- Decide whether the description refers to every surveyed customer or only a particular group. - Which phrase signals that two events must both happen? - In the descriptions starting with “among,” which buyers are already selected?

1. Both purchases are required without restricting the customer group, so the expression is \(P(C\cap D)\). 2. Buying a drink is the condition, and buying popcorn is the event of interest, so the expression is \(P(C\mid D)\). 3. Buying popcorn is the condition, and buying a drink is the event of interest, so the expression is \(P(D\mid C)\).

1. \(P(C\cap D)\). 2. \(P(C\mid D)\). 3. \(P(D\mid C)\).

- Look at the event written after the vertical bar. - Conditional probability changes the reference group rather than the event being counted.

1. The event after the vertical bar is the condition, so the restricted group is \(B\). 2. \(P(A\mid B)\) is the probability that \(A\) occurs among outcomes for which \(B\) has occurred.

The restricted group is \(B\). \(P(A\mid B)\) is the probability of \(A\) given that \(B\) has occurred.

- Use only the members in the group named after the vertical bar. - Within that group, count how many also satisfy the first event.

1. Conditioning on \(C\) means the reference group is the \(12\) chess-club members. 2. Within that group, \(3\) are also in \(R\), so \(P(R\mid C)=\frac{3}{12}=\frac{1}{4}\).

\(P(R\mid C)=\frac{1}{4}=0.25\).

- Which books can actually be selected? - How many of those eligible books have torn covers?

1. The condition restricts the original group of fifteen books to the nine paperbacks. 2. Four of those nine have torn covers, so the probability is \(\frac{4}{9}\).

\(\frac{4}{9}\)

- List the numbers you could still have after hearing “even.” - Count the numbers greater than five within that shorter list. - For the comparison, return to all ten original numbers.

1. Given that the number is even, the possible outcomes are \(2,4,6,8,10\). 2. Three of these numbers are greater than \(5\), so the conditional probability is \(\frac{3}{5}\). 3. Without the condition, five of the ten original numbers are greater than \(5\), so the probability would be \(\frac{1}{2}\).

Given even: \(\frac{3}{5}\). Without the condition: \(\frac{1}{2}\). They differ.

- First list only the coin outcomes that agree with what you learned. - Then compare with the original set of all ordered coin outcomes.

1. Once the first toss is known to be heads, the possible outcomes are \(HH\) and \(HT\), equally likely. One of these has two heads, giving \(\frac{1}{2}\). 2. Without the information, the four ordered outcomes are \(HH,HT,TH,TT\). Only \(HH\) has two heads, giving \(\frac{1}{4}\).

Given first heads: \(\frac{1}{2}\). Without that information: \(\frac{1}{4}\).

- What does the event after the vertical bar tell you? - Which event is still uncertain?

1. The event after the vertical bar is the known condition, L. 2. The probability asks for the chance a student rides the bus among students who are late.

The probability that a student rides the bus, given that the student is late. We already know the student is late.

- Find the words after “given that.” - Which letter represents the event we want to predict?

1. The known condition is that the student plays tennis, so T goes after the vertical bar. 2. Being on the honor roll is the event whose probability is asked for, so the expression is \(P(H\mid T)\).

\(P(H\mid T)\)

- For a conditional probability, first identify the restricted group named by the condition. - For the interpretation, describe what portion of that restricted group has the other characteristic. - In part c, notice that the condition is reversed from part a.

1. Use the definition of conditional probability: \(P(C\mid I)=\frac{P(I\cap C)}{P(I)}=\frac{0.18}{0.45}=0.40\). 2. This means that \(40\%\) of the students who play an instrument also sing in the choir. 3. Reverse the condition for part c: \(P(I\mid C)=\frac{P(I\cap C)}{P(C)}=\frac{0.18}{0.30}=0.60\).

a) \(P(C\mid I)=0.40\) b) Of the students who play an instrument, \(40\%\) also sing in the choir. c) \(P(I\mid C)=0.60\), or \(60\%\).

- For each part, identify the circle named after the vertical bar. - The overlap supplies the favorable outcomes in both directions. - Compare the sizes of the two conditioned groups.

1. Event \(B\) contains \(6+14=20\) outcomes, of which \(6\) are also in \(A\). Thus, \(P(A\mid B)=\frac{6}{20}=0.30\). 2. Event \(A\) contains \(10+6=16\) outcomes, of which \(6\) are also in \(B\). Thus, \(P(B\mid A)=\frac{6}{16}=0.375\). 3. The numerator is the same overlap count, but the conditioned groups have different sizes, so the denominators differ.

a) \(P(A\mid B)=0.30\) b) \(P(B\mid A)=0.375\) c) Both use the overlap count \(6\), but \(B\) has \(20\) outcomes while \(A\) has \(16\).

- Which group remains after the condition \(B\) is imposed? - The denominator in a conditional probability is the size of the conditioned group. - Compare the denominator Taylor used with the number of bicycle commuters.

1. Conditioning on \(B\) restricts the reference group to the \(18\) bicycle commuters, not all \(60\) commuters. 2. Within that group, \(12\) leave before \(7{:}00\) a.m., so \(P(E\mid B)=\frac{12}{18}=\frac{2}{3}\). 3. Taylor used the original sample size as the denominator instead of the size of the conditioned group.

Taylor used the wrong denominator. \(P(E\mid B)=\frac{12}{18}=\frac{2}{3}\approx 0.6667\).

- Which group is the student chosen from? - How many students belong to both groups?

1. The condition restricts the group to six students who play an instrument. 2. Three of these also play soccer, so \(P(\text{soccer}\mid\text{instrument})=\frac{3}{6}=\frac{1}{2}\).

\(\frac{1}{2}\)

- The requested group is made up only of swimmers. - Which count refers to art students who are also swimmers?

1. The relevant group has \(15\) swimming students. 2. Five of them also take art, so the probability is \(\frac{5}{15}=\frac{1}{3}\). 3. The count \(12\) includes art students who may not take swimming.

\(\frac{1}{3}\). The numerator is the five students in both groups, not all twelve art students.

- Which of the two probabilities describes the condition? - Find the fraction of B that also belongs to A.

1. Conditional probability uses the shared event as the numerator and the condition B as the denominator. 2. \(P(A\mid B)=\frac{0.18}{0.45}=0.40\).

\(P(A\mid B)=0.40\)

- Which portion of U is not inside V? - Keep U as the denominator because U is the condition.

1. Inside U, the part outside V has probability \(P(U)-P(U\cap V)=0.30\). 2. Divide by the probability of U: \(P(V^c\mid U)=\frac{0.30}{0.50}=0.60\).

\(P(V^c\mid U)=0.60\)

- Which probability is used as the denominator for conditioning on B? - How can you rearrange the conditional-probability relationship?

1. The conditional formula is \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\). 2. Multiply by \(P(B)\): \(P(A\cap B)=0.30\cdot0.40=0.12\).

\(P(A\cap B)=0.12\)

- Identify the set after the bar in the first probability. - Does the second probability select the same group of students? - Find each selected group's total from its circle. - For the last probability, has any smaller group been selected in advance?

1. B contains \(2+9=11\) students, so \(P(A\mid B)=\frac{2}{11}\). 2. A contains \(6+2=8\) students, so \(P(B\mid A)=\frac{2}{8}=\frac{1}{4}\). 3. Since \(\frac{1}{4}>\frac{2}{11}\), \(P(B\mid A)\) is larger. 4. The full sample space contains \(6+2+9+3=20\) students, so \(P(A\cap B)=\frac{2}{20}=\frac{1}{10}\).

a) \(P(A\mid B)=\frac{2}{11}\), \(P(B\mid A)=\frac{1}{4}\); \(P(B\mid A)\) is larger. b) \(P(A\cap B)=\frac{1}{10}\).

- What does a conditional probability of \(1\) tell you about the relationship between two events? - How are a joint probability, a conditional probability, and a marginal probability related? - Look for a formula that contains three of the given or unknown quantities.

1. Use the multiplication rule: \(P(S\cap F)=P(F\mid S)\cdot P(S)=0.25\cdot 0.8=0.20\). 2. Since \(P(S\mid F)=\frac{P(S\cap F)}{P(F)}\), substitute the known values: \(1=\frac{0.20}{P(F)}\). Therefore, \(P(F)=0.20\). 3. A conditional probability of \(1\) means that every student who studies French also studies Spanish.

a) \(P(S\cap F)=0.20\) and \(P(F)=0.20\). b) Every student who studies French also studies Spanish.

- Use the probability of neither defect to find the probability of at least one defect. - Apply the addition rule for two events. - In part b, the scratched phones form the denominator.

1. The probability of at least one defect is \(P(D\cup S)=1-0.85=0.15\). 2. By the addition rule, \(P(D\cap S)=P(D)+P(S)-P(D\cup S)=0.12+0.08-0.15=0.05\). 3. Therefore, \(P(D\mid S)=\frac{P(D\cap S)}{P(S)}=\frac{0.05}{0.08}=0.625\).

a) \(0.05\), or \(5\%\). b) \(P(D\mid S)=0.625\), or \(62.5\%\).

- How many ordered outcomes are possible when a die is rolled twice? - List the ordered pairs whose sum is \(10\). - For a conditional probability, restrict the sample space to outcomes that satisfy the condition. - Determine systematically which outcomes satisfy “the sum is even” and “both rolls are even.”

1. There are \(6\cdot 6=36\) equally likely ordered outcomes. A sum of \(10\) occurs for \((4, 6)\), \((5, 5)\), and \((6, 4)\), so the probability is \(\frac{3}{36}=\frac{1}{12}\). 2. An even sum occurs when both rolls have the same parity, giving \(18\) outcomes. All three outcomes with sum \(10\) are included, so the conditional probability is \(\frac{3}{18}=\frac{1}{6}\). 3. If both rolls are even, the restricted sample space has \(3\cdot 3=9\) outcomes. Only \((4, 6)\) and \((6, 4)\) have sum \(10\), so the conditional probability is \(\frac{2}{9}\).

a) \(\frac{1}{12}\) b) \(\frac{1}{6}\) c) \(\frac{2}{9}\)

- Without replacement, the same label cannot appear twice. - How many ordered pairs are possible? - When is the sum of two integers odd? - Which numbers from \(1\) through \(8\) are prime? - Once the first token is fixed, how many choices remain for the second token?

1. There are \(8\cdot 7=56\) equally likely ordered outcomes. The outcomes with sum \(9\) are \((1, 8)\), \((2, 7)\), \((3, 6)\), \((4, 5)\), \((5, 4)\), \((6, 3)\), \((7, 2)\), and \((8, 1)\). Thus, the probability is \(\frac{8}{56}=\frac{1}{7}\). 2. An odd sum occurs when one number is odd and the other is even. There are \(4\cdot 4+4\cdot 4=32\) such ordered outcomes. All \(8\) outcomes with sum \(9\) satisfy the condition, so the conditional probability is \(\frac{8}{32}=\frac{1}{4}\). 3. The prime labels are \(2, 3, 5, 7\), so there are \(4\cdot 7=28\) outcomes in which the first token is prime. The favorable outcomes are \((2, 7)\), \((3, 6)\), \((5, 4)\), and \((7, 2)\). Therefore, the conditional probability is \(\frac{4}{28}=\frac{1}{7}\).

a) \(\frac{1}{7}\) b) \(\frac{1}{4}\) c) \(\frac{1}{7}\)

- In \(P(X\mid Y)\), first identify what is already known: the event after the vertical bar. - A superscript \(^c\) denotes an event's complement. - For part b, focus on the actual condition after the test result is known. - Ask what the person wants to know about the true allergy status after seeing the result.

1. \(P(T\mid A)\) is the probability of a positive test when the allergy is present. 2. \(P(T^c\mid A)\) is the probability of a negative test even though the allergy is present. 3. \(P(T^c\mid A^c)\) is the probability of a negative test when the allergy is absent. 4. \(P(A\mid T)\) is the probability that the allergy is present when the test is positive. 5. \(P(A^c\mid T)\) is the probability that the allergy is absent even though the test is positive. 6. \(P(A^c\mid T^c)\) is the probability that the allergy is absent when the test is negative. 7. For confidence in a positive result, \(P(A\mid T)\) should be large. For confidence in a negative result, \(P(A^c\mid T^c)\) should be large.

a) (1) A positive test given that the allergy is present. (2) A negative test given that the allergy is present. (3) A negative test given that the allergy is absent. (4) The allergy is present given a positive test. (5) The allergy is absent given a positive test. (6) The allergy is absent given a negative test. b) For a positive result, \(P(A\mid T)\) should be large. For a negative result, \(P(A^c\mid T^c)\) should be large.

- In \(P(X\mid Y)\), first identify the group described after words such as “among,” “given,” or “reported.” - Decide which event describes the group being restricted and place that event after the vertical bar. - For the error-rate question, distinguish an incorrect sensor report from a correct one.

1. In a conditional probability \(P(X\mid Y)\), the event after the vertical bar identifies the group being considered. 2. For a), the group is cracked bottles, and the event of interest is being reported acceptable. This gives \(P(S^c\mid C)\). 3. For b), the group is bottles without cracks, and the event of interest is being reported defective. This gives \(P(S\mid C^c)\). 4. For c), the group is bottles reported defective, and the event of interest is actually having a crack. This gives \(P(C\mid S)\). 5. For d), the group is bottles reported acceptable, and the event of interest is actually having no crack. This gives \(P(C^c\mid S^c)\). 6. For e), the group is cracked bottles, and the event of interest is being reported defective. This gives \(P(S\mid C)\). 7. For f), the group is bottles reported defective, and the event of interest is actually having no crack. This gives \(P(C^c\mid S)\). 8. The sensor's two error rates are \(P(S^c\mid C)\), the false-negative rate, and \(P(S\mid C^c)\), the false-positive rate. Both should be small. 9. \(P(C\mid S)\) asks what proportion of defect reports correspond to cracked bottles. \(P(S\mid C)\) asks what proportion of cracked bottles are reported defective. Their conditioned groups are different, so the probabilities need not be equal.

a) \(P(S^c\mid C)\) b) \(P(S\mid C^c)\) c) \(P(C\mid S)\) d) \(P(C^c\mid S^c)\) e) \(P(S\mid C)\) f) \(P(C^c\mid S)\) g) The error rates are \(P(S^c\mid C)\) and \(P(S\mid C^c)\). Both should be small. h) \(P(C\mid S)\) conditions on bottles reported defective, while \(P(S\mid C)\) conditions on cracked bottles. They answer different questions and need not be equal.

- Which group does the “three out of every four” statement describe? - How is that group related to the entire group who bring lunch? - What is left after removing students who belong to both groups? - For part c), which group is already known?

1. The conditional three-fourths refers to the \(16\) chess-club members. The number in both groups is \(\frac{3}{4}\cdot16=12\). 2. There are \(22\) lunch-bringers altogether, so \(22-12=10\) bring lunch but are not in the chess club. 3. Among all \(22\) lunch-bringers, \(12\) are chess-club members. The fraction is \(\frac{12}{22}=\frac{6}{11}\), not \(\frac{3}{4}\): that fraction uses chess-club members as its starting group.

a) \(12\) students. b) \(10\) students. c) \(\frac{6}{11}\), because it uses all lunch-bringers rather than just chess-club members.