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Two-way tables and conditional relative frequency

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Difficulty: 1 – 5

55588910

The table summarizes \(40\) households by whether they have a dog and whether they live in an apartment. <table><tr><td></td><td>Apartment</td><td>Not apartment</td><td>Total</td></tr><tr><td>Dog</td><td>\(14\)</td><td>\(6\)</td><td>\(20\)</td></tr><tr><td>No dog</td><td>\(8\)</td><td>\(12\)</td><td>\(20\)</td></tr><tr><td>Total</td><td>\(22\)</td><td>\(18\)</td><td>\(40\)</td></tr></table> a) Which entry is the joint count for “dog and apartment”? b) Which entry is the marginal total for households with a dog?

Hints

- A joint cell combines one row category with one column category. - A marginal total summarizes an entire row or column.

Solution

1. A joint count lies at the intersection of the two categories, so “dog and apartment” is the cell containing \(14\). 2. A marginal total is at the edge of the table. The total for the Dog row is \(20\).

Answer

a) \(14\) b) \(20\)

55589010

The table shows whether students play a school sport and whether they have an after-school job. <table><tr><td></td><td>Job</td><td>No job</td><td>Total</td></tr><tr><td>Sport</td><td>\(9\)</td><td>\(6\)</td><td>\(15\)</td></tr><tr><td>No sport</td><td>\(5\)</td><td>\(10\)</td><td>\(15\)</td></tr><tr><td>Total</td><td>\(14\)</td><td>\(16\)</td><td>\(30\)</td></tr></table> Among students who play a school sport, what is the relative frequency of having an after-school job?

Hints

- The phrase “among students who play a school sport” tells you which row is the reference group. - Compare the job count within that row with the row total.

Solution

1. Restrict attention to the Sport row, which contains \(15\) students. 2. Of those \(15\), \(9\) have a job, so the conditional relative frequency is \(\frac{9}{15}=0.60\).

Answer

\(0.60\), or \(60\%\).

52206910

A survey of \(400\) students at a school serving grades \(6\) through \(12\) recorded whether students regularly ride a bicycle to school. <table> <tr><td></td><td>Regular bicycle use</td><td>No regular bicycle use</td><td>Total</td></tr> <tr><td>Grades \(6\)–\(8\)</td><td>\(144\)</td><td>\(96\)</td><td>\(240\)</td></tr> <tr><td>Grades \(9\)–\(12\)</td><td>\(64\)</td><td>\(96\)</td><td>\(160\)</td></tr> <tr><td>Total</td><td>\(208\)</td><td>\(192\)</td><td>\(400\)</td></tr> </table> a) Find the probability that a randomly selected surveyed student regularly rides a bicycle to school. b) Now consider only students in grades \(9\)–\(12\). Find the probability that a randomly selected student from this group regularly rides a bicycle to school. Compare it with the result from part a.

Hints

- Identify the total number of surveyed students. - In part b, identify the size of the restricted grade group. - Use the table cell representing both the grade group and regular bicycle use.

Solution

1. The overall probability of regular bicycle use is \(P(B)=\frac{208}{400}=0.52\). 2. Among students in grades \(9\)–\(12\), \(64\) of \(160\) regularly ride a bicycle, so \(P(B\mid H)=\frac{64}{160}=0.40\). 3. The conditional rate of \(40\%\) for grades \(9\)–\(12\) is lower than the overall rate of \(52\%\).

Answer

a) \(0.52\), or \(52\%\). b) \(0.40\), or \(40\%\). This is lower than the overall rate.

52207310

A company surveyed \(200\) employees about work area and usual commuting method. The results are shown below. <table> <tr><td></td><td>Public transit \((T)\)</td><td>Personal vehicle \((V)\)</td></tr> <tr><td>Production \((P)\)</td><td>\(45\)</td><td>\(75\)</td></tr> <tr><td>Administration \((A)\)</td><td>\(55\)</td><td>\(25\)</td></tr> </table> One employee is selected at random. Find each probability. a) The employee works in administration and uses public transit. b) The employee uses public transit, given that the employee works in administration. c) The employee works in administration, given that the employee uses a personal vehicle.

Hints

- Distinguish a joint probability indicated by “and” from a conditional probability indicated by “given that.” - Identify the appropriate denominator for each part. - Add row and column totals to make the table easier to use.

Solution

1. The administration total is \(55+25=80\), and the personal-vehicle total is \(75+25=100\). 2. For part a, \(P(A\cap T)=\frac{55}{200}=0.275\). 3. For part b, \(P(T\mid A)=\frac{55}{80}=0.6875\). 4. For part c, \(P(A\mid V)=\frac{25}{100}=0.25\).

Answer

a) \(P(A\cap T)=0.275\) b) \(P(T\mid A)=0.6875\) c) \(P(A\mid V)=0.25\)

53867710

A two-way table for \(120\) orders contains incorrect entries. The grand total, the entire first row, and the Express column total of \(42\) have been checked and are correct. Exactly two of the remaining entries are wrong. Find and correct them. <table><tr><th></th><th>Express \(E\)</th><th>Not express \(E^c\)</th><th>Total</th></tr><tr><th>Pickup location \(A\)</th><td>\(24\)</td><td>\(36\)</td><td>\(60\)</td></tr><tr><th>No pickup location \(A^c\)</th><td>\(18\)</td><td>\(44\)</td><td>\(62\)</td></tr><tr><th>Total</th><td>\(42\)</td><td>\(78\)</td><td>\(120\)</td></tr></table>

Hints

- Check row totals and column totals independently. - Find the incorrect interior entry first, then update its row total.

Solution

1. The checked grand total and Express total imply that the Not express total is \(120-42=78\), so that entry is correct. 2. The Express column is consistent because \(24+18=42\), so the interior entry \(18\) is correct. 3. The lower-right interior entry must be \(78-36=42\), not \(44\). 4. The second-row total must then be \(18+42=60\), not \(62\).

Answer

Replace \(44\) with \(42\). Replace the second-row total \(62\) with \(60\).

51010510

A two-way probability table for events \(A\) and \(B\) is partially completed. Fill in the missing entries. Then find \(P(A\mid B^c)\). <table border="1"> <tr><td></td><td>\(A\)</td><td>\(A^c\)</td><td>Total</td></tr> <tr><td>\(B\)</td><td></td><td>\(0.40\)</td><td></td></tr> <tr><td>\(B^c\)</td><td></td><td></td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.30\)</td><td></td><td>\(1\)</td></tr> </table>

Hints

- Use row and column totals to find missing cells by addition or subtraction. - In a conditional probability, identify the group named after the vertical bar. - Which cell is the intersection of \(A\) and \(B^c\)?

Solution

1. Since \(P(B^c)=0.40\), \(P(B)=1-0.40=0.60\). 2. Because \(P(B\cap A^c)=0.40\), \(P(A\cap B)=0.60-0.40=0.20\). 3. Since \(P(A)=0.30\), \(P(A\cap B^c)=0.30-0.20=0.10\). 4. The remaining entries are \(P(A^c)=0.70\) and \(P(A^c\cap B^c)=0.40-0.10=0.30\). 5. The completed table is: <table border="1"> <tr><td></td><td>\(A\)</td><td>\(A^c\)</td><td>Total</td></tr> <tr><td>\(B\)</td><td>\(0.20\)</td><td>\(0.40\)</td><td>\(0.60\)</td></tr> <tr><td>\(B^c\)</td><td>\(0.10\)</td><td>\(0.30\)</td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.30\)</td><td>\(0.70\)</td><td>\(1\)</td></tr> </table> 6. Therefore, \(P(A\mid B^c)=\frac{P(A\cap B^c)}{P(B^c)}=\frac{0.10}{0.40}=0.25\).

Answer

<table border="1"> <tr><td></td><td>\(A\)</td><td>\(A^c\)</td><td>Total</td></tr> <tr><td>\(B\)</td><td>\(0.20\)</td><td>\(0.40\)</td><td>\(0.60\)</td></tr> <tr><td>\(B^c\)</td><td>\(0.10\)</td><td>\(0.30\)</td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.30\)</td><td>\(0.70\)</td><td>\(1\)</td></tr> </table> \(P(A\mid B^c)=0.25\)

51474610

A recreation club has \(100\) members. Sixty members play basketball, \(50\) play tennis, and \(30\) play both sports. a) Create a complete two-way table for the events \(B\): plays basketball and \(T\): plays tennis. b) A member is selected at random. Given that the member plays basketball, find the probability that the member also plays tennis.

Hints

- Enter the three counts stated directly in the problem. - Use row and column totals to find the remaining cells. - The phrase “given that” tells you which subgroup becomes the denominator.

Solution

1. The intersection count is \(|B\cap T|=30\). 2. Basketball only: \(|B\cap T^c|=60-30=30\). 3. Tennis only: \(|B^c\cap T|=50-30=20\). 4. Neither sport: \(100-(30+30+20)=20\). 5. The completed table is: <table> <tr><td></td><td>\(T\)</td><td>\(T^c\)</td><td>Total</td></tr> <tr><td>\(B\)</td><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr> <tr><td>\(B^c\)</td><td>\(20\)</td><td>\(20\)</td><td>\(40\)</td></tr> <tr><td>Total</td><td>\(50\)</td><td>\(50\)</td><td>\(100\)</td></tr> </table> 6. Restricting to the \(60\) basketball players, \(P(T\mid B)=\frac{30}{60}=0.50\).

Answer

a) <table> <tr><td></td><td>\(T\)</td><td>\(T^c\)</td><td>Total</td></tr> <tr><td>\(B\)</td><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr> <tr><td>\(B^c\)</td><td>\(20\)</td><td>\(20\)</td><td>\(40\)</td></tr> <tr><td>Total</td><td>\(50\)</td><td>\(50\)</td><td>\(100\)</td></tr> </table> b) \(P(T\mid B)=0.50\), or \(50\%\).

51475910

A company makes USB drives in two colors: silver \((S)\) and black \((B)\). Some drives are defective \((D)\). Silver drives make up \(70\%\) of production. Of the silver drives, \(5\%\) are defective. Overall, \(4.4\%\) of all drives are defective. a) Create a complete two-way probability table for color and condition. b) Given that a drive is defective, find the probability that it is black.

Hints

- Enter the probabilities stated directly in the problem. - Use row and column totals to complete the table. - Distinguish “a black drive is defective” from “a defective drive is black.”

Solution

1. The silver-and-defective probability is \(P(S\cap D)=P(S)P(D\mid S)=0.70\cdot0.05=0.035\). 2. Since \(P(B)=0.30\), the black-and-defective probability is \(P(B\cap D)=0.044-0.035=0.009\). 3. The remaining cells are \(P(S\cap D^c)=0.70-0.035=0.665\) and \(P(B\cap D^c)=0.30-0.009=0.291\). Also, \(P(D^c)=1-0.044=0.956\). 4. The completed table is: <table> <tr><td></td><td>\(D\)</td><td>\(D^c\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(0.035\)</td><td>\(0.665\)</td><td>\(0.70\)</td></tr> <tr><td>\(B\)</td><td>\(0.009\)</td><td>\(0.291\)</td><td>\(0.30\)</td></tr> <tr><td>Total</td><td>\(0.044\)</td><td>\(0.956\)</td><td>\(1\)</td></tr> </table> 5. Therefore, \(P(B\mid D)=\frac{0.009}{0.044}=\frac{9}{44}\approx0.2045\).

Answer

a) <table> <tr><td></td><td>\(D\)</td><td>\(D^c\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(0.035\)</td><td>\(0.665\)</td><td>\(0.70\)</td></tr> <tr><td>\(B\)</td><td>\(0.009\)</td><td>\(0.291\)</td><td>\(0.30\)</td></tr> <tr><td>Total</td><td>\(0.044\)</td><td>\(0.956\)</td><td>\(1\)</td></tr> </table> b) \(P(B\mid D)=\frac{9}{44}\approx0.2045\), or about \(20.45\%\).

51477310

An automated inspection system checks \(5000\) LED bulbs. Of the bulbs, \(150\) are defective. The system correctly rejects \(140\) defective bulbs, but it also rejects \(200\) bulbs that are not defective. a) Create a complete two-way table. b) Given that a bulb was rejected, find the probability that it is actually not defective. c) Explain one economic consequence of your answer to part b.

Hints

- Enter the counts stated directly, then use subtraction to find the missing cells. - Decide which variable belongs in the rows and which belongs in the columns. - For part b, use only rejected bulbs as the denominator. - Consider the cost of rejecting products that are actually usable.

Solution

1. There are \(5000-150=4850\) nondefective bulbs. 2. The defective-and-rejected count is \(140\), and the nondefective-and-rejected count is \(200\). 3. The defective-and-not-rejected count is \(150-140=10\). The nondefective-and-not-rejected count is \(4850-200=4650\). 4. The total rejected count is \(140+200=340\), and the total not rejected count is \(10+4650=4660\). 5. Therefore, \(P(D^c\mid R)=\frac{200}{340}=\frac{10}{17}\approx0.5882\). 6. Because more than half of the rejected bulbs are usable, the company may lose good products or incur extra inspection costs.

Answer

a) <table> <tr><th></th><th>Rejected \((R)\)</th><th>Not rejected \((R^c)\)</th><th>Total</th></tr> <tr><th>Defective \((D)\)</th><td>\(140\)</td><td>\(10\)</td><td>\(150\)</td></tr> <tr><th>Not defective \((D^c)\)</th><td>\(200\)</td><td>\(4650\)</td><td>\(4850\)</td></tr> <tr><th>Total</th><td>\(340\)</td><td>\(4660\)</td><td>\(5000\)</td></tr> </table> b) \(P(D^c\mid R)=\frac{10}{17}\approx0.5882\), or about \(58.82\%\). c) The company may discard usable bulbs or spend money retesting them.

51477410

A choir director screens \(200\) children to identify those who can sing very high notes accurately. Forty children can sing the notes. The screening correctly identifies \(36\) of them, but it also identifies \(10\) children who cannot actually sing the notes accurately. a) Display the data in a two-way table. b) Given that a child did not pass the screening, find the probability that the child truly cannot sing the high notes accurately. c) Which result is more reliable: a positive screening result or a negative screening result? Justify your answer mathematically.

Hints

- Enter the absolute frequencies before converting to probabilities. - For part b, use the group of children with negative results as the denominator. - Compare the two conditional probabilities that measure correct conclusions after positive and negative results. - Use the row and column totals to check the table.

Solution

1. Of the \(160\) children who cannot sing the notes accurately, \(10\) test positive and \(150\) test negative. 2. Of the \(40\) children who can sing the notes, \(36\) test positive and \(4\) test negative. 3. There are \(46\) positive results and \(154\) negative results. 4. Thus, \(P(T^c\mid -)=\frac{150}{154}=\frac{75}{77}\approx0.9740\). 5. For a positive result, \(P(T\mid +)=\frac{36}{46}=\frac{18}{23}\approx0.7826\). Since \(0.9740>0.7826\), the negative result is more reliable for the corresponding conclusion.

Answer

a) <table> <tr><th></th><th>Positive \((+)\)</th><th>Negative \((- )\)</th><th>Total</th></tr> <tr><th>Can sing the notes \((T)\)</th><td>\(36\)</td><td>\(4\)</td><td>\(40\)</td></tr> <tr><th>Cannot sing the notes \((T^c)\)</th><td>\(10\)</td><td>\(150\)</td><td>\(160\)</td></tr> <tr><th>Total</th><td>\(46\)</td><td>\(154\)</td><td>\(200\)</td></tr> </table> b) \(P(T^c\mid -)=\frac{75}{77}\approx0.9740\), or about \(97.40\%\). c) A negative result is more reliable: \(P(T^c\mid -)\approx97.40\%\), while \(P(T\mid +)\approx78.26\%\).

51477510

A warehouse tested a new fire-alarm system for \(365\) days. Small fires occurred on \(5\) days, and the system sounded an alarm on all \(5\) of those days. The system also sounded a false alarm on \(10\) days when there was no fire. a) Organize the observations in a two-way table. b) Given that the alarm sounds, find the probability that there is actually a fire. c) An employee says, “The system is perfect because it detected every fire.” Evaluate this statement using your answer to part b.

Hints

- Use the statement that every fire was detected to determine the missed-fire cell. - For part b, restrict the table to alarm days. - Compare “probability of an alarm given a fire” with “probability of a fire given an alarm.” - Consider how frequent false alarms affect the usefulness of the system.

Solution

1. There were \(365-5=360\) days without a fire. 2. The fire-and-alarm count is \(5\), and the no-fire-and-alarm count is \(10\). 3. Since every fire was detected, the fire-and-no-alarm count is \(0\). The no-fire-and-no-alarm count is \(360-10=350\). 4. There were \(15\) alarm days. Therefore, \(P(F\mid A)=\frac{5}{15}=\frac{1}{3}\). 5. The system has perfect observed sensitivity in this sample, but only one third of its alarms correspond to a fire. The employee's statement ignores the false alarms.

Answer

a) <table> <tr><th></th><th>Alarm \((A)\)</th><th>No alarm \((A^c)\)</th><th>Total</th></tr> <tr><th>Fire \((F)\)</th><td>\(5\)</td><td>\(0\)</td><td>\(5\)</td></tr> <tr><th>No fire \((F^c)\)</th><td>\(10\)</td><td>\(350\)</td><td>\(360\)</td></tr> <tr><th>Total</th><td>\(15\)</td><td>\(350\)</td><td>\(365\)</td></tr> </table> b) \(P(F\mid A)=\frac{1}{3}\), or about \(33.33\%\). c) The statement is incomplete. The system detected every observed fire, but two out of every three alarms were false alarms.

52207410

A bookstore observed \(150\) visitors during one morning. Of these, \(60\) were returning customers and \(90\) were occasional customers. Forty-eight returning customers made a purchase, while \(30\) occasional customers made a purchase. Find each probability. a) A randomly selected visitor made a purchase. b) A visitor is a returning customer, given that the visitor made a purchase. c) A visitor made no purchase, given that the visitor is an occasional customer.

Hints

- Organize the information in a two-way table. - For each conditional probability, identify the group that forms the denominator. - Decide whether each question uses all visitors or only a stated subgroup.

Solution

1. Returning customers who made no purchase: \(60-48=12\). Occasional customers who made no purchase: \(90-30=60\). 2. There were \(48+30=78\) purchases, so \(P(P)=\frac{78}{150}=0.52\). 3. Of the \(78\) purchasers, \(48\) were returning customers. Thus, \(P(R\mid P)=\frac{48}{78}=\frac{8}{13}\approx0.6154\). 4. Of the \(90\) occasional customers, \(60\) made no purchase. Thus, \(P(P^c\mid O)=\frac{60}{90}=\frac{2}{3}\approx0.6667\).

Answer

a) \(P(P)=0.52\) b) \(P(R\mid P)=\frac{8}{13}\approx0.6154\) c) \(P(P^c\mid O)=\frac{2}{3}\approx0.6667\)

52207510

Lakeshore Heights High School surveys students about interest in volleyball \((V)\) and basketball \((B)\). The results show that \(30\%\) are interested in volleyball, \(20\%\) are interested in both sports, and \(60\%\) are not interested in basketball. a) Create a complete two-way table of relative frequencies. b) Find the probability that a randomly selected student is interested in basketball or volleyball. c) Find the probability that a student is interested in basketball, given that the student is interested in volleyball. d) Find \(P(B\mid V^c)\) and interpret it in context.

Hints

- Enter the probabilities stated directly, then use row and column totals. - For “or,” use the union and avoid counting the intersection twice. - In a conditional probability, identify the group named after the vertical bar. - Interpret the conditional probability within its restricted group.

Solution

1. Since \(P(B^c)=0.60\), \(P(B)=0.40\). The cells are \(P(B\cap V)=0.20\), \(P(B^c\cap V)=0.30-0.20=0.10\), \(P(B\cap V^c)=0.40-0.20=0.20\), and \(P(B^c\cap V^c)=0.60-0.10=0.50\). 2. \(P(B\cup V)=P(B)+P(V)-P(B\cap V)=0.40+0.30-0.20=0.50\). 3. \(P(B\mid V)=\frac{0.20}{0.30}=\frac{2}{3}\approx0.6667\). 4. \(P(B\mid V^c)=\frac{0.20}{0.70}=\frac{2}{7}\approx0.2857\). This is the proportion interested in basketball among students not interested in volleyball.

Answer

a) <table> <tr><td></td><td>\(V\)</td><td>\(V^c\)</td><td>Total</td></tr> <tr><td>\(B\)</td><td>\(0.20\)</td><td>\(0.20\)</td><td>\(0.40\)</td></tr> <tr><td>\(B^c\)</td><td>\(0.10\)</td><td>\(0.50\)</td><td>\(0.60\)</td></tr> <tr><td>Total</td><td>\(0.30\)</td><td>\(0.70\)</td><td>\(1\)</td></tr> </table> b) \(P(B\cup V)=0.50\) c) \(P(B\mid V)=\frac{2}{3}\approx0.6667\) d) \(P(B\mid V^c)=\frac{2}{7}\approx0.2857\). About \(28.57\%\) of students not interested in volleyball are interested in basketball.

52207610

The following two-way probability table for events \(A\) and \(B\) is incomplete. <table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td></td><td>\(0.12\)</td><td></td></tr> <tr><td>\(A^c\)</td><td>\(0.18\)</td><td></td><td>\(0.45\)</td></tr> <tr><td>Total</td><td></td><td></td><td>\(1\)</td></tr> </table> a) Complete the table. b) Find \(P(A\cup B)\). c) Find \(P(A^c\mid B)\) and \(P(B\mid A^c)\). d) Determine whether \(A\) and \(B\) are independent.

Hints

- Complete the table using row and column totals. - In the addition rule, subtract the intersection because it is counted twice. - The condition determines the denominator of each conditional probability. - For independence, compare \(P(A\cap B)\) with \(P(A)P(B)\).

Solution

1. Since \(P(A^c)=0.45\), \(P(A)=0.55\). Then \(P(A\cap B)=0.55-0.12=0.43\) and \(P(A^c\cap B^c)=0.45-0.18=0.27\). Therefore, \(P(B)=0.43+0.18=0.61\) and \(P(B^c)=0.12+0.27=0.39\). 2. \(P(A\cup B)=0.55+0.61-0.43=0.73\). 3. \(P(A^c\mid B)=\frac{0.18}{0.61}\approx0.2951\), and \(P(B\mid A^c)=\frac{0.18}{0.45}=0.40\). 4. The events are not independent because \(P(A)P(B)=0.55\cdot0.61=0.3355\neq0.43=P(A\cap B)\).

Answer

a) <table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td>\(0.43\)</td><td>\(0.12\)</td><td>\(0.55\)</td></tr> <tr><td>\(A^c\)</td><td>\(0.18\)</td><td>\(0.27\)</td><td>\(0.45\)</td></tr> <tr><td>Total</td><td>\(0.61\)</td><td>\(0.39\)</td><td>\(1\)</td></tr> </table> b) \(P(A\cup B)=0.73\) c) \(P(A^c\mid B)\approx0.2951\); \(P(B\mid A^c)=0.40\) d) The events are dependent.

52208110

A survey of \(200\) students records whether each student belongs to a sports club \((S)\) and plays a musical instrument \((M)\). <table> <tr><td></td><td>\(M\)</td><td>\(M^c\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(40\)</td><td>\(80\)</td><td>\(120\)</td></tr> <tr><td>\(S^c\)</td><td>\(40\)</td><td>\(40\)</td><td>\(80\)</td></tr> <tr><td>Total</td><td>\(80\)</td><td>\(120\)</td><td>\(200\)</td></tr> </table> a) Explain the difference between \(P(M\cap S)\) and \(P(S\mid M)\), and find both values. b) Given that a student does not play an instrument, find the probability that the student belongs to a sports club. Write the probability in symbols.

Hints

- Identify the reference group for each probability. - Is the selection made from all students or from a stated subgroup? - The event after the vertical bar is the condition. - Use the appropriate row or column total as the denominator.

Solution

1. \(P(M\cap S)\) uses all \(200\) students as the reference group and represents students with both characteristics. Thus, \(P(M\cap S)=\frac{40}{200}=0.20\). 2. \(P(S\mid M)\) uses only the \(80\) students who play an instrument as the reference group. Thus, \(P(S\mid M)=\frac{40}{80}=0.50\). 3. Among the \(120\) students who do not play an instrument, \(80\) belong to a sports club. Therefore, \(P(S\mid M^c)=\frac{80}{120}=\frac{2}{3}\approx0.6667\).

Answer

a) \(P(M\cap S)=0.20\) refers to all students, while \(P(S\mid M)=0.50\) refers only to students who play an instrument. b) \(P(S\mid M^c)=\frac{2}{3}\approx0.6667\)

52208210

A study of \(500\) people records smoking status \((S)\) and high blood pressure \((H)\). <table> <tr><td></td><td>\(H\)</td><td>\(H^c\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(60\)</td><td>\(90\)</td><td>\(150\)</td></tr> <tr><td>\(S^c\)</td><td>\(40\)</td><td>\(310\)</td><td>\(350\)</td></tr> <tr><td>Total</td><td>\(100\)</td><td>\(400\)</td><td>\(500\)</td></tr> </table> a) Find and interpret \(P(H\mid S)\) and \(P(S\mid H)\). b) Given that a randomly selected person does not have high blood pressure, find the probability that the person smokes. Write the probability in symbols.

Hints

- For each conditional probability, identify the starting group. - The condition often follows phrases such as “given that” or “among.” - Read \(P(A\mid B)\) as the probability of \(A\) within group \(B\). - Use the table margins to choose the denominator.

Solution

1. Among the \(150\) smokers, \(60\) have high blood pressure. Thus, \(P(H\mid S)=\frac{60}{150}=0.40\). This means \(40\%\) of the smokers in the study have high blood pressure. 2. Among the \(100\) people with high blood pressure, \(60\) smoke. Thus, \(P(S\mid H)=\frac{60}{100}=0.60\). This means \(60\%\) of the people with high blood pressure in the study smoke. 3. Among the \(400\) people without high blood pressure, \(90\) smoke. Therefore, \(P(S\mid H^c)=\frac{90}{400}=0.225\).

Answer

a) \(P(H\mid S)=0.40\): \(40\%\) of smokers in the study have high blood pressure. \(P(S\mid H)=0.60\): \(60\%\) of people with high blood pressure in the study smoke. b) \(P(S\mid H^c)=0.225\)

52208710

A warehouse contains \(5000\) light bulbs. Let \(A\) be the event that a bulb is an LED bulb and \(B\) the event that it is defective. <table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td>\(40\)</td><td></td><td></td></tr> <tr><td>\(A^c\)</td><td></td><td></td><td>\(1500\)</td></tr> <tr><td>Total</td><td></td><td>\(4850\)</td><td>\(5000\)</td></tr> </table> Complete the table and find \(P(B\mid A)\), \(P(A\mid B)\), \(P(B\mid A^c)\), and \(P(A^c\mid B^c)\).

Hints

- Find missing row and column totals first. - Use subtraction to complete the inner cells. - For each conditional probability, use the condition's row or column total as the denominator. - Check that all table cells add to \(5000\).

Solution

1. The missing margins are \(n(A)=5000-1500=3500\) and \(n(B)=5000-4850=150\). 2. The missing cells are \(n(A\cap B^c)=3500-40=3460\), \(n(A^c\cap B)=150-40=110\), and \(n(A^c\cap B^c)=1500-110=1390\). 3. \(P(B\mid A)=\frac{40}{3500}=\frac{2}{175}\approx0.0114\). 4. \(P(A\mid B)=\frac{40}{150}=\frac{4}{15}\approx0.2667\). 5. \(P(B\mid A^c)=\frac{110}{1500}=\frac{11}{150}\approx0.0733\). 6. \(P(A^c\mid B^c)=\frac{1390}{4850}=\frac{139}{485}\approx0.2866\).

Answer

<table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td>\(40\)</td><td>\(3460\)</td><td>\(3500\)</td></tr> <tr><td>\(A^c\)</td><td>\(110\)</td><td>\(1390\)</td><td>\(1500\)</td></tr> <tr><td>Total</td><td>\(150\)</td><td>\(4850\)</td><td>\(5000\)</td></tr> </table> \(P(B\mid A)\approx0.0114\); \(P(A\mid B)\approx0.2667\); \(P(B\mid A^c)\approx0.0733\); \(P(A^c\mid B^c)\approx0.2866\).

52208910

The following two-way probability table is incomplete. <table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td>\(0.20\)</td><td></td><td></td></tr> <tr><td>\(A^c\)</td><td></td><td></td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td></td><td>\(1.00\)</td></tr> </table> Complete the table. Then find \(P(B\mid A)\), \(P(B^c\mid A)\), \(P(B\mid A^c)\), \(P(B^c\mid A^c)\), \(P(A\mid B)\), \(P(A^c\mid B)\), \(P(A\mid B^c)\), and \(P(A^c\mid B^c)\).

Hints

- Complete the row and column totals before computing conditional probabilities. - The denominator is always the probability of the condition. - For a fixed condition, the probabilities of an event and its complement must sum to \(1\).

Solution

1. Since \(P(A^c)=0.40\), \(P(A)=0.60\). Thus, \(P(A\cap B^c)=0.60-0.20=0.40\). 2. Since \(P(B)=0.50\), \(P(A^c\cap B)=0.50-0.20=0.30\). Then \(P(A^c\cap B^c)=0.40-0.30=0.10\), and \(P(B^c)=0.50\). 3. Conditioning on \(A\): \(P(B\mid A)=\frac{1}{3}\) and \(P(B^c\mid A)=\frac{2}{3}\). 4. Conditioning on \(A^c\): \(P(B\mid A^c)=0.75\) and \(P(B^c\mid A^c)=0.25\). 5. Conditioning on \(B\): \(P(A\mid B)=0.40\) and \(P(A^c\mid B)=0.60\). 6. Conditioning on \(B^c\): \(P(A\mid B^c)=0.80\) and \(P(A^c\mid B^c)=0.20\).

Answer

<table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td>\(0.20\)</td><td>\(0.40\)</td><td>\(0.60\)</td></tr> <tr><td>\(A^c\)</td><td>\(0.30\)</td><td>\(0.10\)</td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr> </table> \(P(B\mid A)=\frac{1}{3}\); \(P(B^c\mid A)=\frac{2}{3}\); \(P(B\mid A^c)=0.75\); \(P(B^c\mid A^c)=0.25\); \(P(A\mid B)=0.40\); \(P(A^c\mid B)=0.60\); \(P(A\mid B^c)=0.80\); \(P(A^c\mid B^c)=0.20\).

52209010

In a survey, \(70\%\) of teenagers regularly use a gaming app \((G)\), \(60\%\) regularly use a music app \((M)\), and \(50\%\) regularly use both types. Create a two-way probability table and find: a) the probability that a teenager uses a music app, given that the teenager uses a gaming app; b) the probability that a teenager uses a gaming app, given that the teenager does not use a music app.

Hints

- Convert the percentages to probabilities and place the intersection first. - Determine which event is the condition in each part. - Use the complements to complete the table.

Solution

1. The inner cells are \(P(G\cap M)=0.50\), \(P(G\cap M^c)=0.70-0.50=0.20\), \(P(G^c\cap M)=0.60-0.50=0.10\), and \(P(G^c\cap M^c)=1-(0.50+0.20+0.10)=0.20\). 2. \(P(M\mid G)=\frac{0.50}{0.70}=\frac{5}{7}\approx0.7143\). 3. \(P(G\mid M^c)=\frac{0.20}{0.40}=0.50\).

Answer

<table> <tr><td></td><td>\(M\)</td><td>\(M^c\)</td><td>Total</td></tr> <tr><td>\(G\)</td><td>\(0.50\)</td><td>\(0.20\)</td><td>\(0.70\)</td></tr> <tr><td>\(G^c\)</td><td>\(0.10\)</td><td>\(0.20\)</td><td>\(0.30\)</td></tr> <tr><td>Total</td><td>\(0.60\)</td><td>\(0.40\)</td><td>\(1\)</td></tr> </table> a) \(P(M\mid G)=\frac{5}{7}\approx0.7143\), or about \(71.43\%\). b) \(P(G\mid M^c)=0.50\), or \(50\%\).

52209110

A fitness club has \(50\) members. Of these, \(35\) attend training regularly. Thirty members pass an end-of-season fitness test, and \(28\) of those who pass attended training regularly. A member is selected at random. Use a two-way table to answer the questions. a) Find the probability that the selected member attended training regularly and passed the test. b) Given that the selected member passed the test, find the probability that the member attended training regularly. c) Given that the selected member did not attend training regularly, find the probability that the member still passed the test.

Hints

- Place the four combinations of regular/not regular training and pass/not pass in a two-way table. - For each probability, decide whether the denominator is all \(50\) members or only a stated subgroup. - A phrase such as “given that” identifies the restricted group used as the denominator. - Let \(T\) represent regular training and \(B\) represent passing the test.

Solution

1. Let \(T\) mean regular training and \(B\) mean passing the test. Complete the two-way table: <table><thead><tr><th></th><th>Passed</th><th>Did not pass</th><th>Total</th></tr></thead><tbody><tr><th>Trained regularly</th><td>\(28\)</td><td>\(7\)</td><td>\(35\)</td></tr><tr><th>Did not train regularly</th><td>\(2\)</td><td>\(13\)</td><td>\(15\)</td></tr><tr><th>Total</th><td>\(30\)</td><td>\(20\)</td><td>\(50\)</td></tr></tbody></table> 2. For part a, \(P(T\cap B)=\frac{28}{50}=0.56\). 3. For part b, restrict the sample to the \(30\) members who passed: \(P(T\mid B)=\frac{28}{30}=\frac{14}{15}\approx0.9333\). 4. For part c, \(15\) members did not train regularly, and \(2\) of them passed. Thus, \(P(B\mid T^c)=\frac{2}{15}\approx0.1333\).

Answer

a) \(P(T\cap B)=0.56\) b) \(P(T\mid B)=\frac{14}{15}\approx0.9333\) c) \(P(B\mid T^c)=\frac{2}{15}\approx0.1333\)

52209210

A factory inspects \(100\) components. Eight components are defective. A screening test gives an alarm for \(7\) of the defective components and also gives an alarm for \(5\) nondefective components. One component is selected at random. Use a two-way table to answer the questions. a) What is the probability that the screening test gives an alarm? b) Given that the test gives an alarm, what is the probability that the component is defective? c) Given that the test does not give an alarm, what is the probability that the component is defective?

Hints

- Organize the four combinations of defective/nondefective and alarm/no alarm in a two-way table. - How many components produce an alarm in total? - For a conditional probability, use only the cases that satisfy the stated condition as the denominator. - Check that all four cells in the table add to \(100\).

Solution

1. Let \(D\) mean defective and \(A\) mean alarm. Complete the two-way table: <table><thead><tr><th></th><th>Alarm</th><th>No alarm</th><th>Total</th></tr></thead><tbody><tr><th>Defective</th><td>\(7\)</td><td>\(1\)</td><td>\(8\)</td></tr><tr><th>Nondefective</th><td>\(5\)</td><td>\(87\)</td><td>\(92\)</td></tr><tr><th>Total</th><td>\(12\)</td><td>\(88\)</td><td>\(100\)</td></tr></tbody></table> 2. There are \(7+5=12\) alarms, so \(P(A)=\frac{12}{100}=0.12\). 3. Among the \(12\) components that produce an alarm, \(7\) are defective. Therefore, \(P(D\mid A)=\frac{7}{12}\approx0.5833\). 4. There are \(88\) components without an alarm, and \(1\) of them is defective. Therefore, \(P(D\mid A^c)=\frac{1}{88}\approx0.0114\).

Answer

a) \(P(A)=0.12\) b) \(P(D\mid A)=\frac{7}{12}\approx0.5833\) c) \(P(D\mid A^c)=\frac{1}{88}\approx0.0114\)

52212810

A test for a particular allergy has a sensitivity of \(90\%\) and a specificity of \(95\%\). In a group of \(10{,}000\) people, \(2\%\) are expected to have the allergy. a) Create a two-way table of counts for this group. b) Find the probability that a person does not have the allergy, given that the test result is positive. c) Find the probability that a person has the allergy, given that the test result is negative.

Hints

- First divide the full group into people who have the allergy and people who do not. - Use sensitivity and specificity to fill the four inner cells with counts. - In parts b and c, identify the group named after “given that” and use its total as the denominator.

Solution

1. Of the \(10{,}000\) people, \(0.02\cdot 10{,}000=200\) have the allergy, and \(9800\) do not. 2. The sensitivity gives \(0.90\cdot 200=180\) true-positive results and \(200-180=20\) false-negative results. 3. The specificity gives \(0.95\cdot 9800=9310\) true-negative results and \(9800-9310=490\) false-positive results. 4. There are \(180+490=670\) positive results and \(20+9310=9330\) negative results. 5. \(P(\text{no allergy}\mid\text{positive})=\frac{490}{670}=\frac{49}{67}\approx 0.7313\). 6. \(P(\text{allergy}\mid\text{negative})=\frac{20}{9330}=\frac{2}{933}\approx 0.00214\).

Answer

a) <table> <tr><td></td><td>Allergy</td><td>No allergy</td><td>Total</td></tr> <tr><td>Positive test</td><td>\(180\)</td><td>\(490\)</td><td>\(670\)</td></tr> <tr><td>Negative test</td><td>\(20\)</td><td>\(9310\)</td><td>\(9330\)</td></tr> <tr><td>Total</td><td>\(200\)</td><td>\(9800\)</td><td>\(10{,}000\)</td></tr> </table> b) \(P(\text{no allergy}\mid\text{positive})=\frac{49}{67}\approx 0.7313\), or about \(73.13\%\). c) \(P(\text{allergy}\mid\text{negative})=\frac{2}{933}\approx 0.00214\), or about \(0.214\%\).

52213910

A company has \(200\) employees. A survey recorded whether each employee commutes by bicycle \((B)\) and whether each employee commutes by train \((T)\). One employee is selected at random. <table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(T\)</td><td></td><td>\(80\)</td><td>\(120\)</td></tr> <tr><td>\(T^c\)</td><td>\(50\)</td><td></td><td></td></tr> <tr><td>Total</td><td></td><td></td><td>\(200\)</td></tr> </table> a) Complete the two-way table. b) Find the probability that the selected employee: (1) commutes by bicycle; (2) commutes by neither train nor bicycle; (3) commutes by train, given that the employee commutes by bicycle.

Hints

- Use row and column totals to find the missing counts. - Decide whether each part asks for a joint, marginal, or conditional probability. - For the conditional probability, restrict the sample space to employees who commute by bicycle.

Solution

1. Complete the counts: \(n(T\cap B)=120-80=40\), \(n(T^c)=200-120=80\), and \(n(T^c\cap B^c)=80-50=30\). The column totals are \(n(B)=40+50=90\) and \(n(B^c)=80+30=110\). 2. \(P(B)=\frac{90}{200}=0.45\). 3. \(P(T^c\cap B^c)=\frac{30}{200}=0.15\). 4. \(P(T\mid B)=\frac{40}{90}=\frac{4}{9}\approx0.4444\).

Answer

a) <table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(T\)</td><td>\(40\)</td><td>\(80\)</td><td>\(120\)</td></tr> <tr><td>\(T^c\)</td><td>\(50\)</td><td>\(30\)</td><td>\(80\)</td></tr> <tr><td>Total</td><td>\(90\)</td><td>\(110\)</td><td>\(200\)</td></tr> </table> b) (1) \(P(B)=0.45\). (2) \(P(T^c\cap B^c)=0.15\). (3) \(P(T\mid B)=\frac{4}{9}\approx0.4444\).

52214010

A school has \(250\) students. In a survey, \(40\%\) of the students said they are vegetarian \((V)\). Of the vegetarian students, \(60\%\) regularly drink milk \((M)\). In all, \(160\) students regularly drink milk. a) Create a complete two-way table of counts. b) Find the probability that a randomly selected student: (1) is not vegetarian and does not regularly drink milk; (2) is vegetarian, given that the student regularly drinks milk; (3) does not regularly drink milk, given that the student is not vegetarian.

Hints

- Convert the percentages to student counts first. - Begin with the vegetarian row, then use the total number of students who regularly drink milk. - For each conditional probability, use the total of the conditioned group as the denominator. - The event “not vegetarian” is \(V^c\).

Solution

1. There are \(0.40\cdot250=100\) vegetarian students and \(250-100=150\) students who are not vegetarian. 2. Of the vegetarian students, \(0.60\cdot100=60\) regularly drink milk. Since \(160\) students regularly drink milk, \(160-60=100\) nonvegetarian students regularly drink milk. 3. The remaining counts are \(100-60=40\) vegetarian students who do not regularly drink milk and \(150-100=50\) nonvegetarian students who do not regularly drink milk. Therefore, \(40+50=90\) students do not regularly drink milk. 4. \(P(V^c\cap M^c)=\frac{50}{250}=0.20\). 5. \(P(V\mid M)=\frac{60}{160}=0.375\). 6. \(P(M^c\mid V^c)=\frac{50}{150}=\frac{1}{3}\).

Answer

a) <table> <tr><td></td><td>\(M\)</td><td>\(M^c\)</td><td>Total</td></tr> <tr><td>\(V\)</td><td>\(60\)</td><td>\(40\)</td><td>\(100\)</td></tr> <tr><td>\(V^c\)</td><td>\(100\)</td><td>\(50\)</td><td>\(150\)</td></tr> <tr><td>Total</td><td>\(160\)</td><td>\(90\)</td><td>\(250\)</td></tr> </table> b) (1) \(P(V^c\cap M^c)=0.20\). (2) \(P(V\mid M)=0.375\). (3) \(P(M^c\mid V^c)=\frac{1}{3}\).

52287910

In a grade level of \(200\) students, \(60\%\) own a tablet. Sixty students do not own a laptop, and \(10\%\) of all students own neither a tablet nor a laptop. One student is selected at random. Find the probability that the student owns a laptop, given that the student owns a tablet.

Hints

- Organize the information in a two-way table. - Convert each percentage to a count. - Separate the students without a laptop into those with and without a tablet. - The condition restricts the sample space to tablet owners.

Solution

1. The number of tablet owners is \(0.60\cdot 200=120\). 2. The number of students who own neither device is \(0.10\cdot 200=20\). 3. Of the \(60\) students without a laptop, \(20\) also do not have a tablet. Therefore, \(60-20=40\) students have a tablet but no laptop. 4. Of the \(120\) tablet owners, \(120-40=80\) also have a laptop. 5. Thus, \(P(\text{laptop}\mid\text{tablet})=\frac{80}{120}=\frac{2}{3}\).

Answer

\(P(\text{laptop}\mid\text{tablet})=\frac{2}{3}\approx 0.6667\), or about \(66.67\%\).

52288010

At a school with \(500\) students, \(40\%\) take music. Of the students who do not take music, one-fourth take art. In all, \(125\) students take art. One student is selected at random. Find the probability that the student takes music, given that the student takes art.

Hints

- Find the number of students who do not take music. - Determine how many of those students take art. - Subtract from the total number of art students to find the overlap. - The condition restricts the sample space to art students.

Solution

1. The number of students who take music is \(0.40\cdot 500=200\), so \(500-200=300\) do not take music. 2. Of the students who do not take music, \(\frac{1}{4}\cdot 300=75\) take art. 3. Since \(125\) students take art in total, \(125-75=50\) students take both music and art. 4. Therefore, \(P(\text{music}\mid\text{art})=\frac{50}{125}=0.40\).

Answer

\(P(\text{music}\mid\text{art})=0.40\), or \(40\%\).

52289710

In a class of \(30\) students, \(18\) play an instrument and \(12\) sing in the choir. Five students do neither. Let \(I\) be the event “plays an instrument” and \(C\) the event “sings in the choir.” One student is selected at random. a) Create a complete two-way table of counts. b) Find the probability that the selected student both plays an instrument and sings in the choir. c) Find the probability that the selected student sings in the choir but does not play an instrument.

Hints

- Place the given totals and the “neither” count in a two-way table. - Use the number who do at least one activity to find the overlap. - Each row and column total is the sum of its two inner cells. - “Neither” corresponds to \(I^c\cap C^c\).

Solution

1. Since \(5\) students do neither activity, \(30-5=25\) students do at least one. 2. Use the addition rule for counts: \(n(I\cap C)=n(I)+n(C)-n(I\cup C)=18+12-25=5\). 3. Then \(n(I\cap C^c)=18-5=13\) and \(n(I^c\cap C)=12-5=7\). The remaining inner cell is the given count \(n(I^c\cap C^c)=5\). 4. \(P(I\cap C)=\frac{5}{30}=\frac{1}{6}\). 5. \(P(I^c\cap C)=\frac{7}{30}\).

Answer

a) <table border="1"> <tr><td></td><td>\(C\)</td><td>\(C^c\)</td><td>Total</td></tr> <tr><td>\(I\)</td><td>\(5\)</td><td>\(13\)</td><td>\(18\)</td></tr> <tr><td>\(I^c\)</td><td>\(7\)</td><td>\(5\)</td><td>\(12\)</td></tr> <tr><td>Total</td><td>\(12\)</td><td>\(18\)</td><td>\(30\)</td></tr> </table> b) \(P(I\cap C)=\frac{1}{6}\approx0.1667\), or about \(16.67\%\). c) \(P(I^c\cap C)=\frac{7}{30}\approx0.2333\), or about \(23.33\%\).

52289810

A recreation club has \(120\) members. Of the members, \(80\) use the fitness center \((F)\), \(50\) use the swimming pool \((S)\), and \(30\) use neither facility. a) Create a two-way table of counts. b) Find the probability that a randomly selected member uses exactly one of the two facilities. c) Given that a member uses the swimming pool, find the probability that the member also uses the fitness center.

Hints

- First find how many members use at least one facility. - “Exactly one” means fitness without swimming or swimming without fitness. - In part c, restrict the sample space to members who use the swimming pool.

Solution

1. Since \(30\) members use neither facility, \(120-30=90\) use at least one. 2. Use the addition rule for counts: \(n(F\cap S)=80+50-90=40\). 3. Therefore, \(n(F\cap S^c)=80-40=40\) and \(n(F^c\cap S)=50-40=10\). 4. Exactly one facility is used by \(40+10=50\) members, so the probability is \(\frac{50}{120}=\frac{5}{12}\). 5. \(P(F\mid S)=\frac{n(F\cap S)}{n(S)}=\frac{40}{50}=0.80\).

Answer

a) <table border="1"> <tr><td></td><td>\(S\)</td><td>\(S^c\)</td><td>Total</td></tr> <tr><td>\(F\)</td><td>\(40\)</td><td>\(40\)</td><td>\(80\)</td></tr> <tr><td>\(F^c\)</td><td>\(10\)</td><td>\(30\)</td><td>\(40\)</td></tr> <tr><td>Total</td><td>\(50\)</td><td>\(70\)</td><td>\(120\)</td></tr> </table> b) \(\frac{5}{12}\approx0.4167\), or about \(41.67\%\). c) \(P(F\mid S)=0.80\), or \(80\%\).

53083310

A transportation survey of \(1000\) people recorded whether each person is a commuter and whether the person prefers traveling by car or train. <table> <thead> <tr> <th></th> <th>Car (\(A\))</th> <th>Train (\(B\))</th> </tr> </thead> <tbody> <tr> <td>Commuter (\(P\))</td> <td>\(420\)</td> <td>\(180\)</td> </tr> <tr> <td>Noncommuter (\(P^c\))</td> <td>\(80\)</td> <td>\(320\)</td> </tr> </tbody> </table> a) Find each relative frequency. (1) The person prefers the train. (2) The person is a commuter and prefers the train. (3) The person prefers the train, given that the person is a commuter. b) Under what condition can these relative frequencies be interpreted as probabilities for selecting one person from the surveyed group? c) Find the conditional relative frequency of preferring the train among noncommuters. Compare it with part a(3) and state which group has the higher train preference rate.

Hints

- Use the grand total for marginal and joint relative frequencies. - For a conditional relative frequency, restrict the denominator to the named group. - Compare the two train-preference rates only after using the correct group total for each.

Solution

1. The train total is \(180+320=500\), so the relative frequency of preferring the train is \(\frac{500}{1000}=0.50\). 2. The commuter-and-train count is \(180\), so the joint relative frequency is \(\frac{180}{1000}=0.18\). 3. There are \(420+180=600\) commuters, so the conditional relative frequency of preferring the train among commuters is \(\frac{180}{600}=0.30\). 4. These relative frequencies are selection probabilities for the surveyed group if each of the \(1000\) people is equally likely to be selected. 5. There are \(80+320=400\) noncommuters, so the train preference rate among noncommuters is \(\frac{320}{400}=0.80\). 6. Since \(0.80>0.30\), the surveyed noncommuters have the higher train preference rate.

Answer

a) (1) \(0.50\) (2) \(0.18\) (3) \(0.30\) b) Each surveyed person must be equally likely to be selected. c) The noncommuter train preference rate is \(0.80\), which is higher than the commuter rate \(0.30\).

53092910

A university summarizes student field of study and gender in the partially completed two-way table below. <table border="1"> <tr><td></td><td>STEM</td><td>Not STEM</td><td>Total</td></tr> <tr><td>Women</td><td>\(2499\)</td><td></td><td>\(12{,}740\)</td></tr> <tr><td>Men</td><td></td><td></td><td></td></tr> <tr><td>Total</td><td>\(8330\)</td><td></td><td>\(24{,}500\)</td></tr> </table> a) Complete the table. b) Find \(P(\text{STEM}\mid\text{woman})\) and \(P(\text{woman}\mid\text{STEM})\). Report each as a percentage to the nearest tenth of a percent. c) Which conditional percentage is larger? Explain why the two denominators are different.

Hints

- Use each row and column total to determine the missing cell in that row or column. - For a conditional relative frequency, the event after the vertical bar determines the denominator. - The two conditional probabilities use the same intersection count but different group totals.

Solution

1. Women not in STEM: \(12{,}740-2499=10{,}241\). 2. Men in STEM: \(8330-2499=5831\). 3. The number of men is \(24{,}500-12{,}740=11{,}760\), so men not in STEM: \(11{,}760-5831=5929\). 4. The not-STEM total is \(24{,}500-8330=16{,}170\). 5. Thus, \(P(\text{STEM}\mid\text{woman})=\frac{2499}{12740}\approx0.1962\), or \(19.6\%\). 6. Also, \(P(\text{woman}\mid\text{STEM})=\frac{2499}{8330}=0.3000\), or \(30.0\%\). 7. The second percentage is larger. The first conditions on all \(12{,}740\) women, while the second conditions on all \(8330\) STEM students.

Answer

a) <table border="1"> <tr><td></td><td>STEM</td><td>Not STEM</td><td>Total</td></tr> <tr><td>Women</td><td>\(2499\)</td><td>\(10{,}241\)</td><td>\(12{,}740\)</td></tr> <tr><td>Men</td><td>\(5831\)</td><td>\(5929\)</td><td>\(11{,}760\)</td></tr> <tr><td>Total</td><td>\(8330\)</td><td>\(16{,}170\)</td><td>\(24{,}500\)</td></tr> </table> b) \(P(\text{STEM}\mid\text{woman})\approx19.6\%\); \(P(\text{woman}\mid\text{STEM})=30.0\%\). c) \(P(\text{woman}\mid\text{STEM})\) is larger. Its denominator is the \(8330\) STEM students, whereas \(P(\text{STEM}\mid\text{woman})\) uses the \(12{,}740\) women.

53093010

A distribution center processes \(15{,}000\) packages per day. Of these, \(10{,}800\) are domestic shipments. Overall, \(5\%\) of all packages are damaged, and exactly \(126\) damaged packages are international. a) Complete the two-way table. <table border="1"> <tr><td></td><td>Damaged</td><td>Undamaged</td><td>Total</td></tr> <tr><td>Domestic</td><td></td><td></td><td>\(10{,}800\)</td></tr> <tr><td>International</td><td>\(126\)</td><td></td><td></td></tr> <tr><td>Total</td><td></td><td></td><td>\(15{,}000\)</td></tr> </table> b) Find the damage rate among domestic packages and the damage rate among international packages. Report each to the nearest tenth of a percent. c) Which shipment group has the higher conditional damage rate?

Hints

- Convert the overall damage percentage into a total damaged count before completing the table. - A conditional damage rate uses the total for that shipment group as its denominator. - Compare rates rather than raw damaged-package counts.

Solution

1. The international total is \(15{,}000-10{,}800=4200\). 2. The total damaged count is \(0.05\cdot15{,}000=750\), so the damaged domestic count is \(750-126=624\). 3. Undamaged domestic packages: \(10{,}800-624=10{,}176\). Undamaged international packages: \(4200-126=4074\). The undamaged total is \(15{,}000-750=14{,}250\). 4. The domestic damage rate is \(P(\text{damaged}\mid\text{domestic})=\frac{624}{10800}\approx0.0578\), or \(5.8\%\). 5. The international damage rate is \(P(\text{damaged}\mid\text{international})=\frac{126}{4200}=0.030\), or \(3.0\%\). 6. The domestic shipment group has the higher conditional damage rate.

Answer

a) <table border="1"> <tr><td></td><td>Damaged</td><td>Undamaged</td><td>Total</td></tr> <tr><td>Domestic</td><td>\(624\)</td><td>\(10{,}176\)</td><td>\(10{,}800\)</td></tr> <tr><td>International</td><td>\(126\)</td><td>\(4074\)</td><td>\(4200\)</td></tr> <tr><td>Total</td><td>\(750\)</td><td>\(14{,}250\)</td><td>\(15{,}000\)</td></tr> </table> b) Domestic: \(5.8\%\). International: \(3.0\%\). c) Domestic packages have the higher conditional damage rate.

52209710

A choir has \(40\) singers. Let \(H\) represent high voices, \(L\) low voices, \(P\) professional singers, and \(A\) amateur singers. Three choir members make these observations: Lucas: “Most professional singers have low voices.” Julia: “Most singers with high voices are amateurs.” Morgan: “Most singers with low voices are professionals.” a) Express each statement as an inequality involving a conditional probability. b) Create a two-way table of whole-number counts for the \(40\) singers that makes all three statements true. Verify your table using the inequalities from part a. c) For your table, find: (1) the probability that a randomly selected professional singer has a high voice; (2) the probability that a randomly selected low-voice singer is an amateur.

Hints

- Translate “most” as a proportion greater than \(0.50\). - Pay attention to which group is the condition in each statement. - Try whole-number values for the four inner cells and test all three inequalities. - The four inner cells must sum to \(40\).

Solution

1. The statements require \(P(L\mid P)>0.50\), \(P(A\mid H)>0.50\), and \(P(P\mid L)>0.50\). 2. One valid table has \(5\) high-voice professionals, \(15\) high-voice amateurs, \(12\) low-voice professionals, and \(8\) low-voice amateurs. 3. The checks are \(P(L\mid P)=\frac{12}{17}\approx 0.7059>0.50\), \(P(A\mid H)=\frac{15}{20}=0.75>0.50\), and \(P(P\mid L)=\frac{12}{20}=0.60>0.50\). 4. For this table, \(P(H\mid P)=\frac{5}{17}\approx 0.2941\), and \(P(A\mid L)=\frac{8}{20}=0.40\).

Answer

a) Lucas: \(P(L\mid P)>0.50\). Julia: \(P(A\mid H)>0.50\). Morgan: \(P(P\mid L)>0.50\). b) One possible table is: <table border="1"> <tr><td></td><td>\(P\)</td><td>\(A\)</td><td>Total</td></tr> <tr><td>\(H\)</td><td>\(5\)</td><td>\(15\)</td><td>\(20\)</td></tr> <tr><td>\(L\)</td><td>\(12\)</td><td>\(8\)</td><td>\(20\)</td></tr> <tr><td>Total</td><td>\(17\)</td><td>\(23\)</td><td>\(40\)</td></tr> </table> c) (1) \(P(H\mid P)=\frac{5}{17}\approx 0.2941\). (2) \(P(A\mid L)=0.40\).

52287410

A sports club has \(120\) members. Of the members, \(60\%\) are teenagers and the rest are adults. Exactly \(20\) more members play tennis than hockey, and every member plays exactly one of these two sports. One-fourth of the teenagers play hockey. Find the probability that a randomly selected member is an adult, given that the member plays tennis.

Hints

- First find the numbers of teenagers and adults. - Write an equation using the difference between the numbers of tennis and hockey players. - Organize the four group counts in a two-way table. - For the final probability, use all tennis players as the conditioned group.

Solution

1. There are \(0.60\cdot 120=72\) teenagers and \(120-72=48\) adults. 2. Let \(x\) be the number of hockey players. Then \(x+(x+20)=120\), so \(x=50\). Therefore, \(50\) members play hockey and \(70\) play tennis. 3. One-fourth of the teenagers play hockey, so \(\frac{1}{4}\cdot 72=18\) teenage members play hockey. Thus, \(72-18=54\) teenagers play tennis. 4. The number of adults who play tennis is \(70-54=16\). 5. Therefore, \(P(\text{adult}\mid\text{tennis})=\frac{16}{70}=\frac{8}{35}\approx 0.2286\).

Answer

\(P(\text{adult}\mid\text{tennis})=\frac{8}{35}\approx 0.2286\), or about \(22.86\%\).

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