Counting and Probability Practice Questions — GMAT — Page 3

Question 41Question →

A market research firm surveyed $300$ consumers regarding their subscriptions to three streaming services: Service X, Service Y, and Service Z.

- $40$ consumers subscribed to none of the three services.
- $160$ consumers subscribed to Service X.
- $140$ consumers subscribed to Service Y.
- $130$ consumers subscribed to Service Z.
- Exactly $60$ consumers subscribed to exactly two of the services.

How many consumers subscribed to all three services?

35

50

55

65

110

Show answer & explanation

Answer: 55

Answer

55 consumers subscribed to all three services.

The total number of consumers subscribing to at least one service is

300 - 40 = 260

. Expressing the total union in terms of non-overlapping region counts gives

e_1 + e_2 + e_3 = 260

, where

e_1

is the number subscribing to exactly one service,

e_2 = 60

is the number subscribing to exactly two, and

e_3

is the number subscribing to all three. Thus,

e_1 + e_3 = 200

. Summing the individual totals gives

|X| + |Y| + |Z| = e_1 + 2e_2 + 3e_3 = 430

, which simplifies to

e_1 + 3e_3 = 310

. Subtracting

e_1 + e_3 = 200

from

e_1 + 3e_3 = 310

yields

2e_3 = 110

, so

e_3 = 55

.

Step-by-Step Solution

1

Calculate the total number of consumers who subscribed to at least one streaming service.

Total in union

|X \cup Y \cup Z| = 300 - 40 = 260

.

Consumers subscribing to at least one service represent the total sample minus those subscribing to none.

2

Set up the region equations for 3-set inclusion-exclusion in terms of elements in exactly 1 set (

e_1

), exactly 2 sets (

e_2

), and all 3 sets (

e_3

).

Equation 1:

e_1 + e_2 + e_3 = 260

; Equation 2:

|X| + |Y| + |Z| = e_1 + 2e_2 + 3e_3 = 160 + 140 + 130 = 430

.

Summing individual set sizes counts elements in exactly one set once, elements in exactly two sets twice, and elements in all three sets three times.

3

Substitute

e_2 = 60

into both equations.

Equation 1:

e_1 + 60 + e_3 = 260 \implies e_1 + e_3 = 200

; Equation 2:

e_1 + 2(60) + 3e_3 = 430 \implies e_1 + 3e_3 = 310

.

Simplifying both equations yields a system of two linear equations in terms of

e_1

and

e_3

.

4

Subtract the simplified Equation 1 from Equation 2 to solve for

e_3

.

(e_1 + 3e_3) - (e_1 + e_3) = 310 - 200 \implies 2e_3 = 110 \implies e_3 = 55

.

Subtracting eliminates

e_1

, directly yielding the number of consumers subscribing to all three services.

Key Concept

3-Set Inclusion-Exclusion Principle using Exact Region Counts

Question 42Question →

A jar contains $3$ green marbles and $2$ red marbles. If two marbles are selected at random one after another without replacement, what is the probability that both selected marbles are green?

Show answer & explanation

Answer: 0.3

Answer

The probability that both selected marbles are green is

0.3

.

Because selection is done without replacement, the outcome of the first draw changes the composition of the jar for the second draw. The probability of selecting a green marble first is

3/5

. With one green marble removed,

2

green marbles remain among

4

total marbles, so the probability of selecting green on the second draw is

2/4

. Multiplying these sequential probabilities gives

(3/5) \times (2/4) = 6/20 = 0.3

.

Step-by-Step Solution

1

Calculate the probability of drawing a green marble on the first selection.

P(\text{First Green}) = \frac{3}{5}

There are

3

green marbles out of

5

total marbles in the jar.

2

Calculate the conditional probability of drawing a green marble on the second selection.

P(\text{Second Green} \mid \text{First Green}) = \frac{2}{4} = \frac{1}{2}

Because the first marble was not replaced,

2

green marbles remain out of a total of

4

remaining marbles.

3

Multiply the probabilities of the dependent sequential events.

P(\text{Both Green}) = \frac{3}{5} \times \frac{1}{2} = \frac{3}{10} = 0.3

The probability of both dependent events occurring is the product of the first event's probability and the conditional probability of the second event.

Key Concept

Probability of Dependent Events Without Replacement

Question 43Question →

A research group evaluated 150 renewable energy projects. Each project utilizes at least one of three primary energy technologies: solar, wind, or hydro. Exactly 85 projects utilize solar power, 60 projects utilize wind power, and 45 projects utilize hydro power. Furthermore, 20 projects utilize both solar and wind power, 15 projects utilize both solar and hydro power, and 10 projects utilize both wind and hydro power. How many of the surveyed projects utilize all three energy technologies?

Show answer & explanation

Answer: 5

Answer

5 projects utilize all three energy technologies.

Applying the three-set inclusion-exclusion principle Total = |A| + |B| + |C| - (|A ∩ B| + |A ∩ C| + |B ∩ C|) + |A ∩ B ∩ C| + Neither gives 150 = 85 + 60 + 45 - (20 + 15 + 10) + |A ∩ B ∩ C| + 0. Simplifying yields 150 = 145 + |A ∩ B ∩ C|, which gives |A ∩ B ∩ C| = 5.

Step-by-Step Solution

1

Identify known set sizes and recall the Principle of Inclusion-Exclusion for three sets.

Total = 150, |S| = 85, |W| = 60, |H| = 45, |S ∩ W| = 20, |S ∩ H| = 15, |W ∩ H| = 10, Neither = 0.

Setting up known quantities ensures proper application of set intersection rules.

2

Substitute the set sizes into the formula: Total = |S| + |W| + |H| - (|S ∩ W| + |S ∩ H| + |W ∩ H|) + |S ∩ W ∩ H| + Neither.

150 = 85 + 60 + 45 - (20 + 15 + 10) + |S ∩ W ∩ H| + 0.

Pairwise overlaps are double-counted when individual sets are summed and must be subtracted; the central triple overlap is over-subtracted and must be added back.

3

Combine known terms and isolate the unknown triple intersection.

150 = 190 - 45 + |S ∩ W ∩ H| => 150 = 145 + |S ∩ W ∩ H| => |S ∩ W ∩ H| = 5.

Subtracting 145 from 150 yields the exact count of projects belonging to all three sets.

Key Concept

Three-Set Inclusion-Exclusion Principle

Estimated Time:2m 0s

Question 44Question →

A container holds $4$ red tokens and $6$ blue tokens. If two tokens are selected at random one after another without replacement, what is the probability that both selected tokens are red?

\frac{4}{25}

\frac{8}{45}

\frac{2}{15}

\frac{2}{5}

\frac{1}{3}

Show answer & explanation

Answer: $\frac{2}{15}$

Answer

\frac{2}{15}

The option stating

\frac{2}{15}

is correct because selecting without replacement makes the second event dependent on the first. The first probability is

\frac{4}{10}

and the second probability is

\frac{3}{9}

. Multiplying these gives

\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}

.

Step-by-Step Solution

1

Calculate the probability of selecting a red token on the first draw.

P(\text{First Red}) = \frac{4}{10} = \frac{2}{5}

There are

4

red tokens out of a total of

10

tokens.

2

Calculate the conditional probability of selecting a red token on the second draw given the first was red.

P(\text{Second Red} \mid \text{First Red}) = \frac{3}{9} = \frac{1}{3}

Because the selection is without replacement,

1

red token and

1

total token have been removed, leaving

3

red tokens out of

9

remaining tokens.

3

Multiply the sequential probabilities to find the combined probability.

P(\text{Both Red}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}

The probability of both dependent events occurring sequentially is the product of their individual conditional probabilities.

Key Concept

Probability of Dependent Events Without Replacement

Question 45Question →

In a group of $100$ students, $60$ students study Spanish and $40$ students study French. Among the $60$ students studying Spanish, $15$ also study French. If a student who studies Spanish is selected at random, what is the probability that the selected student also studies French?

Show answer & explanation

Answer: 0.25

Answer

The probability that a randomly selected Spanish-studying student also studies French is 0.25.

Since the student is chosen from the group of

60

Spanish-studying students, the sample space is restricted to

60

. Within this subset,

15

students study French. The conditional probability is therefore

\frac{15}{60} = 0.25

.

Step-by-Step Solution

1

Identify the total number of outcomes in the restricted sample space.

The sample space is restricted to students studying Spanish:

n(\text{Spanish}) = 60

.

The question specifies that the student is chosen from those who study Spanish.

2

Identify the number of favorable outcomes within this restricted sample space.

The number of students studying both Spanish and French is

n(\text{Spanish} \cap \text{French}) = 15

.

We need the count of students who satisfy both the given condition and the target event.

3

Calculate the conditional probability.

P(\text{French} | \text{Spanish}) = \frac{15}{60} = 0.25

.

Conditional probability is calculated by dividing the intersection count by the given condition's total count.

Key Concept

Conditional Probability:

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

Question 46Question →

A quality control inspector tests items from a batch containing $n$ high-grade items and $4$ low-grade items, where $n > 1$ . Two items are selected sequentially. If the first item selected is high-grade, it is kept out of the batch for testing, and additionally $1$ low-grade item is also removed from the batch before the second selection. If the probability of selecting two high-grade items in succession is $\frac{3}{8}$ , what is the value of $n$ ?

Show answer & explanation

Answer: 6

Answer

The initial number of high-grade items

n

is

6

.

Because the outcome of the first draw alters both the count of high-grade items and the count of low-grade items, the two draws are dependent events. The probability of drawing a high-grade item on the first draw is

\frac{n}{n+4}

. Following this successful outcome, the remaining pool consists of

n-1

high-grade items and

3

low-grade items, yielding a total of

n+2

items. The probability of a second high-grade selection is therefore

\frac{n-1}{n+2}

. Multiplying these probabilities gives

\frac{n(n-1)}{(n+4)(n+2)} = \frac{3}{8}

. Solving the resulting quadratic equation

5n^2 - 26n - 24 = 0

gives

n = 6

(since

n

must be a positive integer).

Step-by-Step Solution

1

Calculate the probability of selecting a high-grade item on the first draw.

P(H_1) = \frac{n}{n+4}

The initial batch contains

n

high-grade items out of a total of

n+4

items.

2

Determine the conditional sample space for the second draw after event

H_1

occurs.

High-grade items remaining =

n-1

; low-grade items remaining =

4-1=3

; total remaining items =

(n-1)+3 = n+2

.

Drawing a high-grade item removes

1

high-grade item and

1

additional low-grade item from the batch.

3

Calculate the conditional probability of selecting a high-grade item on the second draw.

P(H_2 | H_1) = \frac{n-1}{n+2}

There are now

n-1

high-grade items out of

n+2

total remaining items.

4

Set up and solve the algebraic probability equation.

\frac{n(n-1)}{(n+4)(n+2)} = \frac{3}{8} \implies 8n^2 - 8n = 3(n^2 + 6n + 8) \implies 5n^2 - 26n - 24 = 0 \implies (5n+4)(n-6) = 0 \implies n = 6

.

Because

n > 1

is required, the negative fraction root

-\frac{4}{5}

is discarded.

Key Concept

Probability of dependent events with non-standard sample space modifications

Question 47Question →

A museum cataloged $250$ historical artifacts. Each artifact was subjected to at least one of three preservation procedures: micro-scanning, chemical analysis, or radiocarbon dating. Exactly $140$ artifacts underwent micro-scanning, $120$ underwent chemical analysis, and $100$ underwent radiocarbon dating. If $40$ artifacts underwent both micro-scanning and chemical analysis, $35$ underwent both chemical analysis and radiocarbon dating, and $45$ underwent both micro-scanning and radiocarbon dating, how many artifacts underwent all three procedures?

Show answer & explanation

Answer: 10

Answer

10 artifacts underwent all three preservation procedures.

Using the Inclusion-Exclusion Principle for three sets, the total union is equal to the sum of individual set sizes (

140 + 120 + 100 = 360

), minus the sum of two-set overlaps (

40 + 35 + 45 = 120

), plus the three-set overlap. Setting up

250 = 360 - 120 + x

yields

250 = 240 + x

, so

x = 10

.

Step-by-Step Solution

1

Set up the Inclusion-Exclusion formula for three overlapping sets.

|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |A \cap C|) + |A \cap B \cap C|

The total size of the union equals the sum of the individual set sizes minus the pairwise intersections plus the triple intersection.

2

Substitute the given values into the formula.

250 = 140 + 120 + 100 - (40 + 35 + 45) + |A \cap B \cap C|

Since every artifact underwent at least one procedure, the total number of artifacts equals the union of all three sets.

3

Compute the sum of individual sets and pairwise intersections to isolate the unknown.

250 = 360 - 120 + |A \cap B \cap C| \Rightarrow 250 = 240 + |A \cap B \cap C|

Simplifying the arithmetic terms allows direct calculation of the triple intersection.

4

Subtract

240

from

250

to find the final value.

|A \cap B \cap C| = 10

Solving the linear equation yields the number of artifacts undergoing all three procedures.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 48Question →

A box contains 90 tokens, numbered sequentially from $1$ to $90$ , inclusive. If one token is drawn at random from the box, what is the probability that the number on the token is a multiple of $4$ or $6$ , but not a multiple of $8$ ?

\frac{13}{45}

\frac{1}{3}

\frac{19}{90}

\frac{2}{9}

\frac{7}{30}

Show answer & explanation

Answer: $\frac{19}{90}$

Answer

The probability that the selected number is a multiple of 4 or 6, but not a multiple of 8, is

\frac{19}{90}

.

To find the probability, determine the number of favorable outcomes out of 90 total outcomes. The number of multiples of 4 is 22, and the number of multiples of 6 is 15. The numbers that are multiples of both 4 and 6 are multiples of 12, of which there are 7. By inclusion-exclusion, the number of integers that are multiples of 4 or 6 is

22 + 15 - 7 = 30

. Since all multiples of 8 are automatically multiples of 4, all 11 multiples of 8 in the range are included in these 30 numbers. Excluding the multiples of 8 leaves

30 - 11 = 19

favorable integers. Thus, the probability is

\frac{19}{90}

.

Step-by-Step Solution

1

Find the count of multiples of 4 and multiples of 6 in the range from 1 to 90.

Multiples of 4:

\lfloor \frac{90}{4} \rfloor = 22

. Multiples of 6:

\lfloor \frac{90}{6} \rfloor = 15

.

Identify the size of each individual set of multiples.

2

Calculate the number of elements in the union of multiples of 4 or 6 using the Inclusion-Exclusion Principle.

Multiples of both 4 and 6 are multiples of

\text{lcm}(4,6) = 12

. Count of multiples of 12:

\lfloor \frac{90}{12} \rfloor = 7

. Union size:

22 + 15 - 7 = 30

.

Avoid double-counting numbers that are divisible by both 4 and 6.

3

Identify and subtract the multiples of 8 within this union.

Multiples of 8 in the range:

\lfloor \frac{90}{8} \rfloor = 11

. Since every multiple of 8 is also a multiple of 4, all 11 multiples of 8 are contained within the union. Favorable outcomes:

30 - 11 = 19

.

Filter out numbers that satisfy the exclusion condition (multiples of 8).

4

Compute the single-event probability.

P = \frac{19}{90}

.

Divide the number of favorable outcomes (19) by the total sample space (90).

Key Concept

Basic Single-Event Probability with Principle of Inclusion-Exclusion and Set Restriction

Estimated Time:2m 0s

Question 49Question →

A box contains 12 cards numbered consecutively from 1 through 12. Two cards are selected at random from the box without replacement. Given that the sum of the numbers on the two selected cards is even, what is the probability that at least one of the selected cards has a prime number on it?

\frac{19}{30}

\frac{19}{66}

\frac{2}{3}

\frac{7}{15}

\frac{15}{22}

Show answer & explanation

Answer: $\frac{19}{30}$

Answer

The conditional probability that at least one selected card has a prime number on it, given that their sum is even, is

\frac{19}{30}

.

The option offering 19/30 is correct. Given that the sum of the two cards is even, both cards must be even or both cards must be odd. Choosing 2 even cards from 6 available gives 15 pairs, and choosing 2 odd cards from 6 available gives 15 pairs, making 30 possible pairs in total for the restricted sample space. Among the even cards, 2 is prime while 4, 6, 8, 10, and 12 are non-prime (5 numbers). Among the odd cards, 3, 5, 7, and 11 are prime (4 numbers) while 1 and 9 are non-prime (2 numbers). The pairs containing no primes consist of 2 non-prime evens (10 pairs) and 2 non-prime odds (1 pair), giving 11 non-prime pairs. Subtracting from 30 yields 19 pairs with at least one prime. Thus, the conditional probability is 19/30.

Step-by-Step Solution

1

Determine the restricted sample space (Condition B: Sum of two cards is even)

Total outcomes in Condition B = 30

The sum of two integers is even if both are even or both are odd. Among numbers 1 to 12, there are 6 even numbers ({2, 4, 6, 8, 10, 12}) and 6 odd numbers ({1, 3, 5, 7, 9, 11}). The number of ways to pick 2 even cards is

\binom{6}{2} = 15

, and 2 odd cards is

\binom{6}{2} = 15

. Total pairs with an even sum =

15 + 15 = 30

.

2

Categorize the numbers 1 through 12 by parity and primality

Prime evens = {2} (1 number); Non-prime evens = {4, 6, 8, 10, 12} (5 numbers); Prime odds = {3, 5, 7, 11} (4 numbers); Non-prime odds = {1, 9} (2 numbers)

Note that 1 is not a prime number, and 2 is the only even prime number.

3

Count the number of pairs in the restricted sample space with NO prime numbers

11 non-prime pairs

Pairs of two evens with no primes come from non-prime evens:

\binom{5}{2} = 10

pairs. Pairs of two odds with no primes come from non-prime odds:

\binom{2}{2} = 1

pair. Total non-prime pairs =

10 + 1 = 11

.

4

Calculate favorable outcomes (Event A ∩ B) and the conditional probability

P(A|B) = 19/30

Favorable pairs with at least one prime =

30 - 11 = 19

. Therefore,

P(\text{At least one prime} \mid \text{Even sum}) = \frac{19}{30}

.

Key Concept

Conditional Probability with Restricted Sample Space

Estimated Time:2m 0s

Question 50Question →

A diner offers a three-course meal deal where a customer selects $1$ appetizer from a choice of $4$ , $1$ main dish from a choice of $5$ , and $1$ dessert from a choice of $3$ . How many different three-course meal combinations can a customer order?

12

20

35

60

120

Show answer & explanation

Answer: $60$

Answer

The total number of unique three-course meal combinations is

60

.

According to the Fundamental Counting Principle, when a process involves a series of independent decision stages, the total number of possible outcomes is the product of the number of choices available at each stage. Multiplying

4

appetizers by

5

main dishes and

3

desserts yields

4 \times 5 \times 3 = 60

distinct meal combinations.

Step-by-Step Solution

1

Identify the number of independent choices available at each decision stage

Appetizer choices =

4

, Main dish choices =

5

, Dessert choices =

3

Each course selection represents an independent decision stage.

2

Apply the Fundamental Counting Principle by multiplying the number of options for all stages

4 \times 5 \times 3 = 60

The total number of multi-stage outcomes is the product of the number of options at each individual stage.

Key Concept

Fundamental Counting Principle

Question 51Question →

A corporate venture capital firm evaluated $120$ technology startups for investment. Each startup met at least one of three key criteria: strong artificial intelligence capability, established revenue growth, or international market presence. Exactly $65$ startups met the artificial intelligence criteria, $55$ met the revenue growth criteria, and $50$ met the international presence criteria. Furthermore, $25$ startups met both the artificial intelligence and revenue growth criteria, $20$ met both the revenue growth and international presence criteria, and $15$ met both the artificial intelligence and international presence criteria. How many startups met all three investment criteria?

5

10

15

20

25

Show answer & explanation

Answer: $10$

Answer

The number of startups that met all three investment criteria is 10.

According to the Inclusion-Exclusion Principle for three sets, the total population is equal to the sum of the three individual sets minus the sum of the three pairwise intersections, plus the intersection of all three sets. Substituting the given values:

120 = 65 + 55 + 50 - (25 + 20 + 15) + x

, which simplifies to

120 = 110 + x

, giving

x = 10

. Thus,

10

startups met all three criteria.

Step-by-Step Solution

1

State the Principle of Inclusion-Exclusion for three sets

|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |A \cap C|) + |A \cap B \cap C|

To account for startups counted in multiple overlapping categories without double-counting or over-subtracting.

2

Substitute the known values from the problem statement into the formula

120 = 65 + 55 + 50 - (25 + 20 + 15) + x

, where

x = |A \cap B \cap C|

Every startup meets at least one criterion, so the union size equals the total number of startups (

120

).

3

Simplify the sums and solve for

x

120 = 170 - 60 + x \implies 120 = 110 + x \implies x = 10

Subtracting

110

from

120

gives the exact number of startups meeting all three criteria.

Key Concept

Three-Set Inclusion-Exclusion Principle

Question 52Question →

An integer $n$ is chosen at random from the set of all positive integers less than or equal to $120$ . What is the probability that $n$ is a multiple of either $3$ or $5$ , but not a multiple of $15$ ?

Show answer & explanation

Answer: 0.4

Answer

The probability that the selected integer is a multiple of either 3 or 5, but not 15, is 0.4 (or 2/5).

The total number of integers from 1 to 120 is 120. Multiples of 3 up to 120 total 40, multiples of 5 total 24, and multiples of 15 total 8. Integers that are multiples of 3 but not 15 number 40 - 8 = 32. Integers that are multiples of 5 but not 15 number 24 - 8 = 16. The total number of favorable outcomes is 32 + 16 = 48. Thus, the probability is 48/120 = 2/5 = 0.4.

Step-by-Step Solution

1

Count total possible outcomes in the sample space.

The total number of integers from 1 to 120 is 120.

Each integer in the set {1, 2, ..., 120} is equally likely to be selected.

2

Count the number of multiples of 3, 5, and 15 within the range.

Multiples of 3: 40; Multiples of 5: 24; Multiples of 15: 8.

Since 120 is divisible by 3, 5, and 15, the count of multiples of k up to 120 is 120/k.

3

Calculate the number of integers that are multiples of 3 or 5, but not 15.

Number of favorable outcomes = (Multiples of 3 only) + (Multiples of 5 only) = (40 - 8) + (24 - 8) = 32 + 16 = 48.

Multiples of 15 are common multiples of both 3 and 5 and must be excluded completely according to the condition 'not a multiple of 15'.

4

Compute the single-event probability.

Probability = 48 / 120 = 2 / 5 = 0.4.

Probability of a single event is defined as the ratio of favorable outcomes to total possible outcomes.

Key Concept

Basic Single-Event Probability with Set Restrictions

Question 53Question →

A quality control inspector reviews a shipment containing a total of $N$ customized components, of which exactly 3 are defective. The inspector randomly selects 3 components from the shipment one by one without replacement. If the probability that at least one of the selected components is defective is equal to $\frac{31}{35}$ , what is the value of $N$ ?

Show answer & explanation

Answer: 7

Answer

The total number of components in the shipment,

N

, is 7.

To find

N

, apply the complementary probability formula

P(\text{at least 1 defective}) = 1 - P(\text{0 defective})

. Given that

P(\text{at least 1 defective}) = \frac{31}{35}

, the probability of drawing zero defective components is

1 - \frac{31}{35} = \frac{4}{35}

. Out of

N

total components,

N-3

are non-defective. Drawing 3 non-defective components without replacement gives

P(\text{0 defective}) = \frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)}

. Setting this equal to

\frac{4}{35}

and testing integer values starting at

N=6

yields

N=7

, since

\frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{24}{210} = \frac{4}{35}

.

Step-by-Step Solution

1

Calculate the probability of the complementary event (selecting no defective components).

P(\text{0 defective}) = 1 - \frac{31}{35} = \frac{4}{35}

.

Calculating 'at least one' directly requires summing three separate cases (1 defective, 2 defective, 3 defective), whereas using the complement

P(\text{at least 1}) = 1 - P(\text{none})

requires evaluating only one scenario.

2

Formulate the algebraic expression for picking 3 non-defective components without replacement.

P(\text{0 defective}) = \frac{\binom{N-3}{3}}{\binom{N}{3}} = \frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)}

.

There are

N-3

non-defective components out of

N

total components, and 3 are selected without replacement.

3

Equate the algebraic probability to the known complementary probability and solve for

N

.

\frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)} = \frac{4}{35} \implies N = 7.

Testing

N = 7

yields

\frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{24}{210} = \frac{4}{35}

. The function is strictly increasing for

N \ge 6

, making

N = 7

the unique integer solution.

Key Concept

Complementary Probability and Dependent Sampling (Without Replacement)

Question 54Question →

A box contains $x$ blue spheres and $8$ yellow spheres, where $x$ is a positive integer. If two spheres are selected at random one after another without replacement, the probability that both selected spheres are blue is $\frac{5}{17}$ . What is the total number of spheres in the box initially?

10

15

18

20

25

Show answer & explanation

Answer: 18

Answer

The total number of spheres in the box initially is 18.

The probability of drawing two blue spheres sequentially without replacement is given by

\frac{x}{x+8} \times \frac{x-1}{x+7} = \frac{5}{17}

. Expanding and rearranging the equation gives

3x^2 - 23x - 70 = 0

, which factors into

(3x + 7)(x - 10) = 0

. Since

x

must be a positive integer,

x = 10

. The total number of spheres in the box initially is

x + 8 = 10 + 8 = 18

.

Step-by-Step Solution

1

Set up the probability expression for dependent sequential events.

P(\text{both blue}) = \frac{x}{x+8} \times \frac{x-1}{x+7}

Because the draws occur without replacement, the total count decreases from

x+8

to

x+7

and the number of blue spheres decreases from

x

to

x-1

for the second draw.

2

Equate to the given probability and clear denominators to form a quadratic equation.

\frac{x(x-1)}{(x+8)(x+7)} = \frac{5}{17} \implies 17(x^2 - x) = 5(x^2 + 15x + 56) \implies 3x^2 - 23x - 70 = 0

Cross-multiplying converts the rational probability equation into a standard quadratic equation.

3

Factor the quadratic equation to find the positive integer root.

(3x + 7)(x - 10) = 0 \implies x = 10

Since the count of spheres must be a positive integer,

x = 10

is the only valid solution for the number of blue spheres.

4

Calculate the total initial number of spheres.

\text{Total} = x + 8 = 10 + 8 = 18

The question asks for the total initial number of spheres, which includes both the

x

blue spheres and the 8 yellow spheres.

Key Concept

Probability of Dependent Events Without Replacement

Estimated Time:2m 0s

Question 55Question →

The table below categorizes 100 analysts at a consulting firm by their department and experience level:

Department	Junior (1–3 yrs)	Senior (4–7 yrs)	Lead (8+ yrs)	Total
Technology	14	16	10	40
Analytics	12	18	5	35
Operations	9	11	5	25
Total	35	45	20	100

If one analyst is selected at random from this group, what is the probability that the selected analyst works in the Analytics department or has Lead experience, but NOT both?

Show answer & explanation

Answer: 0.45

Answer

0.45

To find the probability of selecting an analyst who is either in the Analytics department or at the Lead level, but not both, we count the analysts in Analytics who are not Leads (

12 + 18 = 30

) and the analysts at the Lead level who are not in Analytics (

10 + 5 = 15

). The total number of favorable outcomes is

30 + 15 = 45

. Dividing by the total pool of

100

analysts yields a probability of

\frac{45}{100} = 0.45

.

Step-by-Step Solution

1

Determine the total size of the sample space.

Total analysts

N = 100

.

Basic probability requires dividing favorable outcomes by total possible outcomes.

2

Calculate the number of analysts satisfying 'Analytics, but NOT Lead'.

Junior Analytics (

12

) + Senior Analytics (

18

) =

30

.

Excludes the

5

Lead analysts in the Analytics department.

3

Calculate the number of analysts satisfying 'Lead, but NOT Analytics'.

Technology Lead (

10

) + Operations Lead (

5

) =

15

.

Excludes the

5

Lead analysts in the Analytics department.

4

Sum the non-overlapping favorable counts and compute probability.

Favorable outcomes

= 30 + 15 = 45

; Probability

= \frac{45}{100} = 0.45

.

Probability of a single event is defined as

\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}

.

Key Concept

Basic Single-Event Probability from Two-Way Tabular Data with Mutually Exclusive Set Conditions

Question 56Question →

A quality control analyst randomly selects one component from a shipment containing components labeled with distinct integer batch numbers from $21$ to $100$ , inclusive. What is the probability that the batch number of the selected component is a prime number?

\frac{1}{5}

\frac{17}{80}

\frac{17}{79}

\frac{9}{40}

\frac{17}{63}

Show answer & explanation

Answer: $\frac{17}{80}$

Answer

\frac{17}{80}

The total number of batch numbers from 21 through 100 inclusive is given by

100 - 21 + 1 = 80

. The prime numbers in this range are 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97, which count to 17 prime numbers. Dividing the 17 favorable outcomes by the 80 total outcomes gives a probability of

\frac{17}{80}

.

Step-by-Step Solution

1

Determine the total number of components (the sample space size).

Total components

N = 100 - 21 + 1 = 80

.

For an inclusive range of integers from

a

to

b

, the number of elements is

b - a + 1

.

2

Identify all prime numbers between

21

and

100

, inclusive.

The prime numbers in this range are:

23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

. There are

17

prime numbers in total.

Each listed number has exactly two distinct positive divisors:

1

and itself.

3

Calculate the probability of selecting a prime batch number.

P(\text{Prime}) = \frac{\text{Number of Prime Batch Numbers}}{\text{Total Number of Batch Numbers}} = \frac{17}{80}

.

Probability of a single event is the ratio of favorable outcomes to total possible outcomes.

Key Concept

Basic Single-Event Probability

Question 57Question →

Five speakers—Alice, Bob, Carol, David, and Eva—are scheduled to give consecutive presentations at a conference. If Alice must present first, in how many different linear orders can all five speakers present?

120

48

24

16

12

Show answer & explanation

Answer: 24

Answer

The total number of different presentation orders is 24.

Since Alice's position is fixed as the first speaker, there is only

1

way to fill the first slot. The remaining

4

speakers can be arranged in the remaining

4

slots in

4! = 4 \times 3 \times 2 \times 1 = 24

ways. Thus, the total number of linear arrangements is

1 \times 24 = 24

.

Step-by-Step Solution

1

Identify the fixed element and available positions.

Alice must fill the 1st position, leaving 1 choice for slot 1.

The question specifies that Alice must present first.

2

Calculate the number of ways to arrange the remaining speakers.

The remaining 4 speakers (Bob, Carol, David, Eva) can be arranged in

4! = 4 \times 3 \times 2 \times 1 = 24

ways.

Linear permutations of

n

distinct objects use

n!

.

3

Apply the Fundamental Counting Principle.

Total arrangements =

1 \times 24 = 24

.

Combine the independent choices for each position.

Key Concept

Permutations with Fixed Positions

Question 58Question →

A research laboratory's secure vault contains $N$ data drives, where $N > 4$ . Exactly $3$ of the drives contain encrypted security keys, while the remaining $N - 3$ drives contain standard operational data. If two data drives are selected at random one after another without replacement, the probability that at least one of the selected drives contains an encrypted security key is $\frac{8}{15}$ . What is the value of $N$ ?

8

9

10

12

15

Show answer & explanation

Answer: 10

Answer

10

The correct option is 10. The probability of choosing at least one encrypted security key drive in two draws without replacement is

1 - P(\text{no security key drives})

. Since there are

N - 3

standard drives out of

N

total drives,

P(\text{no security key drives}) = \frac{N - 3}{N} \times \frac{N - 4}{N - 1}

. Setting

1 - \frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{8}{15}

gives

\frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{7}{15}

. Solving

15(N^2 - 7N + 12) = 7(N^2 - N)

leads to

8N^2 - 98N + 180 = 0

, or

4N^2 - 49N + 90 = 0

, which factors as

(4N - 9)(N - 10) = 0

. Because

N

must be an integer,

N = 10

.

Step-by-Step Solution

1

Express the complementary probability event

The probability of at least one drive having an encrypted key is

1 - P(\text{both drives are standard data drives})

. Thus,

P(\text{both standard}) = 1 - \frac{8}{15} = \frac{7}{15}

.

Using the complementary probability rule is more efficient than calculating the individual probabilities of key-then-standard, standard-then-key, and key-then-key.

2

Set up the probability expression for dependent sequential selection without replacement

For the first draw, the probability of selecting a standard drive is

\frac{N - 3}{N}

. For the second draw, since one standard drive has been removed, the probability is

\frac{N - 4}{N - 1}

. Therefore,

\frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{7}{15}

.

The events are dependent because the selection is made without replacement.

3

Solve the algebraic equation for N

Expanding both sides gives

15(N^2 - 7N + 12) = 7(N^2 - N)

, which simplifies to

15N^2 - 105N + 180 = 7N^2 - 7N \implies 8N^2 - 98N + 180 = 0

. Dividing by 2 yields

4N^2 - 49N + 90 = 0

. Factoring gives

(4N - 9)(N - 10) = 0

.

Since

N

must be a positive integer greater than 4,

N = 10

.

Key Concept

Dependent probability without replacement and complementary probability

Estimated Time:2m 0s

Question 59Question →

A research team needs to select a subcommittee of 3 scientists from a department containing 7 scientists. How many different 3-member subcommittees can be selected?

Show answer & explanation

Answer: 35

Answer

35 different subcommittees can be selected.

The total number of ways to choose a committee of 3 members from a group of 7 without regard to order is given by the combination formula

C(7,3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

.

Step-by-Step Solution

1

Determine whether the selection depends on order.

Order does not matter since all 3 members of the subcommittee have equal roles.

When order does not matter in group selection, combinations (

nCr

) must be used rather than permutations (

nPr

).

2

Apply the combination formula

C(n,k) = \frac{n!}{k!(n-k)!}

with

n=7

and

k=3

.

C(7,3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1}

This counts the unique groups of 3 that can be chosen from a pool of 7.

3

Calculate the numerical result.

35

Dividing

7 \times 6 \times 5 = 210

by

3 \times 2 \times 1 = 6

yields 35.

Key Concept

Combinations and Group Selections

Question 60Question →

A security system requires each employee to create a 6-character access code consisting of 3 non-zero digits followed by 3 letters from the standard 26-letter English alphabet. The access code must satisfy the following restrictions:

1. The first digit must be odd.
2. No two adjacent digits in the code can be identical.
3. The first letter must be a vowel ( $A, E, I, O, U$ ), the second letter must be a consonant, and the third letter cannot be identical to the first letter.

How many distinct access codes can be created under these rules?

806,400

840,000

1,040,000

1,063,125

2,945

Show answer & explanation

Answer: $840,000$

Answer

840,000

The total number of valid codes is obtained by multiplying the number of choices at each stage. For the digits:

5

(odd non-zero)

\times 8

(non-zero digit

\neq

1st digit)

\times 8

(non-zero digit

\neq

2nd digit)

= 320

. For the letters:

5

(vowels)

\times 21

(consonants)

\times 25

(any letter

\neq

1st letter)

= 2,625

. The overall number of codes is

320 \times 2,625 = 840,000

.

Step-by-Step Solution

1

Calculate the number of choices for the first 3 digit positions.

5 choices for the 1st digit, 8 choices for the 2nd digit, and 8 choices for the 3rd digit.

Non-zero odd digits are

\{1, 3, 5, 7, 9\}

(5 choices). The 2nd digit can be any non-zero digit except the 1st digit (

9 - 1 = 8

choices). The 3rd digit can be any non-zero digit except the 2nd digit (

9 - 1 = 8

choices). Thus,

5 \times 8 \times 8 = 320

digit combinations.

2

Calculate the number of choices for the 3 letter positions.

5 choices for the 1st letter, 21 choices for the 2nd letter, and 25 choices for the 3rd letter.

There are 5 vowels and 21 consonants in the 26-letter alphabet. The 1st letter has 5 choices. The 2nd letter has 21 choices (consonant). The 3rd letter can be any letter except the 1st letter (

26 - 1 = 25

choices). Thus,

5 \times 21 \times 25 = 2,625

letter combinations.

3

Apply the Fundamental Counting Principle to find the total number of access codes.

320 \times 2,625 = 840,000

Since choosing the digits and choosing the letters are independent stages of creating a code, multiply the total number of digit combinations by the total number of letter combinations.

Key Concept

Fundamental Counting Principle with Positional Restrictions

Counting and Probability

Answer

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