Exam 1 Answers

Statistics I — Exam 1 — 2026 Fall

This page gives the questions of Exam 1 together with their solutions and the grading guideline for each question. Every question is worth 20 points.

Question 1

20 points. A system has three components. During an inspection, each component is recorded as working or failed. An outcome is written as \((x_1,x_2,x_3)\), where \(x_i=1\) means that component \(i\) is working and \(x_i=0\) means that it has failed. Let \(A\) be the event that exactly two components are working, and let \(B\) be the event that component 1 is working. Answer the following.

  1. Write the sample space \(S\). Then write the event \(A\) as a set of outcomes.

  2. Write \(B\) and \(B^{c}\) as sets of outcomes. Then write \(A\cap B\) as a set of outcomes.

  3. Let \(C\) be the event that at least one component has failed. Write \(C\) and \(C^{c}\) as sets of outcomes.

  4. Let \(D\) be the event that component 2 is working. Write the event “neither component 1 nor component 2 is working” in two different forms using \(B\), \(D\), complements, unions, and intersections. Show how the two forms are related by De Morgan’s law.

  5. Let \(E\) be the event that all three components are working. Determine whether \(C\) and \(E\) are exhaustive events and whether they are mutually exclusive events.

Solution 1.

  1. \(S=\{000,001,010,011,100,101,110,111\}\), \(A=\{011,101,110\}\).

  2. \(B=\{100,101,110,111\}\), \(B^c=\{000,001,010,011\}\), \(A\cap B=\{101,110\}\).

  3. \(C=\{000,001,010,011,100,101,110\}\), \(C^c=\{111\}\).

  4. “Neither 1 nor 2 is working”:

    \[(B\cup D)^c=B^c\cap D^c.\]

  5. \(E=\{111\}\). Since \(C\cup E=S\), they are exhaustive. Since \(C\cap E=\varnothing\), they are mutually exclusive.

Grading Guideline (20 points)

  • Part (a): correct \(S\): 2 pts; correct \(A\): 2 pts

  • Part (b): correct \(B\): 1 pt; \(B^c\): 1 pt; \(A\cap B\): 2 pts

  • Part (c): correct \(C\): 2 pts; correct \(C^c\): 2 pts

  • Part (d): one correct form: 1 pt; second correct form: 1 pt; correct De Morgan relationship: 2 pts

  • Part (e): exhaustive decision with \(C\cup E=S\): 2 pts; mutually exclusive decision with \(C\cap E=\varnothing\): 2 pts

Question 2

20 points. A laboratory has seven different testing jobs, eight students, and a computer system that uses the six symbols \(A,B,C,D,E,F\) for access codes. For each situation, find the number of possible outcomes and show the counting expression you use. You do not need to calculate large factorials. Answer the following.

  1. All seven testing jobs are done one after another. Describe one outcome in words and find the number of possible job sequences.

  2. Three of the eight students are given the different roles of team leader, recorder, and presenter. Find the number of possible assignments.

  3. Three of the eight students are selected to form a project team. The three students have no different roles. Find the number of possible teams.

  4. A four-character access code is made using the six symbols. A symbol can be used more than once. Find the number of possible codes.

  5. Now suppose that a symbol cannot be used more than once in the four-character code. Find the number of possible codes. Compare your counting expression with part (d).

Solution 2.

  1. One outcome is one complete order of the seven jobs. \(7!=5040\).

  2. \(P(8,3)=8\times7\times6=336\).

  3. \(\binom{8}{3}=56\).

  4. \(6^4=1296\).

  5. \(P(6,4)=6\times5\times4\times3=360\). In (d), repetition is allowed; in (e), it is not.

Grading Guideline (20 points)

  • Part (a): correct description of one outcome: 1 pt; \(7!\): 2 pts; correct value: 1 pt

  • Part (b): correct permutation setup: 3 pts; correct value: 1 pt

  • Part (c): correct combination setup: 3 pts; correct value: 1 pt

  • Part (d): correct repeated-choice setup \(6^4\): 3 pts; correct value: 1 pt

  • Part (e): correct setup \(P(6,4)\): 2 pts; correct value: 1 pt; correct comparison with part (d): 1 pt

Question 3

20 points. A company studies how its customers use two services. Let \(S\) be the set of all customers, \(A=\{\text{customer uses Service A}\}\), and \(B=\{\text{customer uses Service B}\}\). Suppose \(P(A)=0.45\), \(P(B)=0.35\), and \(P(A\cap B)=0.15\). Answer the following.

  1. Write the event that a customer uses at least one of the two services using \(A\) and \(B\). Find its probability.

  2. Write the event that a customer uses neither service. Find its probability.

  3. Use De Morgan’s law to write the event in part (b) in another form using \(A^{c}\), \(B^{c}\), and an intersection.

  4. Write the event that a customer uses Service A but not Service B, and find its probability. Then find the probability that the customer uses exactly one of the two services.

  5. Determine whether \(A\) and \(B\) are exhaustive events. Use \(P(A\cup B)\) to show your answer.

Solution 3.

  1. \(A\cup B\),

    \[P(A\cup B)=0.45+0.35-0.15=0.65.\]

  2. \((A\cup B)^c\),

    \[P((A\cup B)^c)=1-0.65=0.35.\]
  3. \[(A\cup B)^c=A^c\cap B^c.\]

  4. \(A\cap B^c\),

    \[P(A\cap B^c)=0.45-0.15=0.30.\]

    Exactly one:

    \[(0.45-0.15)+(0.35-0.15)=0.50.\]

  5. Not exhaustive because \(P(A\cup B)=0.65\neq1\).

Grading Guideline (20 points)

  • Part (a): correct event \(A\cup B\): 1 pt; correct rule/setup: 2 pts; correct probability: 1 pt

  • Part (b): correct event: 2 pts; correct probability: 2 pts

  • Part (c): correct De Morgan form \(A^c\cap B^c\): 4 pts

  • Part (d): correct \(A\cap B^c\): 1 pt; probability \(0.30\): 1 pt; exactly-one calculation: 2 pts

  • Part (e): correct decision: 2 pts; uses \(P(A\cup B)=0.65\neq1\): 2 pts

Question 4

20 points. Customers on a company website are grouped by shipping type and by the number of orders they made last month. The numbers of customers are shown below.

Express

Standard

Total

No orders

25

15

40

One order

65

45

110

More than one order

40

20

60

Total

130

80

210

Customers are selected without replacement, and the order of selection does not matter. Answer the following.

  1. Suppose 10 Express customers are selected. Let \(A\) be the event that exactly one selected customer had more than one order last month. Find the number of possible samples. Then find the number of samples in event \(A\).

  2. Let \(B\) be the event that at least one of the 10 selected Express customers had more than one order last month. Write \(B^{c}\) in words. Then use \(B^{c}\) to find the number of samples in event \(B\).

  3. Use the results above to find \(P(B)\). You may leave your answer in terms of combinations.

  4. Now suppose 10 Express customers and 15 Standard customers are selected. Let \(C\) be the event that exactly one selected Express customer and exactly one selected Standard customer had no orders last month. Find the number of samples in event \(C\).

Solution 4.

  1. \[|S|=\binom{130}{10},\qquad |A|=\binom{40}{1}\binom{90}{9}.\]

  2. \(B^c\): none of the selected customers had more than one order.

    \[|B^c|=\binom{90}{10},\qquad |B|=\binom{130}{10}-\binom{90}{10}.\]
  3. \[P(B)=1-\frac{\binom{90}{10}}{\binom{130}{10}}.\]
  4. \[|C|=\binom{25}{1}\binom{105}{9}\binom{15}{1}\binom{65}{14}.\]

Grading Guideline (20 points)

  • Part (a): sample-space size \(\binom{130}{10}\): 2 pts; correct event count \(\binom{40}{1}\binom{90}{9}\): 3 pts

  • Part (b): correct meaning of \(B^c\): 2 pts; \(\binom{90}{10}\): 1 pt; correct complement count for \(B\): 2 pts

  • Part (c): correct probability denominator: 1 pt; correct numerator/complement expression: 2 pts; correct final form: 1 pt

  • Part (d): correct Express factors: 3 pts; correct Standard factors: 2 pts; correct multiplication of the two selections: 1 pt

Question 5

20 points. A quality inspector has 10 units. Six units are acceptable and four are defective. Three units are selected one at a time. Let \(A=\{\text{at least one selected unit is defective}\}\) and \(B=\) exactly two selected units are acceptable. For parts (a)–(c), the units are selected without replacement. Answer the following.

  1. If the order of selection is ignored, find the number of possible samples of three units. Write \(A^{c}\) in words.

  2. Find \(P(B)\). Show the counting expression you use.

  3. Find \(P(A)\) by using \(A^{c}\). Briefly explain why using \(A^{c}\) is easier than finding \(P(A)\) directly.

  4. For the remaining parts, suppose each selected unit is returned before the next unit is selected. The sampling is now with replacement. Find the probability that all three selected units are acceptable.

  5. Under sampling with replacement, find \(P(A)\). Compare the probability expression with the one used in part (c).

Solution 5.

  1. \[|S|=\binom{10}{3}=120.\]

    \(A^c\): all three selected units are acceptable.

  2. \[P(B)=\frac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}} =\frac{60}{120}=0.5.\]
  3. \[P(A)=1-\frac{\binom{6}{3}}{\binom{10}{3}} =1-\frac{20}{120} =\frac56.\]

    The complement has only one case: all three units are acceptable.

  4. \[P(\text{all acceptable})=\left(\frac{6}{10}\right)^3=0.216.\]
  5. \[P(A)=1-\left(\frac{6}{10}\right)^3=0.784.\]

    Without replacement, the available units change; with replacement, the probability stays the same for each draw.

Grading Guideline (20 points)

  • Part (a): \(\binom{10}{3}=120\): 2 pts; correct description of \(A^c\): 2 pts

  • Part (b): correct favorable count \(\binom{6}{2}\binom{4}{1}\): 2 pts; correct denominator: 1 pt; correct result \(0.5\): 1 pt

  • Part (c): correct \(P(A^c)\): 1 pt; complement setup: 1 pt; correct result \(5/6\): 1 pt; correct brief explanation: 1 pt

  • Part (d): correct repeated probability structure: 2 pts; correct result \(0.216\): 2 pts

  • Part (e): correct complement setup: 1 pt; correct result \(0.784\): 1 pt; correct comparison with part (c): 2 pts