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Conditional probability

Learning goals

  • Computes conditional probabilities using the correct restricted sample space.
  • Uses P(A|B) notation correctly.
  • Interprets conditional probability in context.
  • Connects conditional probability to two-way tables, trees, or multiplication rules.

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Conditional probability 1
Suppose \(P(B)=0\). Is \(P(A\mid B)\) defined by the usual ratio formula? Explain.

Hints

- Inspect the denominator in the conditional-probability definition. - A conditioning event must have positive probability for the elementary ratio. - Do not assign a value merely from the intersection.

Solution

1. The ratio would require division by \(P(B)=0\). 2. Division by zero is undefined. 3. Therefore the elementary conditional probability \(P(A\mid B)\) is not defined when the conditioning event has probability zero.

Answer

No. The usual conditional probability is undefined because its denominator is \(0\).
Conditional probability 2
A fair six-sided die is rolled. Given that the result is greater than \(3\), what is the probability that the result is even?

Hints

- First restrict the sample space using the information after “given that.” - Count the outcomes in the restricted sample space. - Then count how many of those restricted outcomes satisfy the second event.

Solution

1. The condition restricts the possible outcomes to \(\{4,5,6\}\). 2. Within that restricted set, the even outcomes are \(4\) and \(6\). 3. Therefore the conditional probability is \(\frac23\).

Answer

\(\frac23\)
Conditional probability 3
Events \(A\) and \(B\) satisfy \(P(B)=0.50\) and \(P(A\cap B)=0.20\). Find \(P(A\mid B)\).

Hints

- Identify which event is the condition. - The numerator is the probability of both events occurring. - Divide the intersection probability by the probability of the conditioning event.

Solution

1. Conditional probability uses the restricted probability of \(B\) as the denominator. 2. \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.20}{0.50}=0.40\).

Answer

\(0.40\)
Conditional probability 4
A school has \(60\) students in two after-school clubs. Of the \(18\) debate-club members, \(9\) also volunteer at the library. A student is chosen at random **from the debate club**. What is the probability that this student volunteers at the library? Explain which number is used as the total.

Hints

- Who can actually be chosen for this random selection? - Which number counts that smaller group? - How many students in that group have the requested property?

Solution

1. Only debate-club members can be chosen, so the relevant group has \(18\) students, not \(60\). 2. Nine of those students volunteer at the library, giving \(\frac{9}{18}=\frac{1}{2}\).

Answer

\(\frac{1}{2}\). The total is \(18\) because the student is chosen from the debate club.
Conditional probability 5
In a school survey, event A means “a student rides the bus” and event B means “a student arrives before the first bell.” What does \(P(A\mid B)\) mean in words? Does it describe the same group as \(P(B\mid A)\)?

Hints

- What is the purpose of the vertical bar in the notation? - Would knowing a student arrived early limit the possible group of students? - When the order is reversed, is the same information taken as known?

Solution

1. The condition after the vertical bar states which students are considered: those who arrive before the first bell. 2. Thus \(P(A\mid B)\) is the probability that an early-arriving student rides the bus. 3. In \(P(B\mid A)\), the group is bus riders instead, so it is a different conditional probability.

Answer

\(P(A\mid B)\) means the probability of riding the bus among students who arrive before the first bell. No: \(P(B\mid A)\) describes arriving early among bus riders.
Conditional probability 6
Each student in the survey chose exactly one club (drama or band) and exactly one meeting time (before school or after school). <table> <tr><th></th><th>Before school</th><th>After school</th></tr> <tr><td>Drama</td><td>\(6\)</td><td>\(14\)</td></tr> <tr><td>Band</td><td>\(12\)</td><td>\(8\)</td></tr> </table> To find \(P(\text{drama}\mid\text{before school})\), what number should be the denominator? Explain your choice.

Hints

- Which group comes after the vertical bar? - Find all students in that group, not just the drama students. - Does the denominator include students attending only after school?

Solution

1. The condition is “before school,” so include everyone in that column. 2. The column contains \(6+12=18\) students. The denominator is \(18\), not the overall table total.

Answer

\(18\), because all \(18\) students attending before-school activities make up the group being considered.
Conditional probability 7
A ball is drawn from the jar and turns out to be blue. It is not put back. What is the probability that the next ball drawn is red?
Figure for problem 563316

Hints

- First count the balls of each color in the picture. - Which color was removed from the jar? - What is the new total for the second draw?

Solution

1. The jar begins with \(3\) red and \(2\) blue balls. 2. After one blue ball is removed, the jar has \(3\) red balls among \(4\) remaining balls. 3. The required probability is \(\frac{3}{4}\).

Answer

\(\frac{3}{4}\).
Conditional probability 8
Two tokens are drawn without replacement from the urn shown. Given that the first token is red, find the probability that the second token is red.
Figure for problem 547312

Hints

- Conditioning means the sample space has been narrowed to a specified event. - For sequential settings, update what remains after the stated condition occurs. - Explain what knowing the condition changes about the probability question.

Solution

1. After a red token is drawn, \(4\) red tokens remain out of \(11\) total tokens. 2. Thus \(P(\text{second red}\mid\text{first red})=\frac{4}{11}\).

Answer

\(\frac{4}{11}\).
Conditional probability 9
A randomly chosen order is from the weekend with probability \(0.40\). Among weekend orders, \(35\%\) include express shipping. Find the joint probability that an order is from the weekend and includes express shipping.

Hints

- One probability gives the size of the conditioning group. - The other gives the share inside that group. - Convert the subgroup share to a share of all orders.

Solution

1. Use the conditional multiplication relationship. 2. \(P(\text{weekend and express})=0.40\cdot0.35=0.14\).

Answer

The joint probability is \(0.14\).
Conditional probability 10
Use the probability tree shown to find \(P(A\cap B\cap C)\).
Figure for problem 547323

Hints

- Multiply probabilities along one fully specified path. - Each later probability is conditional on the preceding path. - Keep the event order aligned with the given conditions.

Solution

1. Follow the specified path through the stages. 2. \(P(A\cap B\cap C)=0.60\cdot0.50\cdot0.80=0.24\).

Answer

The probability is \(0.24\).
Conditional probability 11
Give a numerical example showing that conditioning can increase an event’s probability. Use a sample of \(100\) observations and state both \(P(A)\) and \(P(A\mid B)\).

Hints

- Choose a subgroup with a higher concentration of \(A\) than the full sample. - Keep the intersection no larger than either event. - Compare the subgroup rate with the marginal rate.

Solution

1. Let \(A\) occur for \(30\) of \(100\) observations, so \(P(A)=0.30\). 2. Let \(B\) contain \(20\) observations, \(15\) of which are in \(A\). 3. Then \(P(A\mid B)=\frac{15}{20}=0.75>0.30\).

Answer

For example, \(P(A)=0.30\) and \(P(A\mid B)=0.75\) when \(15\) of \(20\) observations in \(B\) also satisfy \(A\).
Conditional probability 12
A shipping company finds that \(0.75\) of its packages leave the warehouse on the scheduled day. Among packages that leave on the scheduled day, \(0.92\) arrive by the promised date. What is the probability that a randomly selected package both leaves on the scheduled day and arrives by the promised date?

Hints

- One probability describes the first event, and the other describes the second event within the first event. - The question asks for both events to occur. - Connect a conditional probability to a joint probability using the multiplication relationship.

Solution

1. Let \(S\) be leaving on the scheduled day and \(A\) be arriving by the promised date. 2. The given values are \(P(S)=0.75\) and \(P(A\mid S)=0.92\). 3. Use the multiplication rule: \(P(S\cap A)=P(S)P(A\mid S)=0.75\cdot0.92=0.69\).

Answer

\(0.69\)
Conditional probability 13
The displayed probability tree describes two machine types, R and S, and whether a machine fails inspection. Given that a randomly selected machine failed inspection, what is the probability that it was type S?
Figure for problem 556201

Hints

- Identify all terminal paths that are consistent with the condition “failed inspection.” - Find the probability represented by each relevant path before reversing the conditioning. - Within the total failure probability, determine the share contributed by type S.

Solution

1. From the tree, \(P(R)=0.60\), \(P(S)=0.40\), \(P(F\mid R)=0.05\), and \(P(F\mid S)=0.10\). 2. The failed type-R path has probability \(0.60\cdot0.05=0.03\), and the failed type-S path has probability \(0.40\cdot0.10=0.04\). 3. Thus \(P(F)=0.03+0.04=0.07\). 4. Therefore \(P(S\mid F)=\frac{0.04}{0.07}=\frac47\).

Answer

\(\frac47\)
Conditional probability 14
In a group of \(200\) job applicants, \(50\) have a professional certificate. Of those \(50\), \(30\) are hired. Altogether, \(80\) applicants are hired. a) Find the probability that an applicant is hired given that the applicant has the certificate. b) Find the probability that an applicant has the certificate given that the applicant is hired. c) Explain why the two conditional probabilities are different.

Hints

- Read the event after “given that” first in each part. - The same overlap count can be divided by different totals when the conditioning direction changes. - Compare the two restricted groups rather than assuming reversed conditionals are equal.

Solution

1. For \(P(\text{hired}\mid\text{certificate})\), restrict to the \(50\) certificate holders. The probability is \(\frac{30}{50}=0.60\). 2. For \(P(\text{certificate}\mid\text{hired})\), restrict to the \(80\) hired applicants. The probability is \(\frac{30}{80}=0.375\). 3. The numerators count the same \(30\) applicants, but the conditioning groups—and therefore the denominators—are different.

Answer

a) \(0.60\) b) \(0.375\) c) The conditioning events are different, so the denominators are \(50\) and \(80\), respectively.
Conditional probability 15
At a school, \(40\) students use the library after classes, \(30\) play a school sport, and \(20\) do both. Find \(P(\text{library}\mid\text{sport})\) and \(P(\text{sport}\mid\text{library})\) for a student chosen at random from the school. Are they equal?

Hints

- Both questions involve students who do both activities. - Which number is the total for the group after the vertical bar in each expression? - Compare the two denominators before deciding whether the chances are equal.

Solution

1. Of the \(30\) athletes, \(20\) use the library, so \(P(\text{library}\mid\text{sport})=\frac{20}{30}=\frac{2}{3}\). 2. Of the \(40\) library users, \(20\) are athletes, so \(P(\text{sport}\mid\text{library})=\frac{20}{40}=\frac{1}{2}\). 3. The numerator is the same overlap, but the conditioning groups differ, so the probabilities are not equal.

Answer

\(P(\text{library}\mid\text{sport})=\frac{2}{3}\) and \(P(\text{sport}\mid\text{library})=\frac{1}{2}\). They are not equal.
Conditional probability 16
A community art studio has \(30\%\) of its customers attend a weekend pottery class. Among pottery students, \(40\%\) bring their own tools. What is the probability that a randomly selected customer attends pottery and brings their own tools?

Hints

- If there were 100 customers, how many would take pottery? - Think about what happens within that smaller group. - Would 40% of the pottery students also be 40% of all customers?

Solution

1. The pottery group accounts for \(0.30\) of all customers. 2. Within that group, \(0.40\) bring their own tools. 3. The share of all customers meeting both conditions is \(0.30\cdot0.40=0.12\).

Answer

\(0.12\), or \(12\%\).
Conditional probability 17
A school tracks how students get home and whether they arrive late. Use the probability tree to find the chance that a randomly chosen student arrives late.
Figure for problem 563319

Hints

- Which two paths of the tree end with “Late”? - How can you find the probability of following each entire path? - Can the same student follow both of those paths on one trip home?

Solution

1. The bus-and-late path has probability \(\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}\). 2. The pickup-and-late path has probability \(\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\). 3. These two paths cannot both occur, so the total late probability is \(\frac{1}{8}+\frac{1}{4}=\frac{3}{8}\).

Answer

\(\frac{3}{8}\).
Conditional probability 18
This fictional survival table tracks \(100{,}000\) people born in the same year in each of two imagined regions, Pine and Harbor. The numbers show how many reached each age. <table> <tr><th>Age</th><th>Pine</th><th>Harbor</th></tr> <tr><td>Birth</td><td>\(100{,}000\)</td><td>\(100{,}000\)</td></tr> <tr><td>\(40\)</td><td>\(92{,}400\)</td><td>\(98{,}800\)</td></tr> <tr><td>\(70\)</td><td>\(52{,}100\)</td><td>\(84{,}500\)</td></tr> <tr><td>\(85\)</td><td>\(12{,}300\)</td><td>\(42{,}600\)</td></tr> </table> a) In each region, estimate the chance of reaching age \(70\), given that a person has reached age \(40\). b) In each region, estimate the chance of reaching age \(85\), given that a person has reached age \(40\). c) Explain why these estimates use observed counts instead of counting equally likely outcomes. Round your answers to four decimal places.

Hints

- For each group, the people who reached the later age are part of those who reached age \(40\). - Since you already know the person reached age \(40\), which row should be the denominator? - Think about the difference between frequencies from a table and a fair-die calculation.

Solution

1. For Pine, the fraction reaching \(70\) among those who reached \(40\) is \(\frac{52100}{92400}\approx0.5639\). For Harbor, it is \(\frac{84500}{98800}\approx0.8553\). 2. For Pine, the fraction reaching \(85\) among those who reached \(40\) is \(\frac{12300}{92400}\approx0.1331\). For Harbor, it is \(\frac{42600}{98800}\approx0.4312\). 3. These are conditional proportions estimated from the table's counts, not theoretical probabilities based on equally likely outcomes.

Answer

a) Pine: approximately \(0.5639\); Harbor: approximately \(0.8553\). b) Pine: approximately \(0.1331\); Harbor: approximately \(0.4312\). c) These estimates come from counting people who reached later ages among those who reached age \(40\).
Conditional probability 19
This hypothetical table tracks \(50{,}000\) smartphones from one production batch. It shows how many still work after different numbers of months. <table> <tr><th>Months in use</th><th>Working phones</th></tr> <tr><td>\(0\)</td><td>\(50{,}000\)</td></tr> <tr><td>\(12\)</td><td>\(48{,}500\)</td></tr> <tr><td>\(24\)</td><td>\(42{,}000\)</td></tr> <tr><td>\(36\)</td><td>\(25{,}000\)</td></tr> <tr><td>\(48\)</td><td>\(8000\)</td></tr> </table> a) Estimate the probability that a phone still working after \(12\) months also works after \(36\) months. b) Estimate the probability that a phone working after \(24\) months stops working by month \(36\). c) Why should we not assume that failure is equally likely in every month? Round each probability to four decimal places.

Hints

- For each question, begin with the phones that meet the stated starting condition. - How can you find the number that stopped working between two listed months? - Does the table provide a separate failure chance for every month?

Solution

1. Of the \(48{,}500\) phones working after \(12\) months, \(25{,}000\) still work after \(36\) months: \(\frac{25000}{48500}\approx0.5155\). 2. Between months \(24\) and \(36\), \(42{,}000-25{,}000=17{,}000\) phones stop working. Out of those working at month \(24\), that is \(\frac{17000}{42000}\approx0.4048\). 3. The table does not say that failure occurs with the same chance in each month. The risk may change as phones get older.

Answer

a) Approximately \(0.5155\). b) Approximately \(0.4048\). c) The data do not establish equal monthly chances of failure; that would require another assumption.
Conditional probability 20
A model of museum memberships follows one group of members from year to year. The table shows how many remain enrolled at each date. <table> <tr><th>Date</th><th>Members still enrolled</th></tr> <tr><td>January 2024</td><td>\(98{,}412\)</td></tr> <tr><td>January 2025</td><td>\(95{,}120\)</td></tr> <tr><td>January 2026</td><td>\(80{,}455\)</td></tr> </table> Assume no new members join this group. Estimate each probability to four decimal places. 1. A member enrolled in January 2024 is still enrolled in January 2026. 2. Two names are picked at random from the January 2025 list, with each name equally likely. The first name is put back before the second pick. Both people are still enrolled in January 2026. 3. A member enrolled in January 2024 stays until January 2025 but leaves before January 2026.

Hints

- Which membership date defines the group for each question? - Think of the later year's members as part of the earlier year's group. - How does selecting two memberships independently affect the chance that both stayed enrolled? - For the last event, what happened to the members who appear in the 2025 count but not the 2026 count?

Solution

1. Of the \(98{,}412\) members enrolled in January 2024, \(80{,}455\) remained in 2026: \(\frac{80455}{98412}\approx0.8175\). 2. The probability that one January 2025 membership remains in 2026 is \(\frac{80455}{95120}\). Independent selections with replacement give \(\left(\frac{80455}{95120}\right)^2\approx0.7154\) for both. 3. The number who left between 2025 and 2026 is \(95{,}120-80{,}455=14{,}665\). Out of the January 2024 group, the probability is \(\frac{14665}{98412}\approx0.1490\).

Answer

1. Approximately \(0.8175\). 2. Approximately \(0.7154\). 3. Approximately \(0.1490\).
Conditional probability 21
A school newspaper chooses either the Events photo folder or the Sports photo folder with equal probability. It then chooses one photo, with every photo in that folder equally likely. Events contains two photos (pep rally and parade); Sports contains four (soccer, tennis, swimming, and track). a) Find the probability of each of the six photos being chosen. b) Explain why the six photos are not equally likely, even though each choice is made at random. c) What is the probability of choosing either a Sports photo or the pep-rally photo?

Hints

- Think about the two choices made in order: folder first, then photo. - Would a photo in the smaller folder be more or less likely than a photo in the larger one? - Can a selected photo be both a Sports photo and the pep-rally photo?

Solution

1. Choosing either folder has probability \(\frac{1}{2}\). 2. Each Events photo then has probability \(\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\); each Sports photo has probability \(\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}\). 3. The folder sizes differ, so the six final photos are not equally likely. 4. The Sports-folder event has probability \(\frac{1}{2}\) and is disjoint from pep rally, which has probability \(\frac{1}{4}\). The combined probability is \(\frac{3}{4}\).

Answer

a) Pep rally and parade: \(\frac{1}{4}\) each. Soccer, tennis, swimming, and track: \(\frac{1}{8}\) each. b) There are fewer choices in Events, so an Events photo has a greater chance of being selected. c) \(\frac{3}{4}\).
Conditional probability 22
At a school fair, a player first chooses one of two prize wheels at random. Wheel One is chosen \(40\%\) of the time and has prize probabilities: badge \(0.50\), pencil \(0.30\), sticker \(0.20\). Wheel Two is chosen \(60\%\) of the time and has prize probabilities: badge \(0.10\), pencil \(0.20\), sticker \(0.70\). a) Find the overall chance of each prize. b) Check that your three answers form a valid probability distribution. c) What is the chance of winning a badge or sticker?

Hints

- Which wheel might the player use before any prize is selected? - Each prize can come from either wheel; how could those two paths be combined? - Check the total of the three prize chances before finding the chance of two prize types.

Solution

1. Badge: \(0.40\cdot0.50+0.60\cdot0.10=0.26\). 2. Pencil: \(0.40\cdot0.30+0.60\cdot0.20=0.24\). Sticker: \(0.40\cdot0.20+0.60\cdot0.70=0.50\). 3. Each chance lies between \(0\) and \(1\), and \(0.26+0.24+0.50=1\). 4. One spin awards one prize, so badge or sticker has probability \(0.26+0.50=0.76\).

Answer

a) Badge: \(0.26\); pencil: \(0.24\); sticker: \(0.50\). b) The probabilities are nonnegative and sum to \(1\). c) \(0.76\).
Conditional probability 23
In a randomized-response survey, a participant flips a fair coin. On heads, the participant answers a sensitive yes/no question truthfully. On tails, the participant must answer “yes.” In a large survey, \(62\%\) of responses are “yes.” Estimate the true proportion \(p\) whose truthful answer is yes. Explain how the randomization protects individual responses while allowing population estimation.

Hints

- Split a reported “yes” into the two disjoint ways it can occur. - One branch depends on the unknown prevalence; the other does not. - Use the known randomization probability to solve for the population rate.

Solution

1. A “yes” response occurs either from a heads flip followed by a truthful yes or from any tails flip. 2. Therefore \(0.62=0.50p+0.50\). 3. Solving gives \(0.50p=0.12\), so \(p=0.24\). 4. A particular yes response may be truthful or forced by the coin, so it does not reveal the individual’s status. Across many responses, the known coin mechanism can be removed algebraically.

Answer

The estimated truthful-yes proportion is \(p=0.24\), or \(24\%\). An individual “yes” may be truthful or forced by the coin, protecting the person’s status, while the known randomization rate allows the population proportion to be recovered.
Conditional probability 24
Let \(A\subseteq C\) and let \(P(B)>0\). Prove that \(P(A\mid B)\le P(C\mid B)\). Explain why ordinary event inclusion remains monotone after conditioning on the same event.

Hints

- Intersect both nested events with the conditioning event. - Inclusion is preserved under intersection with a fixed set. - The conditional probabilities share the same positive denominator.

Solution

1. From \(A\subseteq C\), intersecting both sets with \(B\) gives \(A\cap B\subseteq C\cap B\). 2. Probability monotonicity gives \(P(A\cap B)\le P(C\cap B)\). 3. Divide both sides by the same positive number \(P(B)\) to obtain \(P(A\mid B)\le P(C\mid B)\).

Answer

\(P(A\mid B)\le P(C\mid B)\). Intersecting both events with the same conditioning event preserves inclusion, and dividing by the same positive probability preserves the inequality.
Conditional probability 25
Suppose \(P(A\mid B)=0.40\). Is \(P(A\cap B)\) determined? Give two probability models with different positive values of \(P(B)\) that both satisfy the conditional probability, and find the corresponding joint probabilities.

Hints

- Rearrange the conditional-probability relationship without assuming the size of the conditioning event. - Choose two valid positive probabilities for the conditioning event. - Check the ratio in each constructed model.

Solution

1. The relationship is \(P(A\cap B)=0.40P(B)\). 2. If \(P(B)=0.50\), then \(P(A\cap B)=0.20\). 3. If \(P(B)=0.20\), then \(P(A\cap B)=0.08\). 4. Both models have \(P(A\mid B)=0.40\), so the conditional probability alone does not determine the joint probability.

Answer

No. For example: - If \(P(B)=0.50\), then \(P(A\cap B)=0.20\). - If \(P(B)=0.20\), then \(P(A\cap B)=0.08\). Both produce \(P(A\mid B)=0.40\).
Conditional probability 26
Assume \(P(A)>0\) and \(P(B)>0\). Starting from the definition of conditional probability, prove the two forms of the multiplication rule: \(P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A)\). Explain why the two products describe the same joint event even though they condition in opposite directions.

Hints

- Write each conditional probability as a ratio. - Clear the denominator in each equation. - Compare the joint event appearing in both definitions.

Solution

1. From \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\), multiply by \(P(B)\) to obtain \(P(A\cap B)=P(A\mid B)P(B)\). 2. From \(P(B\mid A)=\frac{P(A\cap B)}{P(A)}\), multiply by \(P(A)\) to obtain \(P(A\cap B)=P(B\mid A)P(A)\). 3. Both expressions calculate the probability of the same intersection; they only factor it using different starting events.

Answer

\(P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A)\). The factors differ, but both products equal the same intersection probability.
Conditional probability 27
Use the probability tree shown. Given that a randomly selected item is defective, find the probability that it came from Machine 3.
Figure for problem 547319

Hints

- Find each machine’s contribution to the defective items. - Add those contributions to form the conditioning group. - Compare Machine 3’s defective contribution with the total defective probability.

Solution

1. The joint defective shares are \(0.50\cdot0.01=0.005\), \(0.30\cdot0.02=0.006\), and \(0.20\cdot0.05=0.010\). 2. The total defective probability is \(0.005+0.006+0.010=0.021\). 3. \(P(\text{Machine 3}\mid\text{defective})=\frac{0.010}{0.021}=\frac{10}{21}\approx 0.476\).

Answer

\(\frac{10}{21}\approx 0.476\).
Conditional probability 28
A point is selected uniformly from the square shown. Given that the point lies in the shaded region, find the probability that \(x<\frac12\).
Figure for problem 547324

Hints

- Treat the condition as a restricted geometric region. - Find the area satisfying both conditions. - Divide intersection area by conditioning-region area.

Solution

1. The conditioning region is the triangle above \(y=x\), with area \(\frac12\). 2. In the half-strip \(x<\frac12\), the region above the line has area \(\frac12-\frac12\cdot\frac12\cdot\frac12=\frac38\). 3. The conditional probability is \(\frac{\frac38}{\frac12}=\frac34\).

Answer

The probability is \(\frac34\).
Conditional probability 29
A fair six-sided die is rolled repeatedly until the first \(6\). Given that the first \(6\) occurs within the first three rolls, find the probability that it occurs on the second roll.

Hints

- Describe the exact sequence required for the first success to occur on the specified roll. - Find the probability of the entire stopping-time condition. - The favorable event is contained inside the conditioning event.

Solution

1. The probability that the first \(6\) occurs on roll \(2\) is \(\frac56\cdot\frac16=\frac{5}{36}\). 2. The probability that a \(6\) occurs within three rolls is \(1-(\frac56)^3=\frac{91}{216}\). 3. The conditional probability is \(\frac{\frac{5}{36}}{\frac{91}{216}}=\frac{30}{91}\).

Answer

The probability is \(\frac{30}{91}\approx 0.330\).
Conditional probability 30
Two fair dice are rolled. Given that the product is even, find the probability that the sum is even.

Hints

- Characterize parity patterns for product and sum. - Remove patterns excluded by the condition. - Count ordered die outcomes for the surviving pattern.

Solution

1. Product even excludes only odd-odd outcomes, so the conditioning set has \(36-9=27\) outcomes. 2. An even sum occurs for even-even or odd-odd. Within the conditioning set, only even-even remains, with \(9\) outcomes. 3. The conditional probability is \(\frac{9}{27}=\frac13\).

Answer

The probability is \(\frac13\).
Conditional probability 31
Within event \(B\), the odds in favor of event \(A\) are \(3\) to \(7\). Also, \(P(B)=0.40\). Find \(P(A\cap B)\) and \(P(A^c\cap B)\).

Hints

- Convert the two parts of the odds into shares of the conditioning event. - The two requested intersections partition event \(B\). - Check that the two joint probabilities add to the given probability of \(B\).

Solution

1. Odds of \(3\) to \(7\) within \(B\) mean \(P(A\mid B)=\frac{3}{10}\) and \(P(A^c\mid B)=\frac{7}{10}\). 2. \(P(A\cap B)=P(B)P(A\mid B)=0.40\cdot0.30=0.12\). 3. \(P(A^c\cap B)=0.40\cdot0.70=0.28\).

Answer

\(P(A\cap B)=0.12\) and \(P(A^c\cap B)=0.28\).
Conditional probability 32
Assume \(P(A)>0\) and \(P(B)>0\). Can both statements hold? \(P(A\mid B)=1\) and \(P(B\mid A)=0\). Prove your conclusion using intersection probabilities.

Hints

- Express both conditionals through the same intersection. - Use the positivity of each conditioning event. - Compare the two consequences for the intersection probability.

Solution

1. \(P(A\mid B)=1\) implies \(P(A\cap B)=P(B)>0\). 2. \(P(B\mid A)=0\) implies \(P(A\cap B)=0\). 3. The same intersection cannot have both positive probability and probability \(0\). Therefore the two statements are incompatible under the positive-margin assumptions.

Answer

No. The first statement forces \(P(A\cap B)=P(B)>0\), while the second forces \(P(A\cap B)=0\).
Conditional probability 33
Suppose \(P(A)=0.55\), \(P(B)=0.40\), and \(P(A\cap B)=0.25\). Find \(P(A\mid B^c)\).

Hints

- Split event \(A\) into its parts inside and outside \(B\). - Use the complement of the conditioning event as the denominator. - Verify that the intersection part does not exceed the complement.

Solution

1. \(P(A\cap B^c)=P(A)-P(A\cap B)=0.55-0.25=0.30\). 2. \(P(B^c)=1-0.40=0.60\). 3. \(P(A\mid B^c)=\frac{0.30}{0.60}=0.50\).

Answer

\(P(A\mid B^c)=0.50\).
Conditional probability 34
Assume \(P(A)>0\), \(P(B)>0\), and \(P(A\cap B)>0\). Prove that \(P(A\mid B)=P(B\mid A)\) if and only if \(P(A)=P(B)\).

Hints

- Write both reverse conditionals with their common numerator. - The positive intersection allows cancellation. - Identify what must be equal after the common factor is removed.

Solution

1. The two conditionals are \(\frac{P(A\cap B)}{P(B)}\) and \(\frac{P(A\cap B)}{P(A)}\). 2. Because the common numerator is positive, equality of the ratios is equivalent to equality of the denominators. 3. Thus the reverse conditionals are equal exactly when \(P(A)=P(B)\).

Answer

\(P(A\mid B)=P(B\mid A)\) exactly when \(P(A)=P(B)\), under the stated positivity assumptions.
Conditional probability 35
A discrete random variable \(X\) has the distribution \(P(X=0)=0.10\), \(P(X=1)=0.20\), \(P(X=2)=0.30\), and \(P(X=3)=0.40\). Condition on the event \(X\ge1\). a) Find the conditional distribution of \(X\). b) Find \(P(X\ge2\mid X\ge1)\).

Hints

- Add the probabilities of all values retained by the condition. - Divide each retained original probability by that conditioning probability. - For part b, combine the retained probabilities for \(X=2\) and \(X=3\).

Solution

1. The conditioning event has probability \(P(X\ge1)=0.20+0.30+0.40=0.90\). 2. Renormalizing the retained values gives \(P(X=1\mid X\ge1)=\frac{0.20}{0.90}=\frac29\), \(P(X=2\mid X\ge1)=\frac{0.30}{0.90}=\frac13\), and \(P(X=3\mid X\ge1)=\frac{0.40}{0.90}=\frac49\). The conditional probability at \(X=0\) is \(0\). 3. \(P(X\ge2\mid X\ge1)=\frac{0.30+0.40}{0.90}=\frac79\).

Answer

a) \(P(X=0\mid X\ge1)=0\), \(P(X=1\mid X\ge1)=\frac29\), \(P(X=2\mid X\ge1)=\frac13\), and \(P(X=3\mid X\ge1)=\frac49\). b) \(P(X\ge2\mid X\ge1)=\frac79\).
Conditional probability 36
Hana claims that \(P(A^c\mid B^c)=1-P(A\mid B)\) for all events with positive conditioning probabilities. Use the uniform sample space \(S=\{1,2,3,4\}\), with \(A=\{1\}\) and \(B=\{1,2\}\), to refute the claim. State the correct conditional-complement identity.

Hints

- Compute each conditional probability from its own restricted sample space. - Changing the event and changing the condition are different operations. - A conditional complement must use the same conditioning event.

Solution

1. \(P(A\mid B)=\frac12\), so \(1-P(A\mid B)=\frac12\). 2. \(B^c=\{3,4\}\) and \(A^c=\{2,3,4\}\). Thus \(P(A^c\mid B^c)=\frac22=1\). 3. Since \(1\ne\frac12\), the claim is false. 4. The correct identity keeps the conditioning event fixed: \(P(A^c\mid B)=1-P(A\mid B)\).

Answer

The proposed identity is false: \(P(A^c\mid B^c)=1\), while \(1-P(A\mid B)=\frac12\). The correct identity is \(P(A^c\mid B)=1-P(A\mid B)\).
Conditional probability 37
In the uniform sample space \(S=\{1,2,3,4\}\), let \(B=\{1,2\}\), \(A=\{1,3\}\), and \(C=\{2,3\}\). a) Are \(A\) and \(C\) mutually exclusive in the original sample space? b) After restricting attention to outcomes in \(B\), can \(A\) and \(C\) occur together? Compute \(P(A\cap C\mid B)\) to justify your answer. c) Explain how the answers to a) and b) can differ.

Hints

- Find the intersection \(A\cap C\) in the original sample space first. - Then check whether that shared outcome is still available after restricting to \(B\). - Express the restricted overlap with the conditional-probability ratio.

Solution

1. In the original sample space, \(A\cap C=\{3\}\), so \(A\) and \(C\) are not mutually exclusive. 2. The conditioning event \(B\) excludes outcome \(3\). Therefore \(A\cap C\cap B=\varnothing\). 3. Since \(P(B)>0\), \(P(A\cap C\mid B)=\frac{P(A\cap C\cap B)}{P(B)}=0\). 4. Conditioning changes the reference set: an overlap that lies entirely outside \(B\) contributes no probability within the restricted sample space.

Answer

a) No; \(A\cap C=\{3\}\). b) No. Within \(B\), they cannot occur together, and \(P(A\cap C\mid B)=0\). c) Their only shared outcome, \(3\), is excluded when attention is restricted to \(B\).
Conditional probability 38
Events \(A\) and \(B\) satisfy \(P(A)=0.50\) and \(P(B)=0.40\). The probability that exactly one of the events occurs is \(0.30\). Find \(P(A\cap B)\) and \(P(A\mid B)\).

Hints

- Divide the outcomes into A only, B only, both, and neither. - What part of each event's probability is counted twice when you add its two margins? - Once you find the overlap, decide which event sets the denominator for the conditional chance.

Solution

1. Let \(x=P(A\cap B)\). The probability of \(A\) but not \(B\) is \(0.50-x\); the probability of \(B\) but not \(A\) is \(0.40-x\). 2. Since exactly one event occurs with probability \(0.30\), \((0.50-x)+(0.40-x)=0.30\). 3. Solving gives \(0.90-2x=0.30\), so \(x=0.30\). 4. Therefore \(P(A\mid B)=\frac{0.30}{0.40}=0.75\).

Answer

\(P(A\cap B)=0.30\) and \(P(A\mid B)=0.75\).
Conditional probability 39
Among \(200\) students, \(80\) are in event \(A\), \(70\) are in event \(B\), and \(50\) are in event \(C\). Also, \(|A\cap B|=30\), \(|A\cap C|=25\), \(|B\cap C|=20\), and \(|A\cap B\cap C|=10\). Find \(P(A\mid B\cup C)\).

Hints

- First determine the size of the conditioning union. - Rewrite the favorable event by distributing the intersection over the union. - Watch for observations counted in both favorable pieces.

Solution

1. \(|B\cup C|=70+50-20=100\). 2. \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\). 3. Its size is \(30+25-10=45\), subtracting the triple intersection once. 4. Therefore \(P(A\mid B\cup C)=\frac{45}{100}=0.45\).

Answer

\(P(A\mid B\cup C)=0.45\).
Conditional probability 40
In a summer survey of \(100\) people, \(40\) bought ice cream, \(30\) got sunburned, and \(25\) did both. a) Find \(P(\text{sunburned}\mid\text{bought ice cream})\). b) A headline says, “Buying ice cream causes sunburn.” Explain why the conditional probability does not justify that causal conclusion, and name one plausible common cause.

Hints

- Use the group named after “given” as the denominator. - Distinguish observing a relationship from assigning a treatment. - Look for a third variable that could influence both events.

Solution

1. The conditional probability is \(\frac{25}{40}=0.625\). 2. The data are observational and establish association, not the effect of an intervention. 3. Hot, sunny weather can increase both ice-cream purchases and sun exposure, creating the observed association without ice cream causing sunburn.

Answer

a) \(P(\text{sunburned}\mid\text{bought ice cream})=0.625\). b) The survey shows association only. Hot, sunny weather is a plausible common cause.
Conditional probability 41
At a community recreation center, \(120\) members attend exercise classes. Of them, \(30\) take a boxing class, and \(18\) of those boxing students arrive by bus. a) What fraction of **boxing students** arrive by bus? b) Find the same fraction by dividing the share of **all members** who take boxing and arrive by bus by the share who take boxing. c) Why do members outside the boxing class not affect either answer?

Hints

- Is the bus-rider count a fraction of every center member or of the boxing class? - What happens when both parts of a ratio are divided by the same number? - Check whether adding new members outside boxing would change your answer.

Solution

1. Of \(30\) boxing students, \(18\) arrive by bus, so the fraction is \(\frac{18}{30}=\frac{3}{5}\). 2. Of all \(120\) members, \(\frac{18}{120}\) take boxing and arrive by bus, while \(\frac{30}{120}\) take boxing. Their ratio is \(\frac{18/120}{30/120}=\frac{3}{5}\). 3. Both fractions use only the boxing group as the final reference group, so other members do not change the result.

Answer

a) \(\frac{3}{5}\). b) \(\frac{18/120}{30/120}=\frac{3}{5}\). c) Both methods ultimately compare the same \(18\) bus riders with the same \(30\) boxing students.
Conditional probability 42
A print shop made \(100\) posters: \(20\) on an older printer and \(80\) on a newer one. In this batch, \(30\%\) of the older printer's posters and \(5\%\) of the newer printer's posters have smudged ink. A poster is chosen at random from **all the smudged posters**. What is the probability that it came from the older printer? Explain why \(30\%\) is not the answer.

Hints

- How many posters from each printer are smudged? - Which posters are eligible to be chosen in the question? - Why is the percentage within one printer's posters a different question?

Solution

1. There are \(0.30\cdot20=6\) smudged posters from the older printer. 2. There are \(0.05\cdot80=4\) smudged posters from the newer printer. 3. Of the \(6+4=10\) smudged posters, \(6\) came from the older printer. The conditional probability is \(\frac{6}{10}=0.60\). 4. The \(30\%\) rate uses all older-printer posters as the denominator; the question uses all smudged posters.

Answer

\(\frac{6}{10}=0.60\), or \(60\%\). The \(30\%\) rate is for smudges among older-printer posters, not older-printer posters among smudged ones.
Conditional probability 43
A theater sold one ticket each to \(80\) customers. Of these, \(30\) bought online. Among online buyers, \(60\%\) also bought a program. Among the remaining buyers, \(10\) bought a program. Complete a two-way table with purchase method (online or at the box office) as rows and program purchase (yes or no) as columns. Then find the probability that a randomly selected ticket buyer bought both an online ticket and a program.

Hints

- Does the \(60\%\) describe all ticket buyers or only online buyers? - How can you use the row totals to fill the no-program cells? - The last question asks about both conditions out of which complete group?

Solution

1. The online group contains \(30\) buyers, so \(0.60\cdot30=18\) bought programs and \(12\) did not. 2. The box-office group contains \(50\) buyers, of whom \(10\) bought programs and \(40\) did not. 3. Program totals are \(28\) yes and \(52\) no. The joint probability of buying online and buying a program is \(\frac{18}{80}=\frac{9}{40}=0.225\).

Answer

<table> <tr><th></th><th>Program</th><th>No program</th><th>Total</th></tr> <tr><td>Online</td><td>\(18\)</td><td>\(12\)</td><td>\(30\)</td></tr> <tr><td>Box office</td><td>\(10\)</td><td>\(40\)</td><td>\(50\)</td></tr> <tr><td>Total</td><td>\(28\)</td><td>\(52\)</td><td>\(80\)</td></tr> </table> The probability is \(\frac{18}{80}=\frac{9}{40}=0.225\).
Conditional probability 44
Events \(A\) and \(B\) are mutually exclusive, and \(C\) is an event with \(P(C)>0\). Show that \(P(A\cup B\mid C)=P(A\mid C)+P(B\mid C)\).

Hints

- Intersect both original events with the conditioning event. - What happens to their overlap after this restriction? - How does restricting every event to \(C\) change the meaning of the probabilities?

Solution

1. Since \(A\cap B=\varnothing\), the events \(A\cap C\) and \(B\cap C\) are also mutually exclusive. 2. Also, \((A\cup B)\cap C=(A\cap C)\cup(B\cap C)\). 3. Add their disjoint probabilities: \(P((A\cup B)\cap C)=P(A\cap C)+P(B\cap C)\). 4. Divide by \(P(C)>0\) to get \(P(A\cup B\mid C)=P(A\mid C)+P(B\mid C)\).

Answer

Since \(A\) and \(B\) have no outcomes in common, neither do \(A\cap C\) and \(B\cap C\). Thus \(P((A\cup B)\cap C)=P(A\cap C)+P(B\cap C)\). Dividing by \(P(C)>0\) gives \(P(A\cup B\mid C)=P(A\mid C)+P(B\mid C)\).
Conditional probability 45
A transit company operates \(10\) small buses carrying \(10\) riders each and \(5\) large buses carrying \(40\) riders each. All buses are full. Two distinct riders are selected uniformly from all riders, conditional on the two riders being on the same bus. Find the probability that they are on a large bus.

Hints

- Represent the random selection in a way that keeps the original elementary outcomes equally likely. - Apply the condition “the two riders are on the same bus” before comparing bus sizes. - Check whether every bus contributes the same number of eligible selections after conditioning.

Solution

1. A small bus contributes \(\binom{10}{2}=45\) same-bus rider pairs, so all small buses contribute \(10\cdot45=450\). 2. A large bus contributes \(\binom{40}{2}=780\) pairs, so all large buses contribute \(5\cdot780=3900\). 3. Under the condition that the pair shares a bus, the eligible pair count is \(450+3900=4350\). 4. The probability of a large bus is \(\frac{3900}{4350}=\frac{26}{29}\).

Answer

The conditional probability is \(\frac{26}{29}\approx0.897\).
Conditional probability 46
In a fictional school survey of \(100\) students, \(60\) play a sport and \(50\) belong to the art club. Students may do both. Among art-club members, what is the **smallest** possible share who play a sport? What is the **largest** possible share? Show how the two extreme cases could be arranged without changing either group total.

Hints

- How many students do **not** play a sport? - Could every art-club member be outside the sport group? Could every one be inside? - Construct the counts in the four sport/art categories for each extreme.

Solution

1. At most \(40\) of the \(50\) art-club members can be outside the sport group, because \(40\) students do not play a sport. At least \(50-40=10\) art-club members play a sport. 2. All \(50\) art-club members could also play a sport, since the sport group contains \(60\) students. 3. The smallest conditional share is \(\frac{10}{50}=0.20\); the largest is \(\frac{50}{50}=1\). 4. The minimum is possible with \(10\) in both, \(50\) sport only, \(40\) art only, and none in neither. The maximum is possible with \(50\) in both, \(10\) sport only, none in art only, and \(40\) in neither.

Answer

The share of art-club members who play a sport can be as low as \(20\%\) or as high as \(100\%\). Minimum example: both \(10\), sport only \(50\), art only \(40\), neither \(0\). Maximum example: both \(50\), sport only \(10\), art only \(0\), neither \(40\).
Conditional probability 47
Suppose \(P(B)>0\) and \(P(A\mid B)=1\). What does this imply about \(P(B\setminus A)\)? Does it necessarily imply the strict set inclusion \(B\subseteq A\) when probability-zero outcomes are possible? Explain.

Hints

- Decompose the conditioning event into the part inside and outside \(A\). - Use the fact that the conditional probability equals \(1\). - Distinguish an empty set from a nonempty probability-zero set.

Solution

1. \(P(A\mid B)=1\) gives \(P(A\cap B)=P(B)\). 2. Since \(B=(A\cap B)\cup(B\setminus A)\) is a disjoint union, \(P(B\setminus A)=P(B)-P(A\cap B)=0\). 3. Thus \(B\) is contained in \(A\) up to a probability-zero part. 4. Strict set inclusion need not follow if \(B\setminus A\) is nonempty but has probability \(0\).

Answer

\(P(B\setminus A)=0\). The condition implies \(B\subseteq A\) only up to probability-zero outcomes, not necessarily as a literal set inclusion.
Conditional probability 48
There is no general monotonicity rule for the conditioning event. In the uniform sample space \(S=\{1,2,3,4\}\), let \(A=\{1,2\}\). a) Find events \(B\subset C\) such that \(P(A\mid B)>P(A\mid C)\). b) Find events \(D\subset E\) such that \(P(A\mid D)<P(A\mid E)\). c) Explain what the two examples show.

Hints

- In this uniform sample space, conditional probability can be computed from the fraction of the conditioning set that lies in \(A\). - You need two nested-set examples in which that fraction moves in opposite directions. - Focus on how adding outcomes changes both the intersection with \(A\) and the size of the conditioning event.

Solution

1. Take \(B=\{1\}\) and \(C=\{1,3\}\). Then \(P(A\mid B)=1\) and \(P(A\mid C)=\frac12\). 2. Take \(D=\{3\}\) and \(E=\{1,3\}\). Then \(P(A\mid D)=0\) and \(P(A\mid E)=\frac12\). 3. Enlarging the conditioning event can either decrease or increase a conditional probability, depending on which new outcomes are added.

Answer

a) One example is \(B=\{1\}\subset C=\{1,3\}\), giving \(1>\frac12\). b) One example is \(D=\{3\}\subset E=\{1,3\}\), giving \(0<\frac12\). c) Conditional probability is not monotone in the conditioning event.
Conditional probability 49
Suppose events \(B\) and \(C\) satisfy \(P(B\triangle C)=0\) and have positive probability. Prove that for every event \(A\), \(P(A\mid B)=P(A\mid C)\).

Hints

- Break the symmetric difference into the two one-sided differences. - Compare the probabilities of the conditioning events and of their intersections with \(A\). - Probability-zero discrepancies do not change either conditional ratio.

Solution

1. Since \(B\triangle C\) has probability \(0\), the differences \(B\setminus C\) and \(C\setminus B\) each have probability \(0\). 2. Therefore \(P(B)=P(B\cap C)=P(C)\). 3. The sets \(A\cap B\) and \(A\cap C\) can differ only inside \(B\triangle C\), so their probabilities are equal. 4. Equal numerators and equal positive denominators give \(P(A\mid B)=P(A\mid C)\).

Answer

For every event \(A\), \(P(A\mid B)=P(A\mid C)\). Conditioning is unchanged when the conditioning events differ only on a null set.
Conditional probability 50
Let \(B\) and \(C\) be mutually exclusive events with \(P(B)>0\) and \(P(C)>0\). Prove that \(P(A\mid B\cup C)=P(A\mid B)P(B\mid B\cup C)+P(A\mid C)P(C\mid B\cup C)\).

Hints

- Split the conditioned intersection across the two disjoint parts. - Insert each subgroup probability as a factor and its reciprocal. - Interpret the result as a weighted average within the union.

Solution

1. Since \(B\) and \(C\) are disjoint, \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\) is a disjoint union. 2. Divide by \(P(B\cup C)\): \(P(A\mid B\cup C)=\frac{P(A\cap B)}{P(B\cup C)}+\frac{P(A\cap C)}{P(B\cup C)}\). 3. Factor each term as \(\frac{P(A\cap B)}{P(B)}\cdot\frac{P(B)}{P(B\cup C)}\) and similarly for \(C\). 4. These factors are the conditional probabilities in the stated identity.

Answer

\(P(A\mid B\cup C)=P(A\mid B)P(B\mid B\cup C)+P(A\mid C)P(C\mid B\cup C)\).
Conditional probability 51
Events satisfy \(P(A\mid B)=0.60\), \(P(B\mid A)=0.40\), and \(P(A\cup B)=0.70\). Find \(P(A)\), \(P(B)\), and \(P(A\cap B)\).

Hints

- Express both marginal probabilities in terms of the shared intersection. - Substitute those expressions into the union formula. - Solve for the intersection before recovering the marginals.

Solution

1. Let \(I=P(A\cap B)\). Then \(P(B)=\frac{I}{0.60}\) and \(P(A)=\frac{I}{0.40}\). 2. The union equation is \(\frac{I}{0.40}+\frac{I}{0.60}-I=0.70\). 3. This gives \(\frac{19}{6}I=0.70\), so \(I=\frac{21}{95}\), \(P(A)=\frac{21}{38}\), and \(P(B)=\frac{7}{19}\).

Answer

\(P(A\cap B)=\frac{21}{95}\), \(P(A)=\frac{21}{38}\), and \(P(B)=\frac{7}{19}\).
Conditional probability 52
Assume events \(A,B,C\) have positive probabilities and positive pairwise intersections. A report claims \(P(A\mid B)=0.60\), \(P(B\mid A)=0.30\), \(P(B\mid C)=0.50\), \(P(C\mid B)=0.50\), \(P(C\mid A)=0.40\), and \(P(A\mid C)=0.40\). Can all six claims hold simultaneously?

Hints

- For one pair of events, write both directions of the conditional probability definition and notice what quantity appears in both numerators. - What necessary relationship between the two marginal probabilities can be recovered from one pair of reverse conditionals? - Try to combine the analogous necessary relationships for all three pairs and ask what must cancel in any internally consistent model.

Solution

1. For positive intersections, \(\frac{P(A\mid B)}{P(B\mid A)}=\frac{P(A)}{P(B)}\), and similarly around the cycle. 2. Multiplying the three ratios must give \(\frac{P(A)}{P(B)}\cdot\frac{P(B)}{P(C)}\cdot\frac{P(C)}{P(A)}=1\). 3. The reported ratios give \(\frac{0.60}{0.30}\cdot\frac{0.50}{0.50}\cdot\frac{0.40}{0.40}=2\), not \(1\). 4. Therefore the claims are inconsistent.

Answer

No. The cycle of reverse-conditional ratios must multiply to \(1\), but the reported values give \(2\).
Conditional probability 53
A normal variable has mean \(70\) and standard deviation \(10\). Given that an observation is above \(70\), find the conditional probability that it is above \(80\). Use a standard normal table or calculator.
Figure for problem 548855

Hints

- Identify the new reference group created by the given condition. - Express the conditional probability as a ratio of nested tail areas. - Use symmetry to find the area above the mean.

Solution

1. The event \(X>80\) is contained in the event \(X>70\), so \(P(X>80\mid X>70)=\frac{P(X>80)}{P(X>70)}\). 2. The standard score for \(80\) is \(z=\frac{80-70}{10}=1\), so \(P(X>80)=P(Z>1)\approx0.1587\). 3. By symmetry, \(P(X>70)=0.5\). 4. Using the unrounded tail probability, the conditional probability is approximately \(\frac{0.158655}{0.5}\approx0.3173\).

Answer

\(P(X>80\mid X>70)\approx0.3173\), or about \(31.73\%\).

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