Data Analysis Practice Questions — GRE General Test — Page 6

Question 101Question →

An academic conference has 7 consecutive presentation time slots. A committee must assign 7 presentations—3 in Biology ( $B$ ), 2 in Chemistry ( $C$ ), and 2 in Physics ( $P$ )—to these slots subject to the following conditions:

1. Presentations of the same discipline are indistinguishable (only the subject sequence matters).
2. No two Biology presentations may be scheduled in consecutive time slots.
3. The two Physics presentations must be scheduled in consecutive time slots.

How many different subject-sequence schedules for the 7 presentation time slots satisfy all of these conditions?

24

12

30

60

72

Show answer & explanation

Answer: $12$

Answer

12

To satisfy the condition that the two Physics presentations are consecutive, we combine them into a single block

(PP)

. We then arrange the non-Biology items—two Chemistry presentations

C

and one block

(PP)

—which can be ordered in

\frac{3!}{2!} = 3

distinct ways. These 3 items create 4 potential gaps (before the first item, between items, and after the last item). To ensure no two Biology presentations are adjacent, we place one Biology presentation into each of 3 chosen gaps out of the 4 available spaces, which can be done in

\binom{4}{3} = 4

ways. By the Fundamental Counting Principle, the total number of valid subject-sequence schedules is

3 \times 4 = 12

.

Step-by-Step Solution

1

Group the adjacent Physics presentations into a single block

The two Physics presentations

(P, P)

form a single block

(PP)

. The set of non-Biology items consists of two indistinguishable

C

's and one

(PP)

block, making 3 items in total.

Since the two Physics presentations must be scheduled in consecutive slots, treating them as a single unit guarantees they stay adjacent.

2

Calculate the number of distinct arrangements of the non-Biology items

The number of distinct arrangements of

\{C, C, (PP)\}

is

\frac{3!}{2!1!} = 3

ways.

The 2 Chemistry presentations are identical, so we divide the total permutations (

3!

) by

2!

to account for indistinguishability.

3

Apply the gap method to place the non-adjacent Biology presentations

Any sequence of the 3 non-Biology items creates 4 available spaces (gaps), including the two ends:

\_ \text{item}_1 \_ \text{item}_2 \_ \text{item}_3 \_

. Choosing 3 gaps out of 4 yields

\binom{4}{3} = 4

ways to insert the 3 identical

B

presentations.

Placing at most one Biology presentation in each gap guarantees no two Biology presentations are consecutive.

4

Apply the Fundamental Counting Principle

Total valid schedules =

3 \times 4 = 12

.

The choices of arranging non-Biology items and selecting gaps for Biology items are independent.

Key Concept

Combinatorics with Adjacency Restrictions and Indistinguishable Objects

Question 102Question →

An executive education program surveyed $400$ professional candidates regarding their enrollment in three specialized tracks: Artificial Intelligence ( $A$ ), Financial Technology ( $F$ ), and Sustainable Energy ( $S$ ).

The survey revealed the following data:
- $40$ candidates were enrolled in none of the three tracks.
- $210$ candidates were enrolled in Artificial Intelligence.
- $180$ candidates were enrolled in Financial Technology.
- $150$ candidates were enrolled in Sustainable Energy.
- $90$ candidates were enrolled in both Artificial Intelligence and Financial Technology.
- $70$ candidates were enrolled in both Financial Technology and Sustainable Energy.
- $60$ candidates were enrolled in both Artificial Intelligence and Sustainable Energy.

How many candidates were enrolled in exactly one of the three tracks?

100

140

180

220

260

Show answer & explanation

Answer: $220$

Answer

The number of candidates enrolled in exactly one of the three tracks is

220

.

To find the number of candidates enrolled in exactly one track, we first apply the Principle of Inclusion-Exclusion to find the triple intersection

|A \cap F \cap S|

:

|A \cup F \cup S| = |A| + |F| + |S| - (|A \cap F| + |F \cap S| + |A \cap S|) + |A \cap F \cap S|

. Substituting known values gives

360 = 210 + 180 + 150 - 220 + |A \cap F \cap S|

, so

|A \cap F \cap S| = 40

. Isolating mutually exclusive regions yields

100

candidates in Artificial Intelligence only,

60

in Financial Technology only, and

60

in Sustainable Energy only. Summing these gives

100 + 60 + 60 = 220

.

Step-by-Step Solution

1

Determine the number of candidates enrolled in at least one track.

|A \cup F \cup S| = 400 - 40 = 360

Candidates in none of the tracks must be subtracted from the total survey population.

2

Calculate the number of candidates enrolled in all three tracks (

x = |A \cap F \cap S|

) using the Principle of Inclusion-Exclusion.

360 = 210 + 180 + 150 - (90 + 70 + 60) + x \implies 360 = 320 + x \implies x = 40

The sum of individual sets overcounts pairwise intersections and undercounts the triple intersection.

3

Calculate the counts for regions representing enrollment in exactly two tracks.

A \cap F \text{ only} = 90 - 40 = 50

,

F \cap S \text{ only} = 70 - 40 = 30

,

A \cap S \text{ only} = 60 - 40 = 20

Subtracting the triple intersection from each pairwise intersection isolates candidates enrolled in precisely two tracks.

4

Calculate the counts for candidates enrolled in exactly one track and sum them.

A \text{ only} = 210 - (50 + 20 + 40) = 100

;

F \text{ only} = 180 - (50 + 30 + 40) = 60

;

S \text{ only} = 150 - (20 + 30 + 40) = 60

. Total

= 100 + 60 + 60 = 220

.

Subtracting all overlap regions from each set total yields the single-track enrollment.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 103Question →

A fair six-sided die with faces numbered $1$ through $6$ is rolled twice. What is the probability of rolling a $4$ on the first roll and an odd number on the second roll?

\frac{1}{12}

\frac{2}{3}

\frac{1}{3}

\frac{1}{4}

\frac{1}{36}

Show answer & explanation

Answer: $\frac{1}{12}$

Answer

\frac{1}{12}

Because the outcome of the first die roll does not affect the outcome of the second die roll, the two events are independent. The probability of rolling a

4

on the first roll is

\frac{1}{6}

, and the probability of rolling an odd number (

1, 3,

or

5

) on the second roll is

\frac{3}{6} = \frac{1}{2}

. Multiplying these individual probabilities together yields

\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}

.

Step-by-Step Solution

1

Determine the probability of the first event (rolling a 4).

There is

1

favorable outcome out of

6

possible outcomes, so

P(\text{First roll is } 4) = \frac{1}{6}

.

Each face of a fair six-sided die is equally likely to land face up.

2

Determine the probability of the second event (rolling an odd number).

The odd numbers are

1, 3,

and

5

, giving

3

favorable outcomes out of

6

total outcomes, so

P(\text{Second roll is odd}) = \frac{3}{6} = \frac{1}{2}

.

Half of the outcomes on a standard six-sided die are odd.

3

Calculate the joint probability of both independent events occurring.

P(\text{4 on first AND odd on second}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}

.

For independent events

A

and

B

, the multiplication rule states that

P(A \text{ and } B) = P(A) \times P(B)

.

Key Concept

Probability of Independent Events

Estimated Time:1m 0s

Question 104Question →

A cybersecurity firm audited a sample of $300$ corporate networks for compliance across three security standards: Network Encryption ( $E$ ), Multi-Factor Authentication ( $M$ ), and Access Logging ( $L$ ). The audit revealed that $160$ networks met standard $E$ , $140$ met standard $M$ , and $120$ met standard $L$ . Exactly $25$ networks met all three standards, while $45$ networks met none of the three standards. How many of the audited networks met exactly two of the three security standards?

90

115

140

165

190

Show answer & explanation

Answer: 115

Answer

115 networks met exactly two of the three security standards.

The total number of networks meeting at least one standard is

300 - 45 = 255

. By the three-set inclusion-exclusion principle,

255 = 160 + 140 + 120 - S_2 + 25

, where

S_2

is the sum of the pairwise intersections

|E \cap M| + |M \cap L| + |E \cap L|

. Solving for

S_2

yields

S_2 = 190

. Since each pairwise intersection includes the 25 networks that met all three standards, the number of networks meeting exactly two standards is

190 - 3(25) = 115

.

Step-by-Step Solution

1

Calculate the total number of networks that met at least one security standard.

|E \cup M \cup L| = 300 - 45 = 255

Subtracting networks that met none of the standards from the total sample size gives the union of the three sets.

2

Apply the Principle of Inclusion-Exclusion for three sets to find the sum of pairwise overlapping regions.

255 = 160 + 140 + 120 - (|E \cap M| + |M \cap L| + |E \cap L|) + 25

, so

|E \cap M| + |M \cap L| + |E \cap L| = 445 - 255 = 190

The formula

|E \cup M \cup L| = |E| + |M| + |L| - (|E \cap M| + |M \cap L| + |E \cap L|) + |E \cap M \cap L|

relates all known quantities.

3

Calculate the number of networks that met exactly two standards.

\text{Exactly 2} = (|E \cap M| + |M \cap L| + |E \cap L|) - 3|E \cap M \cap L| = 190 - 3(25) = 190 - 75 = 115

Each of the three pairwise intersections includes the triple intersection. Subtracting three times the triple intersection isolates the regions representing membership in exactly two sets.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 105Question →

A retail analyst recorded the number of online orders fulfilled per day over a period of $9$ days. The data points, listed in non-decreasing order, are:

3, 5, 8, x, y, 17, 21, 24, z

The median of the $9$ daily order counts is $14$ , the arithmetic mean is $16$ , and the dataset has a unique mode of $8$ . If $z$ represents the highest number of orders fulfilled in a single day, what is the value of $z$ ?

28

38

44

52

54

Show answer & explanation

Answer: 44

Answer

44

For a sorted set of 9 values, the median is the 5th value, giving

y = 14

. To make 8 the unique mode, 8 must appear at least twice, which forces

x = 8

. With an arithmetic mean of 16 across 9 numbers, the total sum is

9 \times 16 = 144

. Subtracting the sum of the eight known numbers (

3+5+8+8+14+17+21+24 = 100

) gives

z = 44

.

Step-by-Step Solution

1

Determine the value of

y

using the median definition.

y = 14

For a dataset of

n = 9

numbers arranged in non-decreasing order, the median is the

\frac{9+1}{2} = 5\text{th}

element. Thus,

y = 14

.

2

Determine the value of

x

using the unique mode condition.

x = 8

The dataset is ordered as

3 \le 5 \le 8 \le x \le 14 \le 17 \le 21 \le 24 \le z

. For

8

to be a unique mode, it must appear more than once. Since all other given numbers are distinct,

x

must equal

8

so that

8

occurs twice.

3

Calculate the required total sum of all 9 data points from the given arithmetic mean.

\text{Total Sum} = 144

\text{Mean} = \frac{\text{Total Sum}}{9} \implies \text{Total Sum} = 9 \times 16 = 144.

4

Sum the known values and solve for

z

.

z = 44

3 + 5 + 8 + 8 + 14 + 17 + 21 + 24 + z = 144 \implies 100 + z = 144 \implies z = 44

.

Key Concept

Combining mean, median, and mode definitions to solve for unknown elements in a ordered dataset.

Question 106Question →

A reliability study recorded the operating lifespan, $t$ (in thousands of hours), for a sample of $250$ semiconductor laser diodes. The results are summarized in the grouped frequency table below.

Lifespan $t$ (thousands of hours)	Frequency
$0 \le t < 4$	$35$
$4 \le t < 8$	$55$
$8 \le t < 12$	$80$
$12 \le t < 16$	$50$
$16 \le t < 20$	$30$

Laser diodes with an operating lifespan of at least $6{}000$ hours ( $t \ge 6$ ) but less than $14{}000$ hours ( $t < 14$ ) are designated as high-efficiency units. Assuming that the values within each class interval are uniformly distributed, what percentage of the $250$ laser diodes in the sample are designated as high-efficiency units?

Show answer & explanation

Answer: 53

Answer

53%

To find the percentage of diodes with lifespans between

6{}000

and

14{}000

hours (

6 \le t < 14

), evaluate the relevant intervals. For

4 \le t < 8

, the sub-interval

[6, 8)

represents

\frac{8-6}{8-4} = 0.5

of the interval width, containing

0.5 \times 55 = 27.5

diodes. The interval

[8, 12)

is fully contained, contributing

80

diodes. For

12 \le t < 16

, the sub-interval

[12, 14)

represents

\frac{14-12}{16-12} = 0.5

of the interval width, containing

0.5 \times 50 = 25

diodes. Totaling these gives

27.5 + 80 + 25 = 132.5

diodes. Expressed as a percentage of the total

250

diodes,

\frac{132.5}{250} \times 100\% = 53\%

.

Step-by-Step Solution

1

Determine the estimated number of diodes in the partial interval

6 \le t < 8

.

The target range

[6, 8)

covers half of the interval

[4, 8)

width of

4

units. With uniform distribution, the count is

0.5 \times 55 = 27.5

diodes.

Linear interpolation estimates frequencies for sub-intervals within grouped data.

2

Include the count for the complete interval

8 \le t < 12

.

All

80

diodes in this interval fall within

6 \le t < 14

.

The entire interval is fully contained within the target upper and lower bounds.

3

Determine the estimated number of diodes in the partial interval

12 \le t < 14

.

The target range

[12, 14)

covers half of the interval

[12, 16)

width of

4

units. With uniform distribution, the count is

0.5 \times 50 = 25

diodes.

Linear interpolation estimates frequencies for the upper partial boundary.

4

Sum the target diode counts and convert to a percentage of the total sample.

Total target count =

27.5 + 80 + 25 = 132.5

diodes. Percentage =

\frac{132.5}{250} \times 100\% = 53\%

.

Divide the calculated frequency sum by the total sample size of 250 and multiply by 100.

Key Concept

Grouped Frequency Distribution and Linear Interpolation

Estimated Time:2m 0s

Question 107Question →

A community health center conducted a study of $250$ adults regarding their participation in three wellness programs: Nutrition Counseling ( $N$ ), Exercise Coaching ( $E$ ), and Stress Management ( $S$ ). The survey revealed the following data:

• $125$ adults participate in Nutrition Counseling.
• $105$ adults participate in Exercise Coaching.
• $85$ adults participate in Stress Management.
• $15$ adults participate in all three programs.
• $30$ adults participate in none of the three programs.

How many adults participate in exactly two of the three programs?

Show answer & explanation

Answer: 65

Answer

65

The total number of surveyed adults is

250

, and

30

participate in none of the programs, meaning

220

adults participate in at least one program. Let

x_1

be the number of adults in exactly one program,

x_2

in exactly two programs, and

x_3 = 15

in all three programs. We have

x_1 + x_2 + 15 = 220

, which simplifies to

x_1 + x_2 = 205

. Additionally, summing the individual program participants yields

|N| + |E| + |S| = 125 + 105 + 85 = 315

. By region expansion,

|N| + |E| + |S| = x_1 + 2x_2 + 3x_3

. Substituting

x_3 = 15

gives

x_1 + 2x_2 + 45 = 315

, or

x_1 + 2x_2 = 270

. Subtracting

x_1 + x_2 = 205

from

x_1 + 2x_2 = 270

gives

x_2 = 65

.

Step-by-Step Solution

1

Determine the number of adults in the union of all three set categories

|N \cup E \cup S| = 250 - 30 = 220

Subtracting the individuals participating in none of the programs from the total surveyed yields the total count of individuals participating in at least one program.

2

Set up an equation for the total unique participants using disjoint region variables

x_1 + x_2 = 205

The union equals

x_1 + x_2 + x_3 = 220

, where

x_1

represents adults in exactly 1 program,

x_2

in exactly 2, and

x_3 = 15

in all 3 programs.

3

Set up an equation using the sum of the individual program totals

x_1 + 2x_2 = 270

The sum

|N| + |E| + |S| = 125 + 105 + 85 = 315

counts single-program participants once, double-program participants twice, and triple-program participants three times (

x_1 + 2x_2 + 3(15) = 315

).

4

Solve the system of linear equations for

x_2

x_2 = 270 - 205 = 65

Subtracting

(x_1 + x_2 = 205)

from

(x_1 + 2x_2 = 270)

isolates

x_2

, which is the exact number of adults participating in exactly two programs.

Key Concept

3-Set Inclusion-Exclusion Principle & Venn Diagram Region Partitioning

Estimated Time:1m 45s

Question 108Question →

A committee of $8$ people consists of $4$ men and $4$ women. A subcommittee of $4$ people is to be selected from this group such that the subcommittee contains at least one man and at least one woman. If two specific members, one man and one woman, refuse to serve together on the same subcommittee, how many different valid subcommittees of $4$ people can be formed?

Show answer & explanation

Answer: 53

Answer

53

The total number of ways to choose 4 people out of 8 is

\binom{8}{4} = 70

. Removing the 2 single-gender subcommittees (4 men or 4 women) leaves 68 gender-valid subcommittees. Among these 68 subcommittees, exactly

\binom{6}{2} = 15

contain both of the two conflicting individuals. Subtracting these 15 forbidden subcommittees gives

68 - 15 = 53

valid subcommittees.

Step-by-Step Solution

1

Calculate total ways to pick 4 people out of 8 without restrictions

\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70

This establishes the total baseline sample space of possible 4-person groups.

2

Exclude single-gender groups to satisfy the gender balance constraint

70 - \binom{4}{4} - \binom{4}{4} = 70 - 1 - 1 = 68

Groups with 0 men or 0 women are invalid.

3

Count the forbidden groups that contain both of the conflicting individuals

\binom{6}{2} = 15

Fixing the 2 specific individuals in the subcommittee requires selecting 2 additional members from the remaining 6 people.

4

Subtract forbidden groups from gender-valid groups

68 - 15 = 53

Every group containing both conflicting individuals already satisfies the gender constraint, so exactly 15 invalid groups must be removed from the 68 gender-valid groups.

Key Concept

Combinations with multiple overlapping constraints (complementary counting)

Question 109Question →

A dataset $S$ consists of $5$ positive integers. The dataset has a unique mode of $15$ , a median of $15$ , and an arithmetic mean of $12$ .

Which of the following could be the range of dataset $S$ ? Indicate all such values.

Select all that apply

8

10

18

26

28

Show answer & explanation

Answer: 10; 18; 26

Answer

The values 10, 18, and 26 are all possible values for the range of dataset

S

.

The range of dataset S is defined as x_5 - x_1. Based on the mean, median, and unique mode conditions, the 5 positive integers must take the form x_1 < x_2 < 15 = 15 < x_5 with x_1 + x_2 + x_5 = 30. Testing valid integer assignments shows the range can be any integer from 10 to 26 inclusive. Therefore, 10, 18, and 26 are all valid possible range values.

Step-by-Step Solution

1

Express the sum of elements using the mean

Sum of 5 elements = 5 × 12 = 60

Since the mean of 5 numbers is 12, their total sum must equal 60.

2

Set up the ordered elements and apply the median and unique mode conditions

Elements in ascending order: x_1 ≤ x_2 ≤ x_3 ≤ x_4 ≤ x_5, with x_3 = 15 and x_4 = 15.

The median (middle element x_3) is 15. For 15 to be the unique mode, 15 must appear at least twice. Since x_3 = 15, either x_2 = 15 or x_4 = 15. If x_2 = x_3 = x_4 = 15, then x_1 + x_5 = 15, which forces x_1 < 1 and violates positive integer constraints. Thus, 15 appears exactly twice: x_3 = 15 and x_4 = 15 (with x_1 < x_2 < 15 and x_5 > 15).

3

Determine the constraints on x_1, x_2, and x_5

x_1 + x_2 + x_5 = 30, where 1 ≤ x_1 < x_2 ≤ 14 and x_5 > 15.

Substituting x_3 = 15 and x_4 = 15 into the sum gives x_1 + x_2 + 15 + 15 + x_5 = 60, simplifying to x_1 + x_2 + x_5 = 30.

4

Calculate the bounds for the Range R = x_5 - x_1

Minimum range = 10, Maximum range = 26

To maximize R = x_5 - x_1 = 30 - 2x_1 - x_2, choose minimum x_1 = 1 and minimum x_2 = 2, yielding x_5 = 27 and R = 26. To minimize R, maximize x_1 = 6 and x_2 = 8, yielding x_5 = 16 and R = 10.

Key Concept

Measures of Central Tendency (Mean, Median, Mode) and Data Range Constraints

Question 110Question →

A container holds 5 red marbles, 4 blue marbles, and 3 green marbles. Two marbles are drawn sequentially at random without replacement. Let event $A$ be the event that at least one of the drawn marbles is red, and let event $B$ be the event that the second marble drawn is green. What is the conditional probability $P(B \mid A)$ ?

\frac{1}{6}

\frac{1}{4}

\frac{3}{22}

\frac{3}{19}

\frac{5}{22}

Show answer & explanation

Answer: $\frac{1}{6}$

Answer

The conditional probability

P(B \mid A)

is

\frac{1}{6}

.

To find

P(B \mid A)

, we evaluate

\frac{P(A \cap B)}{P(A)}

. The probability of at least one red marble

P(A)

is

1 - P(\text{no red}) = 1 - \frac{7 \times 6}{12 \times 11} = \frac{90}{132}

. The intersection event

A \cap B

requires the second marble to be green and at least one marble to be red, which means the first marble must be red and the second green. The probability of this is

\frac{5 \times 3}{12 \times 11} = \frac{15}{132}

. Taking the ratio

\frac{15/132}{90/132}

yields

\frac{15}{90} = \frac{1}{6}

.

Step-by-Step Solution

1

Calculate the total number of ordered outcomes and the probability of event A using the complementary event.

Total outcomes drawing 2 marbles from 12 without replacement is

12 \times 11 = 132

. The complement

A^c

(no red marbles selected from the 7 non-red marbles) has

7 \times 6 = 42

outcomes. Thus,

P(A^c) = \frac{42}{132} = \frac{7}{22}

, which means

P(A) = 1 - \frac{7}{22} = \frac{15}{22} = \frac{90}{132}

.

Using the complement rule is the most efficient way to compute 'at least one' probabilities.

2

Determine the intersection event

A \cap B

and calculate its probability.

Event

B

specifies that the second marble is green. For event

A

(at least one marble is red) to also occur, the first marble must be red. Thus,

A \cap B

is equivalent to 'the first marble is red AND the second marble is green'. The number of favorable outcomes is

5 \times 3 = 15

. So

P(A \cap B) = \frac{15}{132}

.

Mutual exclusivity between red and green on the second draw simplifies the intersection logic.

3

Apply the conditional probability formula

P(B \mid A) = \frac{P(A \cap B)}{P(A)}

.

P(B \mid A) = \frac{\frac{15}{132}}{\frac{90}{132}} = \frac{15}{90} = \frac{1}{6}

.

Evaluating the ratio yields the exact conditional probability requested.

Key Concept

Conditional Probability of Dependent Events

Estimated Time:3m 0s

Question 111Question →

A jar contains $3$ red marbles and $7$ blue marbles. A marble is drawn at random from the jar, its color is noted, and it is returned to the jar. A second marble is then drawn at random. What is the probability that both drawn marbles are red?

\frac{9}{100}

\frac{3}{10}

\frac{3}{5}

\frac{9}{49}

\frac{6}{7}

Show answer & explanation

Answer: $\frac{9}{100}$

Answer

The probability that both drawn marbles are red is

\frac{9}{100}

.

The option stating

\frac{9}{100}

is correct because the two draws are independent due to replacement. The probability of getting a red marble on any single draw is

\frac{3}{10}

. By the multiplication rule for independent events, the probability of both events occurring is

\frac{3}{10} \times \frac{3}{10} = \frac{9}{100}

.

Step-by-Step Solution

1

Calculate the total number of marbles in the jar.

3 \text{ red} + 7 \text{ blue} = 10 \text{ total marbles}

.

Probability requires knowing the size of the full sample space.

2

Find the probability of drawing a red marble on a single draw.

P(\text{Red}) = \frac{3}{10}

.

There are

3

favorable outcomes (red marbles) out of

10

total possible outcomes.

3

Apply the multiplication rule for independent events.

P(\text{Red}_1 \text{ and } \text{Red}_2) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100}

.

Because the first marble is replaced, the second draw is independent of the first, so joint probability is the product of individual probabilities.

Key Concept

Multiplication Rule for Independent Events:

P(A \text{ and } B) = P(A) \times P(B)

when events

A

and

B

are independent.

Estimated Time:45s

Question 112Question →

An aerospace engineering team evaluated telemetry logs from a constellation of $150$ satellites to monitor three types of sensor anomalies: Power Fluctuation ( $P$ ), Thermal Spike ( $T$ ), and Signal Attenuation ( $S$ ). The evaluation revealed the following:

- $65$ satellites exhibited Power Fluctuation.
- $58$ satellites exhibited Thermal Spike.
- $42$ satellites exhibited Signal Attenuation.
- $22$ satellites exhibited both Power Fluctuation and Thermal Spike.
- $18$ satellites exhibited both Thermal Spike and Signal Attenuation.
- $15$ satellites exhibited both Power Fluctuation and Signal Attenuation.
- $8$ satellites exhibited all three anomalies.

How many satellites exhibited exactly one of these three anomalies?

55

79

110

118

125

Show answer & explanation

Answer: 79

Answer

79 satellites exhibited exactly one anomaly.

To find the number of satellites exhibiting exactly one anomaly, break down each set into its mutually exclusive regions. First, find the satellites in exactly two sets by subtracting the triple intersection (8) from each pairwise intersection: Power and Thermal only = 22 - 8 = 14; Thermal and Signal only = 18 - 8 = 10; Power and Signal only = 15 - 8 = 7. Next, subtract all overlapping regions from each individual set total: Power only = 65 - (14 + 7 + 8) = 36; Thermal only = 58 - (14 + 10 + 8) = 26; Signal only = 42 - (7 + 10 + 8) = 17. Adding these single-anomaly counts together gives 36 + 26 + 17 = 79.

Step-by-Step Solution

1

Determine the number of satellites exhibiting ONLY pairwise anomalies (exactly two anomalies).

P and T only = 22 - 8 = 14; T and S only = 18 - 8 = 10; P and S only = 15 - 8 = 7.

The given pairwise intersection values include the 8 satellites that exhibited all three anomalies.

2

Calculate the number of satellites exhibiting exactly one anomaly for each set.

P only = 65 - (14 + 7 + 8) = 36; T only = 58 - (14 + 10 + 8) = 26; S only = 42 - (7 + 10 + 8) = 17.

Subtracting all elements that belong to two or three sets from each total set size leaves only those in the single set.

3

Sum the single-anomaly totals.

36 + 26 + 17 = 79.

The regions corresponding to exactly one anomaly are mutually exclusive.

Key Concept

Principle of Inclusion-Exclusion and Venn Diagram Region Decomposition

Question 113Question →

A software security audit evaluated $500$ open-source repositories for three specific vulnerability types: SQL Injection ( $S$ ), Cross-Site Scripting ( $X$ ), and Buffer Overflow ( $B$ ). Exactly $80$ repositories had none of these vulnerabilities. The audit revealed the following data:

- $220$ repositories contained SQL Injection vulnerabilities.
- $190$ repositories contained Cross-Site Scripting vulnerabilities.
- $210$ repositories contained Buffer Overflow vulnerabilities.
- $75$ repositories contained both SQL Injection and Cross-Site Scripting vulnerabilities.
- $80$ repositories contained both Cross-Site Scripting and Buffer Overflow vulnerabilities.
- $30$ repositories contained all three vulnerability types.

How many of the audited repositories contained Buffer Overflow vulnerabilities ONLY?

55

85

100

130

155

Show answer & explanation

Answer: $85$

Answer

85 repositories contained Buffer Overflow vulnerabilities only.

The total number of repositories containing at least one vulnerability is

500 - 80 = 420

. Using the inclusion-exclusion formula

|S \cup X \cup B| = |S| + |X| + |B| - (|S \cap X| + |X \cap B| + |S \cap B|) + |S \cap X \cap B|

, we substitute the known values:

420 = 220 + 190 + 210 - (75 + 80 + |S \cap B|) + 30

. Solving gives

420 = 650 - 155 - |S \cap B|

, so

|S \cap B| = 75

. The repositories containing Buffer Overflow ONLY are given by subtracting the overlapping regions from the total Buffer Overflow set:

|B| - (|S \cap B| \text{ only}) - (|X \cap B| \text{ only}) - |S \cap X \cap B| = 210 - (75 - 30) - (80 - 30) - 30 = 210 - 45 - 50 - 30 = 85

.

Step-by-Step Solution

1

Calculate the total number of repositories containing at least one vulnerability

|S \cup X \cup B| = 500 - 80 = 420

The total population is 500 and 80 repositories have no vulnerabilities.

2

Apply the 3-Set Inclusion-Exclusion Principle to determine the sum of pairwise intersections

|S \cap X| + |X \cap B| + |S \cap B| = 230

From

|S \cup X \cup B| = |S| + |X| + |B| - (|S \cap X| + |X \cap B| + |S \cap B|) + |S \cap X \cap B|

, we get

420 = 220 + 190 + 210 - \sum |\text{intersections}| + 30

, so

\sum |\text{intersections}| = 650 - 420 = 230

.

3

Find the missing pairwise intersection

|S \cap B|

|S \cap B| = 75

Since

|S \cap X| = 75

and

|X \cap B| = 80

, we have

75 + 80 + |S \cap B| = 230

, which yields

|S \cap B| = 75

.

4

Calculate the number of repositories with Buffer Overflow ONLY

|B \text{ only}| = 85

Subtract the exclusive two-set overlaps and the three-set overlap from

|B|

:

|B \text{ only}| = |B| - (|S \cap B| - |S \cap X \cap B|) - (|X \cap B| - |S \cap X \cap B|) - |S \cap X \cap B| = 210 - (75 - 30) - (80 - 30) - 30 = 210 - 45 - 50 - 30 = 85

.

Key Concept

Three-Set Inclusion-Exclusion Principle and Venn Diagram Region Analysis

Question 114Question →

A quality control manager at a pharmaceutical manufacturing facility evaluated the disintegration time, $t$ (in seconds), for a batch of $400$ coated tablets. The results are summarized in the frequency distribution table below:

Disintegration Time $t$ (seconds)	Frequency
$10 \le t < 20$	$40$
$20 \le t < 30$	$110$
$30 \le t < 40$	$160$
$40 \le t < 50$	$60$
$50 \le t < 60$	$30$

Which of the following statements regarding the distribution of disintegration times must be true? Select all that apply.

Select all that apply

The estimated mean disintegration time of the batch, calculated using class midpoints, is

33.25

seconds.

The median disintegration time of the batch lies within the interval

30 \le t < 40

.

Exactly

40\%

of the tablets in the batch have a disintegration time of less than

30

seconds.

The ratio of the number of tablets with a disintegration time of at least

30

seconds to the number of tablets with a disintegration time of less than

30

seconds is

5

to

3

.

More than

25\%

of the tablets in the batch have a disintegration time of

50

seconds or greater.

Show answer & explanation

Answer: The estimated mean disintegration time of the batch, calculated using class midpoints, is $33.25$ seconds.; The median disintegration time of the batch lies within the interval $30 \le t < 40$ .; The ratio of the number of tablets with a disintegration time of at least $30$ seconds to the number of tablets with a disintegration time of less than $30$ seconds is $5$ to $3$ .

Answer

The statements asserting that the estimated mean is

33.25

seconds, that the median lies in the interval

30 \le t < 40

, and that the ratio of tablets taking at least

30

seconds to those taking less than

30

seconds is

5

to

3

are all correct.

The estimated mean of

33.25

seconds is correctly computed from class midpoints

(13,300 / 400)

. The median interval

30 \le t < 40

correctly encompasses the middle values (

200^{\text{th}}

and

201^{\text{st}}

observations out of

400

). The part-to-part ratio of tablets with disintegration time

\ge 30

seconds (

250

) to

< 30

seconds (

150

) simplifies to

250:150 = 5:3

.

Step-by-Step Solution

1

Calculate the estimated mean using class midpoints.

Midpoints are

15, 25, 35, 45, 55

. Total product sum

\sum f \cdot m = (40 \times 15) + (110 \times 25) + (160 \times 35) + (60 \times 45) + (30 \times 55) = 13,300

. Mean

= 13,300 / 400 = 33.25

seconds.

The mean of grouped data is estimated by taking the weighted sum of interval midpoints divided by total sample size.

2

Determine the interval containing the sample median.

Cumulative frequencies:

10 \le t < 20 \rightarrow 40

;

20 \le t < 30 \rightarrow 150

;

30 \le t < 40 \rightarrow 310

. The

200^{\text{th}}

and

201^{\text{st}}

data points fall between cumulative counts

150

and

310

, putting the median in

30 \le t < 40

.

The median corresponds to the middle position (

N/2 = 200

) of ordered data.

3

Evaluate the percentage of tablets with disintegration time less than

30

seconds.

Count

= 40 + 110 = 150

. Percentage

= (150 / 400) \times 100\% = 37.5\%

.

Part-to-whole percentage requires dividing the sum of frequencies below

30

seconds by the total sample size

400

.

4

Compute the part-to-part ratio of tablets with

t \ge 30

seconds versus

t < 30

seconds.

Tablets with

t \ge 30

:

160 + 60 + 30 = 250

. Tablets with

t < 30

:

40 + 110 = 150

. Ratio

= 250 : 150 = 5 : 3

.

Formulating the part-to-part ratio comparing the upper three intervals to the lower two intervals.

5

Evaluate the percentage of tablets with disintegration time

t \ge 50

seconds.

Count

= 30

. Percentage

= (30 / 400) \times 100\% = 7.5\%

.

Checking if

7.5\%

exceeds

25\%

shows the statement is false.

Key Concept

Grouped Data Analysis: Mean Estimation, Median Interval Identification, and Relative Frequency Calculations

Question 115Question →

A machine operates using two independent components, Component $A$ and Component $B$ . The probability that Component $A$ functions properly on a given day is $0.90$ , and the probability that Component $B$ functions properly on that same day is $0.80$ . What is the probability that at least one of the components functions properly on a given day?

Show answer & explanation

Answer: 0.98

Answer

The probability that at least one component functions properly is

0.98

.

To determine the probability that at least one component functions properly, use the complement rule:

P(\text{at least one}) = 1 - P(\text{neither})

. Since Component

A

and Component

B

operate independently, the probability that

A

fails is

1 - 0.90 = 0.10

and the probability that

B

fails is

1 - 0.80 = 0.20

. The probability of both components failing simultaneously is

0.10 \times 0.20 = 0.02

. Subtracting this probability from

1

gives

1 - 0.02 = 0.98

.

Step-by-Step Solution

1

Find the probability of failure for each component.

P(A^c) = 1 - 0.90 = 0.10

and

P(B^c) = 1 - 0.80 = 0.20

The event that a component fails is the complement of the event that it functions properly.

2

Calculate the joint probability of both components failing.

P(A^c \text{ and } B^c) = 0.10 \times 0.20 = 0.02

Because the components operate independently, their failure events are independent, so their individual probabilities are multiplied.

3

Calculate the probability that at least one component functions properly.

P(\text{at least one functions}) = 1 - 0.02 = 0.98

The event 'at least one component functions' is the exact complement of 'both components fail'.

Key Concept

Probability of Independent Events and Complement Rule

Estimated Time:45s

Question 116Question →

A dataset $W$ consists of $10$ distinct positive integers. The arithmetic mean of the dataset is $24$ , and its median is $22$ . A new dataset $V$ is created by replacing every integer $x$ in $W$ that is strictly less than the median of $W$ with $2x + 1$ , while leaving all other integers in $W$ unchanged. If the sum of the $5$ smallest integers in $W$ is $60$ , what is the arithmetic mean of the integers in dataset $V$ ?

28.5

30.0

30.5

31.0

36.5

Show answer & explanation

Answer: 30.5

Answer

30.5

The total sum of the original dataset

W

is

10 \times 24 = 240

. Since

W

consists of

10

distinct integers, exactly

5

elements are strictly less than the median. The sum of these

5

elements is given as

60

, which leaves the sum of the remaining

5

elements as

240 - 60 = 180

. When each of the

5

lower elements

x

is replaced with

2x + 1

, their new sum becomes

2(60) + 5(1) = 125

. The sum of the new dataset

V

is

125 + 180 = 305

, and its mean is

305 / 10 = 30.5

.

Step-by-Step Solution

1

Calculate the total sum of the original dataset

W

.

Since dataset

W

has

10

elements with a mean of

24

, its total sum is

10 \times 24 = 240

.

The sum of elements in any set is equal to the number of elements multiplied by the arithmetic mean.

2

Determine how many elements are strictly less than the median of

W

.

Because all

10

elements in

W

are distinct positive integers, ordering them as

x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10}

places the median between

x_5

and

x_6

. Thus, exactly

5

elements (

x_1

through

x_5

) are strictly less than the median.

In an even-sized set of distinct values, exactly half of the elements lie strictly below the median position.

3

Find the sum of the remaining

5

elements (the upper half) in

W

.

Given that the sum of the

5

smallest elements is

60

, the sum of the remaining

5

elements is

240 - 60 = 180

.

The total sum of the dataset is the sum of its lower

5

elements plus the sum of its upper

5

elements.

4

Calculate the new sum of the transformed

5

smallest elements in dataset

V

.

Each of the

5

elements

x_i

is replaced by

2x_i + 1

. The new sum is

\sum_{i=1}^5 (2x_i + 1) = 2 \sum_{i=1}^5 x_i + 5(1) = 2(60) + 5 = 125

.

Linear transformations applied to individual elements scale their sum by the multiplier and add the constant term multiplied by the number of elements.

5

Compute the total sum and arithmetic mean of dataset

V

.

The total sum of

V

is

125 + 180 = 305

. The new mean is

\frac{305}{10} = 30.5

.

The mean of

V

is its total sum divided by the number of elements (

10

).

Key Concept

Linear transformation of dataset subsets and median position in distinct ordered sets

Question 117Question →

A research committee must select a delegation of $6$ members from a pool of $4$ senior fellows and $4$ junior analysts to sit around a circular conference table with $6$ evenly spaced seats. The delegation must consist of exactly $3$ senior fellows and $3$ junior analysts. If no two senior fellows may sit in adjacent seats around the table, how many distinct delegation seating arrangements are possible? (Two seating arrangements are considered identical if one can be rotated to match the other.)

Show answer & explanation

Answer: 192

Answer

The total number of distinct delegation seating arrangements is 192.

To find the total number of distinct delegation seating arrangements, we first determine how many ways the 6 delegates can be selected, and then multiply by the number of valid circular seating arrangements for those delegates.

1. Selection of Delegates:
- Ways to select 3 senior fellows from 4:

\binom{4}{3} = 4

- Ways to select 3 junior analysts from 4:

\binom{4}{3} = 4

- Total delegate combinations:

4 \times 4 = 16

2. Circular Seating Arrangements:
- With 3 seniors and 3 juniors at a 6-seat table, no two seniors can sit adjacent if and only if seniors and juniors alternate seats.
- Fix one senior fellow to eliminate rotational symmetry.
- The remaining 2 seniors can be arranged in

2! = 2

ways.
- The 3 junior analysts can be arranged in the 3 intermediate seats in

3! = 6

ways.
- Seating arrangements per delegation =

2 \times 6 = 12

3. Total Arrangements:
- Total =

16 \times 12 = 192

.

Step-by-Step Solution

1

Calculate combinations of senior fellows and junior analysts to form the 6-person delegation.

Number of ways to choose 3 seniors out of 4 is

\binom{4}{3} = 4

. Number of ways to choose 3 juniors out of 4 is

\binom{4}{3} = 4

. Total selection combinations =

4 \times 4 = 16

.

Choosing members from distinct pools uses combinations because member order within the selection does not matter.

2

Analyze the seating constraint for 3 seniors and 3 juniors around a 6-seat circular table.

The senior fellows must occupy alternating seats around the table (e.g., seats 1, 3, 5), leaving seats 2, 4, 6 for the junior analysts.

Placing 3 seniors among 6 circular seats with no two adjacent forces seniors to occupy every second seat.

3

Calculate the number of distinct circular seating arrangements for any specific set of 6 selected people.

Arrangements =

(3 - 1)! \times 3! = 2 \times 6 = 12

.

To account for rotational symmetry at a circular table, fix one senior fellow's seat. The remaining 2 senior fellows can be seated in

2! = 2

ways, and the 3 junior analysts can be seated in

3! = 6

ways in the remaining open seats.

4

Apply the Fundamental Counting Principle to combine selection and seating steps.

Total arrangements =

16 \text{ (selections)} \times 12 \text{ (seating arrangements)} = 192

.

Each of the 16 unique delegations can be seated around the circular table in 12 distinct relative orders.

Key Concept

Combinations and Circular Permutations with Adjacency Restrictions

Estimated Time:2m 0s

Question 118Question →

A survey of $280$ culinary school graduates evaluated their expertise across three specialized culinary disciplines: Molecular Gastronomy ( $M$ ), Pastry Arts ( $P$ ), and Sous-Vide Cooking ( $S$ ). The survey gathered the following data:
- $130$ graduates have expertise in Molecular Gastronomy.
- $140$ graduates have expertise in Pastry Arts.
- $120$ graduates have expertise in Sous-Vide Cooking.
- $50$ graduates have expertise in both Molecular Gastronomy and Pastry Arts.
- $45$ graduates have expertise in both Pastry Arts and Sous-Vide Cooking.
- $55$ graduates have expertise in both Molecular Gastronomy and Sous-Vide Cooking.
- $20$ graduates have expertise in all three disciplines.
- $20$ graduates have expertise in none of the three disciplines.

Which of the following statements MUST be true? Select all that apply.

Select all that apply

Exactly

45

graduates have expertise in Molecular Gastronomy only.

The total number of graduates with expertise in exactly two disciplines is

150

.

The number of graduates who have expertise in Sous-Vide Cooking but not Molecular Gastronomy is

65

.

More than

25\%

of all surveyed graduates have expertise in Pastry Arts only.

The total number of graduates who have expertise in at least one of the three disciplines is

240

.

Show answer & explanation

Answer: Exactly $45$ graduates have expertise in Molecular Gastronomy only.; The number of graduates who have expertise in Sous-Vide Cooking but not Molecular Gastronomy is $65$ .

Answer

The statements confirming that exactly 45 graduates have expertise in Molecular Gastronomy only and that 65 graduates have expertise in Sous-Vide Cooking but not Molecular Gastronomy are correct.

The calculated region for Molecular Gastronomy only yields

130 - (30 + 35 + 20) = 45

, making the statement about Molecular Gastronomy only correct. Furthermore, the number of graduates with expertise in Sous-Vide Cooking but not Molecular Gastronomy equals the sum of the Sous-Vide only region (

40

) and the Pastry-Sous-Vide only region (

25

), which totals

65

, making that statement correct as well.

Step-by-Step Solution

1

Determine the number of graduates in all three disciplines and in exactly two disciplines.

Triple intersection

|M \cap P \cap S| = 20

. Pairwise intersections excluding the triple intersection:

M \cap P \text{ only} = 50 - 20 = 30

,

P \cap S \text{ only} = 45 - 20 = 25

,

M \cap S \text{ only} = 55 - 20 = 35

. Total in exactly two disciplines =

30 + 25 + 35 = 90

.

Each given pairwise intersection contains the central triple intersection, so we must subtract 20 to find the exclusive double-overlap regions.

2

Calculate single-discipline exclusive counts.

Molecular Gastronomy only =

130 - (30 + 35 + 20) = 45

. Pastry Arts only =

140 - (30 + 25 + 20) = 65

. Sous-Vide Cooking only =

120 - (35 + 25 + 20) = 40

.

Subtracting all double-overlap and triple-overlap members from each set total yields the number of people belonging exclusively to that single set.

3

Evaluate the individual option statements.

Molecular Gastronomy only = 45 (True). Exactly two disciplines = 90 (Statement B claims 150, False). Sous-Vide Cooking without Molecular Gastronomy =

40 + 25 = 65

(True). Pastry Arts only percentage =

\frac{65}{280} \approx 23.21\% < 25\%

(Statement D claims

>25\%

, False). Total with at least one discipline =

280 - 20 = 260

(Statement E claims 240, False).

Comparing calculated region sizes against each option statement verifies which conditions must hold true.

Key Concept

3-Set Inclusion-Exclusion Principle and Venn Diagram Region Analysis

Estimated Time:2m 0s

Question 119Question →

A media analytics platform surveyed a group of $240$ subscribers regarding the genres of content they watched over the past month: Documentary ( $D$ ), Sci-Fi ( $S$ ), and Animation ( $A$ ). Among the surveyed subscribers:
- $110$ watched Documentary
- $120$ watched Sci-Fi
- $95$ watched Animation
- $45$ watched both Documentary and Sci-Fi
- $40$ watched both Sci-Fi and Animation
- $35$ watched both Documentary and Animation
- $15$ watched none of these three genres

How many of the surveyed subscribers watched content from exactly one of the three genres?

85

105

145

120

205

Show answer & explanation

Answer: 145

Answer

145 subscribers watched content from exactly one genre.

The correct answer is 145. First, subtract the 15 subscribers who watched no genres from the total population of 240 to get 225 subscribers in the union of all three sets. Applying the principle of inclusion-exclusion yields

225 = 110 + 120 + 95 - (45 + 40 + 35) + |D \cap S \cap A|

, which solves to

|D \cap S \cap A| = 20

. With 20 subscribers watching all three genres, subscribers watching strictly two genres are 25 (Documentary & Sci-Fi only), 20 (Sci-Fi & Animation only), and 15 (Documentary & Animation only). Isolating subscribers watching only one genre gives

110 - (25 + 15 + 20) = 50

for Documentary,

120 - (25 + 20 + 20) = 55

for Sci-Fi, and

95 - (15 + 20 + 20) = 40

for Animation. Summing these single-genre counts (

50 + 55 + 40

) gives 145.

Step-by-Step Solution

1

Find the number of subscribers who watched at least one genre and apply the 3-set Inclusion-Exclusion Principle to determine the number of subscribers who watched all three genres.

Number watching at least one genre:

240 - 15 = 225

. By inclusion-exclusion:

|D \cup S \cup A| = |D| + |S| + |A| - (|D \cap S| + |S \cap A| + |D \cap A|) + |D \cap S \cap A|

225 = 110 + 120 + 95 - (45 + 40 + 35) + |D \cap S \cap A|

225 = 325 - 120 + |D \cap S \cap A| = 205 + |D \cap S \cap A|

|D \cap S \cap A| = 225 - 205 = 20

Knowing the size of the central intersection region (

|D \cap S \cap A| = 20

) is necessary to isolate the individual non-overlapping Venn diagram regions.

2

Calculate the number of subscribers who watched EXCLUSIVELY two genres.

Documentary and Sci-Fi only:

45 - 20 = 25

Sci-Fi and Animation only:

40 - 20 = 20

Documentary and Animation only:

35 - 20 = 15

Subtracting the triple intersection from pairwise intersections isolates subscribers who belong to exactly two sets.

3

Calculate the number of subscribers who watched EXACTLY ONE genre and sum them.

Documentary only:

110 - (25 + 15 + 20) = 50

Sci-Fi only:

120 - (25 + 20 + 20) = 55

Animation only:

95 - (15 + 20 + 20) = 40

Total watching exactly one genre:

50 + 55 + 40 = 145

Subtracting all multi-set overlaps from each genre's total gives the unique single-genre viewers.

Key Concept

Three-Set Inclusion-Exclusion Principle and Regional Segmentation

Estimated Time:1m 45s

Question 120Question →

Let $A$ and $B$ be two events in a sample space such that $0 < P(A) < 1$ and $0 < P(B) < 1$ . Which of the following statements must be true? Select all that apply.

Select all that apply

If

A

and

B

are mutually exclusive, then

A

and

B

cannot be independent.

If

P(A \cup B) = P(A) + P(B) - P(A)P(B)

, then

A

and

B

are independent events.

If

A

and

B

are independent events, then their complements

A^c

and

B^c

must be mutually exclusive.

If

P(A|B) > P(A)

, then

P(B|A) > P(B)

.

If

A

and

B

are independent events, then

P(A \cup B) = P(A) + P(B)

.

Show answer & explanation

Answer: If $A$ and $B$ are mutually exclusive, then $A$ and $B$ cannot be independent.; If $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ , then $A$ and $B$ are independent events.; If $P(A|B) > P(A)$ , then $P(B|A) > P(B)$ .

Answer

The statements asserting that mutually exclusive non-impossible events cannot be independent, that

P(A \cup B) = P(A) + P(B) - P(A)P(B)

implies independence, and that

P(A|B) > P(A)

implies

P(B|A) > P(B)

are all true.

For events with probabilities strictly between 0 and 1: (1) Mutual exclusivity requires

P(A \cap B) = 0

, whereas independence requires

P(A \cap B) = P(A)P(B) > 0

, so mutually exclusive events cannot be independent. (2) Substituting

P(A \cup B) = P(A) + P(B) - P(A)P(B)

into the addition rule yields

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

, which defines independence. (3)

P(A|B) > P(A)

is mathematically equivalent to

P(A \cap B) > P(A)P(B)

, which in turn is equivalent to

P(B|A) > P(B)

.

Step-by-Step Solution

1

Analyze the relationship between mutual exclusivity and independence.

Mutually exclusive events satisfy

P(A \cap B) = 0

. For independent events,

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

. Since

P(A) > 0

and

P(B) > 0

,

P(A)P(B) > 0 \neq 0

. Thus, mutually exclusive non-impossible events can never be independent.

To evaluate structural compatibility between mutual exclusivity and independence.

2

Apply the addition rule of probability to check the union equation.

General addition rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Comparing to the given

P(A \cup B) = P(A) + P(B) - P(A)P(B)

shows

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

, which is the exact condition for independence.

To verify if the given expression for union probability forces independence.

3

Evaluate the joint probability of complementary events

A^c

and

B^c

.

Independence of

A

and

B

implies independence of

A^c

and

B^c

. Thus

P(A^c \cap B^c) = (1-P(A))(1-P(B)) > 0

. Because the joint probability is positive, the complements are not mutually exclusive.

To test whether independence of events implies mutual exclusivity of their complements.

4

Examine the symmetry of conditional probability inequalities.

P(A|B) > P(A) \implies P(A \cap B) > P(A)P(B) \implies P(B|A) = \frac{P(A \cap B)}{P(A)} > P(B)

.

To evaluate directional dependence between conditional probabilities.

5

Check the simple addition rule for independent events.

Simple addition

P(A \cup B) = P(A) + P(B)

applies only when

P(A \cap B) = 0

. Independent events have

P(A \cap B) = P(A)P(B) > 0

, so

P(A \cup B) < P(A) + P(B)

.

To distinguish between addition rules for mutually exclusive vs. independent events.

Key Concept

Theoretical relationships between independent, dependent, mutually exclusive, and conditional events.

Estimated Time:3m 0s

Data Analysis

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Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept