><td>\(S^c\)</td><td>\(0.12\)</td><td>\(0.48\)</td><td>\(0.60\)</td></tr><tr><td>Total</td><td>\(0.42\)</td><td>\(0.58\)</td><td>\(1\)</td></tr></table> 2. \(P(Q)=0.42\), or \(42\%\). 3. \(P(S\mid Q)=\frac{5}{7}\approx 0.7143\), or about \(71.43\%\).

- 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 \(P(L^c\mid M)=0.85\) and \(P(L^c\mid U)=0.75\). 3. The overall probability of not using the library daily is \(P(L^c)=0.42\cdot0.85+0.58\cdot0.75=0.357+0.435=0.792\). 4. Therefore, \(P(U\mid L^c)=\frac{0.58\cdot0.75}{0.792}=\frac{145}{264}\approx0.5492\). 5. The second probability is the direct branch value \(P(L^c\mid U)=0.75\).

1. \(P(U\mid L^c)=\frac{145}{264}\approx0.5492\), or about \(54.92\%\). 2. \(P(L^c\mid U)=0.75\), 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 \(P(F)=0.35\), \(P(F^c)=0.65\), \(P(C\mid F)=0.80\), and \(P(C\mid F^c)=0.10\). 2. The complementary conditional probabilities are \(P(C^c\mid F)=0.20\) and \(P(C^c\mid F^c)=0.90\). 3. The total probability of not using a company car is \(P(C^c)=0.20\cdot 0.35+0.90\cdot 0.65=0.655\). 4. Therefore, \(P(F\mid C^c)=\frac{0.20\cdot 0.35}{0.655}=\frac{0.07}{0.655}\approx 0.1069\). 5. For part 2, \(P(C^c\mid F)=0.20\).

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

- Every pair of outgoing branches from a node must total \(1\). - Multiply first-stage and second-stage probabilities for each complete path. - Among all positive outcomes, what portion belongs to people with the trait?

1. The missing probabilities are \(P(T^c)=0.95\), \(P(-\mid T)=0.05\), and \(P(-\mid T^c)=0.90\). 2. The four path probabilities are \(P(T\cap +)=0.05\cdot0.95=0.0475\), \(P(T\cap -)=0.05\cdot0.05=0.0025\), \(P(T^c\cap +)=0.95\cdot0.10=0.095\), and \(P(T^c\cap -)=0.95\cdot0.90=0.855\). 3. Positive results occur on two paths, giving \(P(+)=0.0475+0.095=0.1425\). 4. Among positive results, the trait path accounts for \(\frac{0.0475}{0.1425}=\frac{1}{3}\).

a) Missing branches: \(0.95\), \(0.05\), and \(0.90\). Path probabilities: \(P(T\cap +)=0.0475\), \(P(T\cap -)=0.0025\), \(P(T^c\cap +)=0.095\), and \(P(T^c\cap -)=0.855\). b) \(P(T\mid +)=\frac{1}{3}\approx33.33\%\).

- A rejected component can come from either machine. - Find the probability of each machine-and-rejected path before reversing the condition. - In a conditional probability, the event after the vertical bar determines the reference group.

1. The two rejected paths have probabilities \(P(M_1\cap R)=0.75\cdot 0.05=0.0375\) and \(P(M_2\cap R)=0.25\cdot 0.10=0.025\). 2. Add the paths: \(P(R)=0.0375+0.025=0.0625\). 3. Reverse the condition: \(P(M_1\mid R)=\frac{P(M_1\cap R)}{P(R)}=\frac{0.0375}{0.0625}=0.60\). 4. \(P(R\mid M_1)=0.05\) conditions on all components from \(M_1\), whereas \(P(M_1\mid R)=0.60\) conditions only on rejected components. The denominators and questions are different.

a) \(P(R)=0.0625\), or \(6.25\%\). b) \(P(M_1\mid R)=0.60\), or \(60\%\). c) \(P(R\mid M_1)\) uses all \(M_1\) components as the conditioned group, while \(P(M_1\mid R)\) uses all rejected components as the conditioned group.

- Each inner table cell combines one transportation mode with one satisfaction level. - 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 mode-and-satisfaction combination: \(P(T\cap S)=0.60\cdot 0.50=0.30\), \(P(T\cap N)=0.60\cdot 0.30=0.18\), \(P(T\cap D)=0.60\cdot 0.20=0.12\), \(P(C\cap S)=0.40\cdot 0.30=0.12\), \(P(C\cap N)=0.40\cdot 0.40=0.16\), 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.

- What must probabilities on branches leaving the same point add to? - Which paths lead to a positive result? - Among people with a positive test, which path includes having the disease?

1. The missing branches are \(P(K^c)=0.9\), \(P(-\mid K)=0.1\), and \(P(+\mid K^c)=0.2\). 2. The two positive-result paths have probabilities \(P(K\cap +)=0.1\cdot0.9=0.09\) and \(P(K^c\cap +)=0.9\cdot0.2=0.18\). 3. The total probability of a positive result is \(P(+)=0.09+0.18=0.27\). 4. Among positive results, the disease path has share \(P(K\mid +)=\frac{0.09}{0.27}=\frac{1}{3}\approx33.3\%\).

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

- 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, \(P(B^c)=0.75\). 3. Reverse the branches from \(B\): \(P(A\mid B)=\frac{0.18}{0.25}=0.72\) and \(P(A^c\mid B)=0.28\). 4. The joint probability \(P(A\cap B^c)=0.30\cdot 0.40=0.12\). Thus, \(P(A\mid B^c)=\frac{0.12}{0.75}=0.16\), and \(P(A^c\mid B^c)=0.84\).

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

- Complete each pair of branches so its probabilities add to \(1\). - Find both paths leading to a flagged chip; either defective or working chips may be flagged. - In each reverse probability, the denominator is the total chance of a flagged chip. - Consider what happens to working chips if every flagged chip is discarded.

1. Missing branches: \(P(D^c)=0.98\), \(P(T^c\mid D)=0.05\), and \(P(T^c\mid D^c)=0.98\). 2. The two flagged paths have probabilities \(0.02\cdot0.95=0.019\) and \(0.98\cdot0.02=0.0196\). Thus, \(P(T)=0.0386\). 3. Of all flagged chips, the defective share is \(P(D\mid T)=\frac{0.019}{0.0386}=\frac{95}{193}\approx0.4922\). 4. The working share among flagged chips is \(P(D^c\mid T)=\frac{0.0196}{0.0386}=\frac{98}{193}\approx0.5078\). Discarding all flagged chips would also discard many good chips; the factory could consider a second check.

a) \(P(D^c)=0.98\), \(P(T^c\mid D)=0.05\), and \(P(T^c\mid D^c)=0.98\). b) \(P(D\mid T)=\frac{95}{193}\approx49.22\%\). c) \(P(D^c\mid T)=\frac{98}{193}\approx50.78\%\). A second check could prevent good chips from being thrown away.

- Branch probabilities leaving the same node sum to \(1\). - A terminal path probability is the product of its edge probabilities; use the given whole-path value to recover a missing edge. - To reverse the condition, find all paths ending in the event after the vertical bar.

1. Since \(P(A)=0.40\), the missing first branch is \(P(A^c)=0.60\). 2. The terminal value gives \(P(A\cap B)=0.10\), so \(P(B\mid A)=\frac{0.10}{0.40}=0.25\), and \(P(B^c\mid A)=0.75\). 3. Since \(P(B^c\mid A^c)=0.30\), the other branch there is \(P(B\mid A^c)=0.70\). 4. The remaining joint probabilities are \(P(A\cap B^c)=0.40\cdot0.75=0.30\), \(P(A^c\cap B)=0.60\cdot0.70=0.42\), and \(P(A^c\cap B^c)=0.60\cdot0.30=0.18\). 5. Thus \(P(B)=0.10+0.42=0.52\) and \(P(B^c)=0.30+0.18=0.48\). 6. Reverse the conditions: \(P(A\mid B)=\frac{0.10}{0.52}=\frac{5}{26}\) and \(P(A\mid B^c)=\frac{0.30}{0.48}=0.625\).

1. Missing branches: \(P(A^c)=0.60\), \(P(B\mid A)=0.25\), \(P(B^c\mid A)=0.75\), and \(P(B\mid A^c)=0.70\). 2. \(P(A\mid B)=\frac{5}{26}\approx0.1923\), and \(P(A\mid B^c)=0.625\).

- Every pair of branches leaving one node must add to \(1\). - Find all paths ending in a negative result and multiply along them. - A joint probability counts both conditions, but a conditional probability restricts the reference group. - For the reverse probability, use all positive results as the denominator.

1. The missing branch probabilities are \(P(A^c)=0.98\), \(P(-\mid A)=0.05\), and \(P(-\mid A^c)=0.96\). 2. The two negative paths give \(P(-)=0.02\cdot0.05+0.98\cdot0.96=0.001+0.9408=0.9418\). 3. The allergy-and-positive joint probability is \(P(A\cap +)=0.02\cdot0.95=0.019\). 4. The probability \(P(+\mid A)\) describes the positive-result rate among people who have the allergy, which is \(95\%\). 5. Positive results have probability \(P(+)=1-0.9418=0.0582\). Among them, the allergy path accounts for \(P(A\mid +)=\frac{0.019}{0.0582}=\frac{95}{291}\approx0.3265\).

a) \(P(A^c)=0.98\), \(P(-\mid A)=0.05\), and \(P(-\mid A^c)=0.96\). b) \(P(-)=0.9418\); \(P(A\cap +)=0.019\). c) Among people who have the allergy, \(95\%\) receive a positive result. d) \(P(A\mid +)=\frac{95}{291}\approx32.65\%\).

- The outgoing branch probabilities from each node must add to \(1\). - For the reverse tree, both old and new orders represent the same four joint outcomes. - Use the combined positive paths for the new positive-test first stage. - For negative results, repeat the process with the two paths ending in \(T^c\).

1. The missing probabilities are \(P(D^c)=0.97\), \(P(T^c\mid D)=0.04\), and \(P(T^c\mid D^c)=0.98\). 2. Add the two positive paths: \(P(T)=0.03\cdot0.96+0.97\cdot0.02=0.0288+0.0194=0.0482\). 3. Reverse the positive condition: \(P(D\mid T)=\frac{0.0288}{0.0482}=\frac{144}{241}\approx0.5975\). 4. The reversed first stage is \(P(T)=0.0482\) and \(P(T^c)=0.9518\). 5. After \(T\), the two branches are \(P(D\mid T)=\frac{144}{241}\) and \(P(D^c\mid T)=\frac{97}{241}\). 6. After \(T^c\), the two branches are \(P(D\mid T^c)=\frac{0.0012}{0.9518}=\frac{6}{4759}\) and \(P(D^c\mid T^c)=\frac{0.9506}{0.9518}=\frac{4753}{4759}\).

a) \(P(D^c)=0.97\), \(P(T^c\mid D)=0.04\), and \(P(T^c\mid D^c)=0.98\). b) \(P(T)=0.0482\), or \(4.82\%\). c) \(P(D\mid T)=\frac{144}{241}\approx59.75\%\). d) First stage: \(P(T)=0.0482\), \(P(T^c)=0.9518\). After \(T\): \(\frac{144}{241}\), \(\frac{97}{241}\). After \(T^c\): \(\frac{6}{4759}\), \(\frac{4753}{4759}\).

- Find the four joint outcomes by multiplying along diagram a). - Add the paths ending in streaming to find the reversed tree's first-stage rates. - Each reverse branch is a joint probability divided by the appropriate new first-stage probability. - Identify which group comes after the vertical bar when interpreting the two conditional probabilities.

1. The four joint probabilities are \(P(L\cap S)=0.40\cdot0.70=0.28\), \(P(L\cap S^c)=0.40\cdot0.30=0.12\), \(P(L^c\cap S)=0.60\cdot0.30=0.18\), and \(P(L^c\cap S^c)=0.60\cdot0.70=0.42\). 2. Add the paths ending in \(S\): \(P(S)=0.28+0.18=0.46\), so \(P(S^c)=0.54\). 3. The reversed branches after \(S\) are \(P(L\mid S)=\frac{0.28}{0.46}=\frac{14}{23}\) and \(P(L^c\mid S)=\frac{9}{23}\). 4. After \(S^c\), the reversed branches are \(P(L\mid S^c)=\frac{0.12}{0.54}=\frac{2}{9}\) and \(P(L^c\mid S^c)=\frac{7}{9}\). 5. \(P(S\mid L)=0.70\) is the streaming share among podcast listeners, whereas \(P(L\mid S)=\frac{14}{23}\approx0.6087\) is the podcast-listening share among streaming subscribers.

a) \(P(S)=0.46\), or \(46\%\). b) First stage: \(P(S)=0.46\), \(P(S^c)=0.54\). After \(S\): \(P(L\mid S)=\frac{14}{23}\), \(P(L^c\mid S)=\frac{9}{23}\). After \(S^c\): \(P(L\mid S^c)=\frac{2}{9}\), \(P(L^c\mid S^c)=\frac{7}{9}\). c) \(P(S\mid L)=0.70\) is the streaming rate among podcast listeners; \(P(L\mid S)=\frac{14}{23}\approx0.6087\) is the podcast-listening rate among streaming subscribers.

- Identify every plant-and-defective path. - Add the probabilities of those paths for part a. - For part b, divide the defective contribution 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 \(P(A\cap B^c)=0.70\cdot 0.20=0.14\). 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 \(P(A\cap B)=0.40\cdot 0.75=0.30\), \(P(A\cap B^c)=0.40\cdot 0.25=0.10\), \(P(A^c\cap B)=0.60\cdot 0.25=0.15\), and \(P(A^c\cap B^c)=0.60\cdot 0.75=0.45\). 2. Since \(P(B)=0.30+0.15=0.45\), \(P(A\mid B)=\frac{0.30}{0.45}=\frac{2}{3}\). 3. Since \(P(B^c)=0.10+0.45=0.55\), \(P(A\mid B^c)=\frac{0.10}{0.55}=\frac{2}{11}\).

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

- 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 \(P(A^c\cap B)=0.40\cdot 0.20=0.08\). 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 \(69.23\%\).

- What share of the boards came from the third shop? - A defective board could have come from any of the shops. How can you count each possible source? - Once a board is known to be defective, which group is the new total?

1. The shops' shares are \(0.50\), \(0.30\), and \(0.20\). 2. The probabilities of a defective board from Lakeside, Bayview, and Summit are \(0.50\cdot0.01=0.005\), \(0.30\cdot0.03=0.009\), and \(0.20\cdot0.05=0.010\). 3. The total probability of a defective board is \(0.005+0.009+0.010=0.024\). 4. The conditional probability for Summit is \(\frac{0.010}{0.024}=\frac{5}{12}\).

\(\frac{5}{12}\approx41.67\%\)

- Read the known counts from the tree before filling any blank. - 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 \(70.59\%\).

- Read the three given probabilities from tree a) and calculate its four terminal probabilities. - Regroup those terminal probabilities by the event that will appear first in the new tree.

1. From tree a), the terminal probabilities are \(P(D\cap +)=0.2\cdot 0.6=0.12\) and \(P(D\cap -)=0.8\cdot 0.1=0.08\). 2. Therefore, \(P(D)=0.12+0.08=0.20\) and \(P(D^c)=0.80\). 3. After \(D\), \(P(+\mid D)=\frac{0.12}{0.20}=0.60\), so \(P(-\mid D)=0.40\). 4. Also, \(P(D^c\cap +)=0.20-0.12=0.08\). Thus, \(P(+\mid D^c)=\frac{0.08}{0.80}=0.10\), so \(P(-\mid D^c)=0.90\).

\(P(D)=0.20\), \(P(D^c)=0.80\); after \(D\): \(P(+\mid D)=0.60\), \(P(-\mid D)=0.40\); after \(D^c\): \(P(+\mid D^c)=0.10\), \(P(-\mid D^c)=0.90\).

- First identify which tree branch is the forward conditional probability in part a. - For part b, find every tree path that ends with an award. - The denominator for \(P(M\mid W)\) must include award recipients from both first-stage branches.

1. The tree gives \(P(W\mid M)=0.30\). 2. The music-and-award path has probability \(P(M\cap W)=0.40\cdot0.30=0.12\). 3. The nonmusic-and-award path has probability \(P(M^c\cap W)=0.60\cdot0.10=0.06\), so \(P(W)=0.12+0.06=0.18\). 4. Therefore, \(P(M\mid W)=\frac{0.12}{0.18}=\frac{2}{3}\approx0.6667\). 5. Part a conditions on music students, while part b conditions on award recipients, so their denominators are different groups.

a) \(P(W\mid M)=0.30\). b) \(P(M\mid W)=\frac{2}{3}\approx66.7\%\). c) Part a uses music students as the conditioned group; part b uses award recipients as the conditioned group.

- Find the probability of each group-and-pass path. - Add those 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.

- Find the total probabilities of positive and negative results separately. - For each conditional probability, restrict attention to the corresponding test-result group. - Compare the relevant joint path with that new group total.

1. The total positive probability is \(P(+)=0.10\cdot0.80+0.90\cdot0.10=0.17\). Therefore, \(P(I\mid +)=\frac{0.08}{0.17}=\frac{8}{17}\approx0.4706\). 2. The total negative probability is \(P(-)=0.10\cdot0.20+0.90\cdot0.90=0.83\). Therefore, \(P(I^c\mid -)=\frac{0.81}{0.83}=\frac{81}{83}\approx0.9759\).

1) \(P(I\mid +)=\frac{8}{17}\approx47.06\%\). 2) \(P(I^c\mid -)=\frac{81}{83}\approx97.59\%\).

- 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 \(1.37\%\).

- How many different paths end at Late? - What is the probability of reaching Late by each path? - Which part of all late buses came from South?

1. The North–Late path has probability \(0.50\cdot0.10=0.05\). 2. Central–Late contributes \(0.30\cdot0.20=0.06\), and South–Late contributes \(0.20\cdot0.40=0.08\). 3. The total late probability is \(0.05+0.06+0.08=0.19\). 4. Among late buses, the South share is \(\frac{0.08}{0.19}=\frac{8}{19}\).

\(\frac{8}{19}\)

- First find how much of the overall canceled group comes from indoor concerts. - What part must therefore come from outdoor concerts? - Does the requested final probability use all concerts or only canceled concerts as its whole?

1. Indoor canceled concerts account for \(0.8\cdot0.125=0.10\) of all concerts. 2. Outdoor cancellations must therefore account for \(0.18-0.10=0.08\). Because outdoor concerts make up \(0.2\) of the total, \(P(\text{Canceled}\mid\text{Outdoor})=\frac{0.08}{0.20}=0.40\). 3. The other outdoor branch is \(1-0.40=0.60\). 4. Among canceled concerts, the outdoor share is \(\frac{0.08}{0.18}=\frac{4}{9}\).

a) Outdoor → Canceled: \(0.40\); Outdoor → Held: \(0.60\). b) \(\frac{4}{9}\).

- In each school, find how many athletes and nonathletes wear running shoes. - The condition limits you to the students wearing those shoes. - Compare the starting numbers of athletes across the two schools.

1. At West High, \(0.90\cdot20=18\) athletes and \(0.40\cdot180=72\) other students wear running shoes. Of the \(90\) wearers, \(\frac{18}{90}=\frac{1}{5}\) are athletes. 2. At East High, \(0.90\cdot100=90\) athletes and \(0.40\cdot100=40\) other students wear running shoes. Of the \(130\) wearers, \(\frac{90}{130}=\frac{9}{13}\) are athletes. 3. Athletes make up different shares of the two schools, so they also make up different shares of the running-shoe wearers.

a) \(\frac{1}{5}\). b) \(\frac{9}{13}\). East High has a larger athlete share, although the within-group rates are the same.

- Before using the sensor rates, how many parts have cracks and how many do not? - How do the two sensor rates apply to these different groups? - Which parts are in the group Maya is making a claim about?

1. The sensor flags \(0.90\cdot10=9\) cracked parts. 2. There are \(1000-10=990\) uncracked parts. Of these, \(0.10\cdot990=99\) are flagged. 3. There are \(9+99=108\) flagged parts, including \(9\) cracked ones. So the fraction is \(\frac{9}{108}=\frac{1}{12}\). 4. Maya reversed the condition: the \(90\%\) rate is among cracked parts, not among all flagged parts.

\(\frac{1}{12}\). Maya uses the percentage flagged among cracked parts instead of the percentage cracked among flagged parts.

- The \(60\%\) statement is about which group of arrivals? - Once you know one delayed count, how can you find the other? - Which arrivals are included when the condition is Route B rather than Delayed?

1. Route B accounts for \(0.60\cdot25=15\) of the delayed arrivals. The remaining \(25-15=10\) delayed arrivals belong to Route A. 2. Route B has \(20+15=35\) arrivals in total, of which \(15\) are delayed. 3. The reversed conditional probability is \(\frac{15}{35}=\frac{3}{7}\).

a) Route A: \(10\) delayed; Route B: \(15\) delayed. b) \(\frac{3}{7}\).

- For each case, who belongs to the group of players who made a three-pointer? - What stays the same between the two cases, and what changes? - Can a single guard-only fraction determine the share among all successful shooters?

1. If \(10\) forwards made one, the group of players who made a three-pointer has \(15+10=25\) people. The guard fraction is \(\frac{15}{25}=\frac{3}{5}\). 2. If \(25\) forwards made one, that group has \(15+25=40\) people. The guard fraction is \(\frac{15}{40}=\frac{3}{8}\). 3. Both situations agree that \(15\) of \(25\) guards made one, but the reverse fraction changes because the number of successful forwards changes.

a) \(\frac{3}{5}\). b) \(\frac{3}{8}\). c) The number of forwards who made a three-pointer changes how many successful players could be selected. The guard-only fraction does not determine the reverse fraction.

- Are the return rates percentages of shipped products or of all returns? - Could a product type with fewer shipments have more returns than another type? - When choosing a returned product, which products could be selected? - Would the different return rates preserve the original mix of shipped products?

1. There are \(0.20\cdot40=8\) mug returns, \(0.30\cdot30=9\) notebook returns, and \(0.10\cdot30=3\) tote-bag returns. 2. Of the \(8+9+3=20\) returns, \(9\) are notebooks. Thus \(P(\text{notebook}\mid\text{returned})=\frac{9}{20}\). 3. Among all shipped products the notebook share was \(\frac{30}{100}=\frac{3}{10}\), but returned products are not distributed in the same proportions because the return rates differ.

a) \(8\) mugs, \(9\) notebooks, and \(3\) tote bags. b) \(\frac{9}{20}\). Among shipments the notebook fraction was \(\frac{3}{10}\); notebook returns make up a larger share because return rates differ.

- 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: \(P(T)=P(A)P(T\mid A)+P(A^c)P(T\mid A^c)=0.01\cdot 0.98+0.99\cdot 0.03=0.0395\). 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.

- How often does a person without the intolerance receive a positive result? - Imagine each group contains the same large number of people. How many in each group would have the intolerance? - In each group, who could receive a positive result?

1. Let \(I\) mean having the intolerance and \(T\) a positive result. In both groups, the false-positive rate is \(P(T\mid I^c)=1-0.96=0.04\). 2. For Group A, \(P(T)=0.01\cdot0.98+0.99\cdot0.04=0.0494\). Thus, \(P(I\mid T)=\frac{0.0098}{0.0494}\approx0.1984\). 3. For Group B, \(P(T)=0.15\cdot0.98+0.85\cdot0.04=0.181\). Thus, \(P(I\mid T)=\frac{0.147}{0.181}\approx0.8122\). 4. A positive result is more likely to be a true positive in the group where the intolerance is more common. In Group A, the many people without the intolerance generate a larger share of positive results despite the same test characteristics.

a) Group A: \(19.84\%\). Group B: \(81.22\%\). b) The intolerance is more common in Group B. The same test therefore produces a larger share of positive results from people who actually have it.

- 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 \(P(E\cap T)=P(E)P(T\mid E)=0.005\cdot 0.98=0.0049\), or \(0.49\%\). 3. The total probability of a positive result is \(P(T)=0.0049+0.995\cdot 0.03=0.03475\). 4. Therefore, \(P(E\mid T)=\frac{P(E\cap T)}{P(T)}=\frac{0.0049}{0.03475}\approx 0.1410\), or about \(14.1\%\). 5. If \(E\) and \(T\) were independent, \(P(T\mid E)\) would equal \(P(T)\). Since \(0.98\ne 0.03475\), 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) \(P(E\mid T)\approx 0.141=14.1\%\). d) The events are dependent because \(P(T\mid E)=0.98\ne P(T)=0.03475\).

- Add the two manufacturer-and-defective paths. - 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.

- Complete each pair of outgoing branches so the probabilities sum to \(1\). - Express orders occur along two complete paths; combine them. - For the nonexpress group, find the joint returning-customer path and divide by all nonexpress orders. - Decide which group is the denominator for each probability and for the displayed fraction.

1. Missing probabilities: \(P(R^c)=0.40\), \(P(E^c\mid R)=0.20\), and \(P(E^c\mid R^c)=0.85\). 2. Express-shipping paths total \(P(E)=0.60\cdot0.80+0.40\cdot0.15=0.54\). 3. Thus \(P(E^c)=0.46\) and \(P(R\cap E^c)=0.60\cdot0.20=0.12\), giving \(P(R\mid E^c)=\frac{0.12}{0.46}=\frac{6}{23}\). 4. The joint \(P(R\cap E)\) is the portion of all orders that are both returning-customer and express orders. The conditional \(P(E\mid R)\) is the express share among returning-customer orders. 5. The numerator of the fraction is \(P(R^c\cap E)\), while its denominator is \(P(E)\). Therefore, it is \(P(R^c\mid E)\), the proportion of express orders from other customers.

a) \(P(R^c)=0.40\), \(P(E^c\mid R)=0.20\), and \(P(E^c\mid R^c)=0.85\). b) \(54\%\). c) \(P(R\mid E^c)=\frac{6}{23}\approx26.09\%\). d) \(P(R\cap E)\) is relative to all orders; \(P(E\mid R)\) is relative only to returning-customer orders. The fraction represents \(P(R^c\mid E)\), the proportion of express orders placed by other customers.

- Follow the same positive-then-negative sequence from each initial disease-status group. - Divide the probability of the disease sequence by the total probability of both matching sequences.

1. The probability of having the disease and receiving the sequence positive, then negative is \(0.02\cdot 0.95\cdot 0.05=0.00095\). 2. The probability of not having the disease and receiving the same sequence is \(0.98\cdot 0.02\cdot 0.98=0.019208\). 3. The total probability of the observed sequence is \(0.00095+0.019208=0.020158\). 4. Therefore, \(P(\text{disease}\mid +,-)=\frac{0.00095}{0.020158}\approx 0.0471\).

Approximately \(4.71\%\).

- Split the joint event by sunny and cloudy weather. - Find the total probability of buying ice cream across both weather conditions. - Look for a third variable that affects both observed events.

1. The probability of buying ice cream and getting sunburned is \(0.60\cdot 0.80\cdot 0.50+0.40\cdot 0.20\cdot 0.05=0.244\). 2. The total probability of buying ice cream is \(0.60\cdot 0.80+0.40\cdot 0.20=0.56\). 3. Therefore, \(P(\text{sunburn}\mid \text{ice cream})=\frac{0.244}{0.56}\approx 0.4357\), 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.

\(P(\text{sunburn}\mid \text{ice cream})\approx 43.6\%\). Weather is a common influencing variable, so the association does not establish causation.

- Start by converting the prevalence into a count out of \(100{,}000\). - Split both the trait and no-trait groups into flagged and not-flagged counts. - The conditioned group for the final percentage is all flagged people.

1. Of \(100{,}000\) people, \(100{,}000\cdot 0.003=300\) have the trait, so \(99{,}700\) do not. 2. Among the \(300\) with the trait, \(300\cdot 0.90=270\) are flagged and \(30\) are not flagged. 3. Among the \(99{,}700\) without the trait, \(99{,}700\cdot 0.02=1994\) are flagged and \(97{,}706\) are not flagged. 4. The total number flagged is \(270+1994=2264\). 5. The proportion of flagged people with the trait is \(\frac{270}{2264}=\frac{135}{1132}\approx 0.1193\), or about \(11.9\%\).

<table><tr><td></td><td>Flagged</td><td>Not flagged</td><td>Total</td></tr><tr><td>Trait</td><td>\(270\)</td><td>\(30\)</td><td>\(300\)</td></tr><tr><td>No trait</td><td>\(1994\)</td><td>\(97{,}706\)</td><td>\(99{,}700\)</td></tr><tr><td>Total</td><td>\(2264\)</td><td>\(97{,}736\)</td><td>\(100{,}000\)</td></tr></table> The percentage of flagged people who have the trait is \(\frac{270}{2264}\approx 11.9\%\).

- 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 sequence from each initial disease-status group. - Divide the probability of the disease sequence by the total probability of both matching sequences.

1. The probability of having the disease and testing positive twice is \(0.02\cdot 0.95^2=0.01805\). 2. The false-positive rate is \(1-0.98=0.02\). The probability of not having the disease and testing positive twice is \(0.98\cdot 0.02^2=0.000392\). 3. The total probability of two positive results is \(0.01805+0.000392=0.018442\). 4. Therefore, \(P(\text{disease}\mid +,+)=\frac{0.01805}{0.018442}\approx 0.9787\).

Approximately \(97.9\%\).

- Follow the two-positive sequence from each initial disease-status group. - The two test stages use different conditional probabilities. - Compare the disease-and-two-positive path with all paths producing two positive results.

1. The probability of having the disease and testing positive on both tests is \(0.10\cdot 0.90\cdot 0.80=0.072\). 2. The probability of not having the disease and testing positive on both tests is \(0.90\cdot 0.05\cdot 0.10=0.0045\). 3. The total probability of two positive results is \(0.072+0.0045=0.0765\). 4. Therefore, \(P(\text{disease}\mid A+,B+)=\frac{0.072}{0.0765}=\frac{16}{17}\approx 0.9412\).

\(\frac{16}{17}\approx 94.1\%\).

- 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, \(P(+\mid D)\), with the reverse conditional probability \(P(D\mid +)\).

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 group. - Compare the two reverse conditional probabilities rather than the sensitivities alone.

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.