Counting and Probability Alıştırma Soruları — GMAT — Sayfa 2

Soru 21Soru →

A financial analyst chooses two investment assets at random, one after another without replacement, from a portfolio consisting of $3$ technology stocks and $2$ utility stocks. The probability of achieving a target return depends on the composition of the selected assets:
- If both selected assets are technology stocks, the probability of achieving the target return is $0.80$ .
- If exactly one selected asset is a technology stock, the probability of achieving the target return is $0.50$ .
- If neither selected asset is a technology stock, the probability of achieving the target return is $0.10$ .

What is the probability, expressed as a decimal, that the analyst achieves the target return?

Cevabı ve açıklamayı göster

Cevap: 0.55

Cevap

The probability that the analyst achieves the target return is 0.55.

To find the overall probability of achieving the target return, calculate the probability of each mutually exclusive composition of assets selected without replacement, and weight each by its conditional probability of success. The probability of selecting two technology stocks is (3/5) * (2/4) = 0.30. The probability of selecting one technology stock and one utility stock in either order is (3/5)*(2/4) + (2/5)*(3/4) = 0.60. The probability of selecting two utility stocks is (2/5) * (1/4) = 0.10. Multiplying each by its respective success rate (0.80, 0.50, and 0.10) yields 0.24 + 0.30 + 0.01 = 0.55.

Adım Adım Çözüm

1

Determine the probabilities of the three mutually exclusive selection scenarios without replacement.

P(TT) = 0.30, P(1T) = 0.60, P(UU) = 0.10.

Since selections are made without replacement, the outcome of the first draw affects the total count and available assets for the second draw.

2

Calculate the joint probability of achieving the target return for each scenario.

P(TT and Target) = 0.24, P(1T and Target) = 0.30, P(UU and Target) = 0.01.

Multiply the probability of drawing each combination by the conditional probability of success for that specific combination.

3

Sum the joint probabilities across all possible scenarios.

Total Probability = 0.24 + 0.30 + 0.01 = 0.55.

The scenarios are mutually exclusive and exhaustive, so their sum gives the total overall probability of achieving the target return.

Anahtar Kavram

Probability of Dependent Events and Law of Total Probability

Tahmini Süre:2m 0s

Soru 22Soru →

A committee of $5$ members is to be formed from a pool of $6$ men and $5$ women. The committee must include at least $2$ men and at least $2$ women. However, two specific individuals in the pool, one man and one woman, refuse to serve on the committee together. How many different valid $5$ -member committees can be formed?

Cevabı ve açıklamayı göster

Cevap: 280

Cevap

The correct answer is 280.

The correct answer is 280, found by taking the total possible committees satisfying the gender constraint (350) and subtracting the invalid committees containing both conflicting members (70).

Adım Adım Çözüm

1

Calculate total combinations satisfying gender constraints without individual restriction

350 valid gender-balanced committees (200 from 3M/2W and 150 from 2M/3W)

Establishes the total baseline number of committee selections before removing forbidden pairings

2

Calculate forbidden combinations containing both specific conflicting individuals

70 forbidden combinations (40 with 3M/2W overall and 30 with 2M/3W overall)

Identifies committee selections that violate the condition that the two individuals cannot serve together

3

Subtract forbidden combinations from total baseline combinations

280 valid committee selections

Applying complementary counting (

350 - 70

) yields the exact number of allowable committees

Anahtar Kavram

Combinations with group constraints and complementary counting

Soru 23Soru →

A quality control engineer inspects a shipment of 25 solar panels, of which 9 are premium grade and 16 are standard grade. The engineer randomly selects 2 panels from the shipment, one after another without replacement. What is the probability, expressed as a decimal, that at least one of the two selected panels is premium grade?

Cevabı ve açıklamayı göster

Cevap: 0.6

Cevap

The probability that at least one of the two selected panels is premium grade is 0.6.

Selection without replacement creates dependent events because the outcome of the first draw alters the pool available for the second draw. Out of 25 panels, 16 are standard grade. The probability that the first drawn panel is standard grade is 16/25. If the first panel is standard grade, 15 standard panels remain out of 24 total panels, giving a probability of 15/24 for the second draw. The probability of selecting two standard panels is (16/25) × (15/24) = 0.40. Using complementary probability, the probability that at least one panel is premium grade is 1 - 0.40 = 0.60.

Adım Adım Çözüm

1

Determine the initial counts of total, premium, and standard grade panels.

Total panels = 25, premium grade panels = 9, standard grade panels = 16.

Establishing the sample space composition is necessary to calculate draw probabilities.

2

Calculate the probability that neither panel selected is premium grade (i.e., both are standard grade).

P(both standard) = (16 / 25) * (15 / 24) = (16 / 25) * (5 / 8) = 0.40.

Because selection is done without replacement, the total pool size and remaining standard panels each decrease by 1 for the second draw.

3

Apply complementary probability to determine the probability of selecting at least one premium panel.

P(at least one premium) = 1 - P(both standard) = 1 - 0.40 = 0.60.

The scenario of selecting at least one premium panel is the exact complementary event of selecting zero premium panels.

Anahtar Kavram

Dependent Events and Complementary Probability

Alternatif Yöntem

Sum the probabilities of mutually exclusive favorable outcomes: P(1st premium, 2nd standard) + P(1st standard, 2nd premium) + P(both premium) = (9/25)(16/24) + (16/25)(9/24) + (9/25)(8/24) = 0.24 + 0.24 + 0.12 = 0.60.

Tahmini Süre:1m 30s

Soru 24Soru →

An automated security gate opens successfully on any given approach with a probability of $0.8$ , independently of other attempts. If a driver approaches the gate $3$ times, the probability that the gate opens successfully at least once is $0.992$ .

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

The statement is true because the probability of failing all

3

independent attempts is

(1 - 0.8)^3 = 0.008

, making the probability of at least one successful opening

1 - 0.008 = 0.992

.

The statement accurately reflects the complementary probability principle

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

. For

3

independent trials with single-trial failure probability

0.2

, the probability of all trials failing is

0.2^3 = 0.008

. Subtracting this from

1

yields

0.992

.

Adım Adım Çözüm

1

Find the probability of failure on a single trial

P(\text{failure}) = 1 - 0.8 = 0.2

The failure of a trial is the complement of its success.

2

Calculate the probability of zero successes across 3 independent trials

P(\text{0 successes}) = (0.2)^3 = 0.008

For independent events, the joint probability is the product of individual probabilities.

3

Apply the complementary probability rule to find the probability of at least one success

P(\text{at least 1 success}) = 1 - 0.008 = 0.992

The event 'at least one success' is the exact complement of 'zero successes'.

Anahtar Kavram

Complementary Probability and At-Least-One Scenarios

Soru 25Soru →

A testing kit contains $8$ electronic sensors, of which exactly $3$ are uncalibrated and $5$ are fully calibrated. If $3$ sensors are chosen at random without replacement, what is the probability that at least $1$ of the chosen sensors is uncalibrated?

\frac{5}{28}

\frac{15}{28}

\frac{23}{28}

\frac{55}{56}

\frac{387}{512}

Cevabı ve açıklamayı göster

Cevap: $\frac{23}{28}$

Cevap

The correct answer is

\frac{23}{28}

.

To find the probability that at least one sensor is uncalibrated, calculate 1 minus the probability that no selected sensors are uncalibrated. Out of 8 sensors, there are

\binom{8}{3} = 56

ways to pick 3. The number of ways to pick 3 calibrated sensors from the 5 available is

\binom{5}{3} = 10

. Thus,

P(\text{no uncalibrated}) = \frac{10}{56} = \frac{5}{28}

. Subtracting this from 1 gives

1 - \frac{5}{28} = \frac{23}{28}

.

Adım Adım Çözüm

1

Calculate the total number of ways to choose 3 sensors out of 8.

\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

total outcomes.

Selection is made without replacement and order does not matter.

2

Calculate the complementary event: selecting 0 uncalibrated sensors (all 3 selected sensors are calibrated).

\binom{5}{3} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10

outcomes with all calibrated sensors.

There are 5 calibrated sensors available, and 3 must be selected from them.

3

Find the probability of the complementary event (no uncalibrated sensors selected).

P(\text{no uncalibrated}) = \frac{10}{56} = \frac{5}{28}

.

Divide the favorable outcomes for selecting all calibrated sensors by total possible outcomes.

4

Apply the complementary probability principle to find

P(\text{at least one uncalibrated})

.

P(\text{at least one uncalibrated}) = 1 - P(\text{no uncalibrated}) = 1 - \frac{5}{28} = \frac{23}{28}

.

The sum of probabilities of complementary events is 1.

Anahtar Kavram

Complementary Probability: P(at least one X) = 1 - P(no X)

Soru 26Soru →

A museum curator is arranging $5$ distinct marble statues and $2$ distinct bronze statues in a single row along a gallery wall. If the $2$ bronze statues must not be placed next to each other, how many different linear arrangements of all $7$ statues are possible?

Cevabı ve açıklamayı göster

Cevap: 3600

Cevap

3,600

Using complementary counting, the total unrestricted arrangements of

7

distinct statues is

7! = 5,040

. The number of arrangements where the

2

bronze statues are placed together is determined by treating them as a single block:

6! \times 2! = 1,440

. Subtracting these forbidden arrangements from the total yields

5,040 - 1,440 = 3,600

valid linear arrangements.

Adım Adım Çözüm

1

Find total arrangements without restriction.

7! = 5,040

There are 7 distinct statues in total to arrange in a line.

2

Find arrangements where the 2 bronze statues are adjacent.

6! × 2! = 1,440

Grouping the 2 bronze statues into 1 block yields 6 items to order (6!), and the 2 bronze statues can swap positions inside the block (2!).

3

Apply complementary counting to find non-adjacent arrangements.

5,040 - 1,440 = 3,600

Subtracting the adjacent arrangements from total arrangements gives all valid arrangements.

Anahtar Kavram

Linear arrangements with non-adjacency restrictions using complementary counting.

Tahmini Süre:1m 30s

Soru 27Soru →

A 6-letter security code is created by arranging all the letters in the word $BANANA$ . How many distinct security codes can be formed such that the two letters $N$ are not adjacent to each other?

20

30

40

60

120

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

40 distinct security codes can be formed such that the two

N

's are not adjacent.

To find the number of arrangements where the two

N

's are not adjacent, use complementary counting. First, compute total unrestricted permutations of

BANANA

(1

B

, 3

A

's, 2

N

's), which gives

\frac{6!}{3! \times 2!} = 60

. Next, count the forbidden arrangements where the two

N

's are together by grouping them into a single block

(NN)

. Arranging

B

,

A

,

A

,

A

, and

(NN)

yields

\frac{5!}{3!} = 20

. Subtracting forbidden arrangements from total arrangements gives

60 - 20 = 40

.

Adım Adım Çözüm

1

Calculate the total number of distinct arrangements of the letters in

BANANA

without restrictions.

The word

BANANA

contains 6 letters in total: 1

B

, 3

A

's, and 2

N

's. The number of distinct permutations is

\frac{6!}{1! \times 3! \times 2!} = \frac{720}{1 \times 6 \times 2} = 60

.

Indistinguishable duplicate letters must be accounted for by dividing the total permutations by the factorial of each letter's count.

2

Calculate the number of arrangements where the two

N

's ARE adjacent.

Treat the pair

NN

as a single block. The items to arrange are

B

,

A

,

A

,

A

, and

(NN)

, which gives 5 items with 3 identical

A

's. The number of arrangements is

\frac{5!}{1! \times 3! \times 1!} = \frac{120}{6} = 20

.

Grouping restricted elements into a single composite element simplifies calculating forbidden configurations.

3

Apply complementary counting to find the number of valid arrangements where the two

N

's are not adjacent.

Total valid arrangements = Total arrangements - Adjacent arrangements =

60 - 20 = 40

.

Complementary counting is the most efficient method when calculating non-adjacent placements.

Anahtar Kavram

Counting with Restrictions and Repetitions (Complementary Counting Principle)

Tahmini Süre:1m 30s

Soru 28Soru →

How many distinct 6-digit positive integers can be formed using all of the digits $1, 1, 2, 2, 3,$ and $4$ such that no two identical digits are adjacent to each other?

36

60

72

84

96

Cevabı ve açıklamayı göster

Cevap: 84

Cevap

84

To find the number of 6-digit integers where no two identical digits are adjacent, apply complementary counting with the Principle of Inclusion-Exclusion. First, calculate the total distinct arrangements of the digits

1, 1, 2, 2, 3, 4

, which is

\frac{6!}{2!2!} = 180

. Next, count the forbidden outcomes. Treating the two 1s as a single block gives

\frac{5!}{2!} = 60

arrangements with adjacent 1s. Similarly, treating the two 2s as a single block gives

\frac{5!}{2!} = 60

arrangements with adjacent 2s. Arrangements where both the 1s are adjacent and the 2s are adjacent use two single blocks along with digits 3 and 4, giving

4! = 24

arrangements. By Inclusion-Exclusion, the total number of forbidden arrangements is

60 + 60 - 24 = 96

. Subtracting these from the total unrestricted arrangements gives

180 - 96 = 84

.

Adım Adım Çözüm

1

Calculate the total number of unrestricted distinct 6-digit arrangements of the multiset {1, 1, 2, 2, 3, 4}.

Total arrangements =

\frac{6!}{2! \times 2!} = \frac{720}{4} = 180

.

Dividing by

2! \times 2!

accounts for the indistinguishable repeated digits 1 and 2.

2

Define the forbidden events and calculate their individual counts.

Let

A

be the set of arrangements with adjacent 1s, and

B

be the set of arrangements with adjacent 2s.
For

|A|

, treat

(11)

as a single block: elements are

\{(11), 2, 2, 3, 4\}

, so

|A| = \frac{5!}{2!} = 60

.
For

|B|

, treat

(22)

as a single block: elements are

\{1, 1, (22), 3, 4\}

, so

|B| = \frac{5!}{2!} = 60

.

Grouping adjacent identical numbers into a single block simplifies counting restriction violations.

3

Calculate the intersection count

|A \cap B|

where both pairs are adjacent.

Treating both

(11)

and

(22)

as single blocks yields elements

\{(11), (22), 3, 4\}

, so

|A \cap B| = 4! = 24

.

Arrangements where both pairs of identical digits are adjacent are counted in both

A

and

B

and must be identified.

4

Apply the Principle of Inclusion-Exclusion to find the total forbidden arrangements and subtract from the total unrestricted arrangements.

|A \cup B| = |A| + |B| - |A \cap B| = 60 + 60 - 24 = 96

.
Valid arrangements =

180 - 96 = 84

.

Complementary counting yields the exact number of arrangements where no two identical digits are adjacent.

Anahtar Kavram

Permutations of Multisets with Restrictions via Inclusion-Exclusion

Soru 29Soru →

A 7-digit security passcode is to be formed using the digits $1, 1, 2, 2, 3, 4,$ and $5$ . How many distinct 7-digit passcodes can be formed such that neither the two $1$ s nor the two $2$ s are adjacent to each other?

540

660

780

900

1,020

Cevabı ve açıklamayı göster

Cevap: 660

Cevap

660 distinct passcodes can be formed.

To find the number of arrangements where neither identical pair is adjacent, we use complementary counting. First, compute total unrestricted arrangements of the digits

1,1,2,2,3,4,5

, which is

\frac{7!}{2!2!} = 1,260

. Next, by treating

(11)

as a single block, there are

\frac{6!}{2!} = 360

arrangements with adjacent

1

s. Similarly, treating

(22)

as a single block yields

\frac{6!}{2!} = 360

arrangements with adjacent

2

s. Arrangements where both pairs are adjacent treat

(11)

and

(22)

as two single blocks alongside

3,4,5

, giving

5! = 120

arrangements. Applying Inclusion-Exclusion gives

360 + 360 - 120 = 600

forbidden arrangements. Subtracting from the total gives

1,260 - 600 = 660

.

Adım Adım Çözüm

1

Calculate the total number of distinct arrangements without any restrictions.

Total arrangements =

\frac{7!}{2! \times 2!} = \frac{5,040}{4} = 1,260

.

There are 7 digits total, with the digit 1 repeated twice and the digit 2 repeated twice.

2

Calculate the number of arrangements where the two

1

s are adjacent.

N(\text{adjacent } 1\text{s}) = \frac{6!}{2!} = \frac{720}{2} = 360

.

Treat the pair

(11)

as a single unit, leaving 6 items:

(11), 2, 2, 3, 4, 5

with two identical

2

s.

3

Calculate the number of arrangements where the two

2

s are adjacent.

N(\text{adjacent } 2\text{s}) = \frac{6!}{2!} = \frac{720}{2} = 360

.

Treat the pair

(22)

as a single unit, leaving 6 items:

1, 1, (22), 3, 4, 5

with two identical

1

s.

4

Calculate the number of arrangements where both the two

1

s and two

2

s are simultaneously adjacent.

N(\text{both adjacent}) = 5! = 120

.

Treat

(11)

and

(22)

each as single distinct units, giving 5 distinct items:

(11), (22), 3, 4, 5

.

5

Apply the Principle of Inclusion-Exclusion to find the total restricted count.

Valid passcodes =

1,260 - (360 + 360 - 120) = 1,260 - 600 = 660

.

Subtract the union of forbidden cases from the total arrangements:

N(\text{total}) - [N(\text{1s adj}) + N(\text{2s adj}) - N(\text{both adj})]

.

Anahtar Kavram

Complementary Counting with Inclusion-Exclusion Principle on Permutations of Repeated Elements

Tahmini Süre:2m 0s

Soru 30Soru →

In a technology firm of $100$ software engineers, $65$ engineers know Python and $45$ know Java. If $20$ engineers know both Python and Java, how many of the engineers know neither Python nor Java?

10

0

15

20

30

Cevabı ve açıklamayı göster

Cevap: $10$

Cevap

The number of engineers who know neither Python nor Java is

10

.

The total number of engineers is

100

. By inclusion-exclusion, the union of Python and Java speakers is

65 + 45 - 20 = 90

. The number of engineers who know neither language is the total minus the union,

100 - 90 = 10

.

Adım Adım Çözüm

1

Calculate the number of engineers who know at least one language using the Inclusion-Exclusion Principle.

\text{At least one} = \text{Python} + \text{Java} - \text{Both} = 65 + 45 - 20 = 90

Engineers who know both languages are counted in both individual totals, so their count must be subtracted once to avoid double counting.

2

Subtract the number of engineers who know at least one language from the total number of engineers to find those who know neither.

\text{Neither} = \text{Total} - \text{At least one} = 100 - 90 = 10

The total group consists of those who know at least one language plus those who know neither.

Anahtar Kavram

Two-Set Inclusion-Exclusion Principle

Tahmini Süre:45s

Soru 31Soru →

An executive board must arrange 7 books side-by-side on a display shelf: 3 identical copies of a Finance textbook, 2 identical copies of an Economics textbook, and 2 distinct novels (Novel A and Novel B). How many distinct linear arrangements of these 7 books are possible such that Novel A and Novel B are not placed adjacent to each other?

Cevabı ve açıklamayı göster

Cevap: 300

Cevap

300

To find the number of distinct arrangements where Novel A and Novel B are not adjacent, use complementary counting. First, compute the total distinct arrangements of the 7 books with identical items: 7! / (3! * 2!) = 420. Next, find the number of arrangements where Novel A and Novel B ARE adjacent by bundling them as one unit [AB], which has 2 internal orderings (AB or BA). Arranging the 6 units ([AB], 3 Finance, 2 Economics) gives 6! / (3! * 2!) = 60 ways. Multiplying by 2 internal orderings gives 120 adjacent arrangements. Subtracting these from the total yields 420 - 120 = 300.

Adım Adım Çözüm

1

Calculate the unrestricted total number of distinct permutations of the 7 books.

420 distinct arrangements

There are 7 total items with 3 identical Finance books and 2 identical Economics books. Using the formula N! / (n1! * n2! ...), we get 7! / (3! * 2!) = 5,040 / 12 = 420.

2

Calculate the number of forbidden arrangements where Novel A and Novel B are adjacent.

120 forbidden arrangements

Treat Novel A and Novel B as a single composite item [AB]. There are 2! = 2 ways to order A and B inside this block (AB or BA). Then, arrange the 6 items ([AB] and the 5 textbooks with repetitions): 6! / (3! * 2!) = 60. Multiplying by 2 internal orderings gives 2 * 60 = 120.

3

Subtract the forbidden (adjacent) arrangements from the total unrestricted arrangements.

300 valid arrangements

By complementary counting, Valid = Total - Forbidden = 420 - 120 = 300.

Anahtar Kavram

Counting permutations with indistinguishable items and non-adjacency restrictions using complementary counting

Tahmini Süre:2m 0s

Soru 32Soru →

How many distinct 8-digit positive integers can be formed by rearranging all of the digits $1, 1, 1, 2, 2, 3, 3,$ and $3$ such that the resulting integer is even and no two $1$ s are adjacent?

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

40

To ensure the 8-digit number is even, the units (8th) digit must be 2. This leaves 7 positions to fill with three 1s, one 2, and three 3s. By arranging the 4 non-1 digits ({2, 3, 3, 3}) first, there are 4! / (1! 3!) = 4 distinct arrangements. Placing 4 digits creates 5 distinct gaps where the 1s can be placed without being adjacent. Choosing 3 gaps out of 5 for the three identical 1s gives C(5, 3) = 10 ways. Thus, the total number of valid integers is 4 * 10 = 40.

Adım Adım Çözüm

1

Fix the last digit to satisfy the even integer restriction

The 8th digit is fixed as 2, leaving 7 positions to fill with the remaining digits {1, 1, 1, 2, 3, 3, 3}.

An integer is even if and only if its units digit is even. The digit 2 is the only even digit in the set.

2

Calculate the arrangements of the non-restricted digits

4! / (1! * 3!) = 4 distinct arrangements.

Arranging the four non-1 digits ({2, 3, 3, 3}) first sets up the framework for placing the restricted 1s.

3

Place the three identical 1s into the gaps using the combination formula

C(5, 3) = 10 distinct gap selections.

The 4 arranged digits create 5 gaps. Selecting 3 distinct gaps ensures no two 1s are adjacent.

4

Multiply the arrangements of non-1 digits by the number of gap choices

4 * 10 = 40.

By the Fundamental Counting Principle, each non-1 arrangement can be combined with any valid gap placement.

Anahtar Kavram

Permutations with Indistinguishable Objects and Non-Adjacency Restrictions (Gap Method)

Tahmini Süre:2m 0s

Soru 33Soru →

A bookshelf is to be arranged using 3 identical Mathematics books, 2 identical Physics books, and 1 Chemistry book. In how many distinct ways can all 6 books be arranged in a single row such that the Chemistry book is not adjacent to any Mathematics book?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

12

The single Chemistry book must not touch any of the 3 Mathematics books. This leaves three mutually exclusive possibilities for the Chemistry book: at the left end preceded by a Physics book (4 ways), at the right end followed by a Physics book (4 ways), or sandwiched between both Physics books as a single block (4 ways). Summing these gives 12 total valid arrangements.

Adım Adım Çözüm

1

Analyze the placement restriction on the Chemistry book

The Chemistry book (C) cannot be placed next to any Mathematics book (M). Thus, C can only be adjacent to Physics books (P) or placed at the boundary of the row touching a P.

Eliminating adjacency to M restricts C to touch only P or the ends of the shelf.

2

Count valid arrangements when C is at the far left (position 1)

The row begins with C-P. The remaining 4 spots must be filled with 3 identical M's and 1 P, yielding 4! / (3! 1!) = 4 distinct ways.

Position 1 has only one neighbor (position 2), which must be P.

3

Count valid arrangements when C is at the far right (position 6)

The row ends with P-C. The remaining 4 spots must be filled with 3 identical M's and 1 P, yielding 4! / (3! 1!) = 4 distinct ways.

Position 6 has only one neighbor (position 5), which must be P.

4

Count valid arrangements when C is in an interior position (positions 2 through 5)

C must be sandwiched between two P's, forming the block (P-C-P). Arranging this single block along with the 3 identical M's (total of 4 items) yields 4! / (3! 1!) = 4 distinct ways.

Any interior placement requires both adjacent neighbors of C to be P.

5

Sum the counts from all mutually exclusive cases

Total valid arrangements = 4 + 4 + 4 = 12.

The three cases cover all possible non-overlapping valid placements for C.

Anahtar Kavram

Counting permutations with identical elements and positional restrictions using case analysis and block formation.

Tahmini Süre:1m 30s

Soru 34Soru →

A conference schedule consists of $6$ consecutive time slots. The organizers must schedule $3$ identical workshops on Artificial Intelligence, $2$ identical workshops on Cybersecurity, and $1$ keynote address on Data Privacy. If the $2$ Cybersecurity workshops cannot be scheduled in consecutive time slots, how many distinct presentation schedules are possible?

20

40

50

60

120

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

40 distinct presentation schedules are possible.

The total number of unrestricted ways to arrange the 6 events (3 identical AI, 2 identical Cybersecurity, 1 Data Privacy) is calculated using multiset permutations as

\frac{6!}{3! \times 2! \times 1!} = 60

. To find the number of ways where the two Cybersecurity workshops are NOT consecutive, we use complementary counting. By treating the two Cybersecurity workshops as one glued block, we arrange 5 items (3 identical AI, 1 Cybersecurity block, 1 Data Privacy), yielding

\frac{5!}{3! \times 1! \times 1!} = 20

forbidden arrangements. Subtracting the 20 forbidden arrangements from the 60 total arrangements gives 40 valid presentation schedules.

Adım Adım Çözüm

1

Calculate the total number of distinct schedules without restrictions.

Total arrangements =

\frac{6!}{3! \times 2! \times 1!} = \frac{720}{6 \times 2 \times 1} = 60

.

There are

6

total slots with

3

identical AI workshops,

2

identical Cybersecurity workshops, and

1

Data Privacy keynote.

2

Calculate the number of forbidden schedules where the

2

Cybersecurity workshops are in consecutive slots.

Forbidden arrangements =

\frac{5!}{3! \times 1! \times 1!} = \frac{120}{6} = 20

.

Treat the

2

identical Cybersecurity workshops as a single combined block. This leaves

5

items to arrange (

3

AI,

1

combined Cybersecurity block,

1

Data Privacy).

3

Subtract the forbidden arrangements from the total arrangements using complementary counting.

Valid schedules =

60 - 20 = 40

.

The number of valid restricted arrangements is total arrangements minus restricted consecutive arrangements.

Anahtar Kavram

Counting with Restrictions and Repetitions (Complementary Counting and Permutations of Multisets)

Tahmini Süre:2m 0s

Soru 35Soru →

A company creates 5-digit employee identification codes using all of the digits $1, 1, 2, 2,$ and $3$ . How many distinct 5-digit identification codes can be formed such that the two $2$ 's are not adjacent to each other?

12

18

24

30

60

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

18

To find the number of distinct 5-digit identification codes where the two 2's are not adjacent, use complementary counting. First, compute the total distinct arrangements of the digits

1, 1, 2, 2, 3

, which is

\frac{5!}{2! \times 2!} = 30

. Next, find the number of arrangements where the two 2's are adjacent by treating

(22)

as a single item. Arranging

(22), 1, 1, 3

yields

\frac{4!}{2!} = 12

arrangements. Subtracting the adjacent arrangements from the total gives

30 - 12 = 18

.

Adım Adım Çözüm

1

Calculate the total number of distinct 5-digit arrangements of the digits

1, 1, 2, 2, 3

without any restrictions.

Total arrangements =

\frac{5!}{2! \times 2! \times 1!} = \frac{120}{4} = 30

.

When arranging elements with repeated indistinguishable items, divide

n!

by the factorials of the counts of each repeated item.

2

Calculate the number of restricted (forbidden) arrangements where the two

2

's are adjacent.

Adjacent arrangements =

\frac{4!}{2! \times 1!} = \frac{24}{2} = 12

.

Treat the two adjacent

2

's as a single block

(22)

. We now arrange 4 items:

(22), 1, 1, 3

, where the digit

1

appears twice.

3

Subtract the forbidden arrangements from the total arrangements using complementary counting.

Non-adjacent arrangements =

30 - 12 = 18

.

Complementary counting gives the number of valid arrangements where the two

2

's are not adjacent.

Anahtar Kavram

Permutations with Repeated Indistinguishable Elements and Complementary Counting

Tahmini Süre:1m 30s

Soru 36Soru →

A box contains 20 cards numbered sequentially from 1 through 20, inclusive. If one card is drawn at random from the box, what is the probability that the number on the drawn card is a prime number? Express your answer as a decimal.

Cevabı ve açıklamayı göster

Cevap: 0.4

Cevap

0.4

The total number of possible outcomes when selecting one card from 20 is 20. The prime numbers between 1 and 20 inclusive are 2, 3, 5, 7, 11, 13, 17, and 19, giving 8 favorable outcomes (remembering that 1 is not prime). The single-event probability is calculated by dividing the number of favorable outcomes by the total number of outcomes, yielding

8 / 20 = 0.4

.

Adım Adım Çözüm

1

Determine the total number of possible outcomes in the sample space.

The sample space consists of 20 equally likely outcomes (integers 1 through 20).

Calculating single-event probability requires establishing the size of the total outcome space

N

.

2

Count the number of prime numbers in the set

\{1, 2, \dots, 20\}

.

There are 8 prime numbers:

\{2, 3, 5, 7, 11, 13, 17, 19\}

.

By definition, a prime number is an integer greater than 1 with exactly two positive divisors: 1 and itself. Thus, 1 is excluded.

3

Compute the probability using

P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}

.

P = \frac{8}{20} = 0.4

.

Directly apply the basic single-event probability formula.

Anahtar Kavram

Basic Single-Event Probability and Prime Number Identification

Tahmini Süre:45s

Soru 37Soru →

At a financial analytics firm, a group of 200 analysts were evaluated on their proficiency in three software tools: Options Analytics, Futures Trader, and Swaps Pricing. Exactly 15% of the analysts had no proficiency in any of the three tools. Among the remaining analysts, 110 were proficient in Options Analytics, 95 were proficient in Futures Trader, and 85 were proficient in Swaps Pricing. If exactly 20 analysts were proficient in all three tools, how many analysts were proficient in exactly one of the three tools?

10

30

50

70

100

Cevabı ve açıklamayı göster

Cevap: 70

Cevap

70 analysts were proficient in exactly one of the three tools.

The total number of analysts is 200, and 15% (30 analysts) have no proficiency, leaving 170 analysts proficient in at least one tool. Applying the 3-set inclusion-exclusion formula

N(A \cup B \cup C) = N(A) + N(B) + N(C) - S_2 + N(A \cap B \cap C)

, we get

170 = 110 + 95 + 85 - S_2 + 20

, which yields

S_2 = 140

. Since

S_2

counts elements in exactly two sets once and elements in all three sets three times, the number of analysts proficient in exactly two tools is

140 - 3(20) = 80

. Finally, subtracting those proficient in exactly two tools (80) and all three tools (20) from the total proficient in at least one tool (170) gives

170 - 80 - 20 = 70

.

Adım Adım Çözüm

1

Calculate the total number of analysts proficient in at least one tool

\text{At least one} = 200 - (0.15 \times 200) = 200 - 30 = 170

Analysts who are not proficient in any tool must be excluded from the total group size to find the union of the three sets.

2

Apply the 3-set inclusion-exclusion formula to find the sum of pairwise intersections

170 = 110 + 95 + 85 - S_2 + 20 \implies 170 = 310 - S_2 \implies S_2 = 140

, where

S_2 = N(\text{Options} \cap \text{Futures}) + N(\text{Futures} \cap \text{Swaps}) + N(\text{Options} \cap \text{Swaps})

The standard inclusion-exclusion principle states that

N(A \cup B \cup C) = N(A) + N(B) + N(C) - S_2 + N(A \cap B \cap C)

.

3

Determine the number of analysts proficient in exactly two tools

\text{Exactly 2} = S_2 - 3 \times N(\text{All 3}) = 140 - 3(20) = 140 - 60 = 80

Each member of the triple intersection is counted 3 times in

S_2

. Subtracting

3 \times N(\text{All 3})

isolates the elements belonging to exactly two sets.

4

Calculate the number of analysts proficient in exactly one tool

\text{Exactly 1} = N(\text{At least 1}) - \text{Exactly 2} - N(\text{All 3}) = 170 - 80 - 20 = 70

The union of the three sets consists of elements proficient in exactly 1 tool, exactly 2 tools, and all 3 tools.

Anahtar Kavram

Three-Set Inclusion-Exclusion Principle and Subset Decomposition

Tahmini Süre:2m 0s

Soru 38Soru →

In a game, a fair wheel lands on a single integer from $1$ to $100$ , inclusive, with each integer being equally likely to be selected. What is the probability that the integer selected is either a prime number or leaves a remainder of $1$ when divided by $5$ , but not both?

\frac{7}{20}

\frac{17}{50}

\frac{2}{5}

\frac{9}{20}

\frac{1}{20}

Cevabı ve açıklamayı göster

Cevap: $\frac{7}{20}$

Cevap

The correct answer is 7/20.

The total number of possible outcomes is 100. There are 25 prime numbers in the range 1 to 100, and 20 numbers that leave a remainder of 1 when divided by 5. Exactly 5 numbers (11, 31, 41, 61, 71) meet both conditions. To find the numbers that meet either condition but NOT both, subtract the 5 overlapping numbers from each category: 20 numbers are prime only, and 15 numbers leave a remainder of 1 only. The number of favorable outcomes is 20 + 15 = 35. Therefore, the single-event probability is 35/100 = 7/20.

Adım Adım Çözüm

1

Determine the total sample space size.

The total number of possible outcomes is

N = 100

.

Each integer from 1 to 100 inclusive is equally likely.

2

Identify and count all prime numbers in the set

\{1, 2, \dots, 100\}

.

There are 25 prime numbers:

\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}

.

Note that 1 is not a prime number.

3

Identify and count all integers in

\{1, 2, \dots, 100\}

that leave a remainder of 1 when divided by 5.

There are 20 such integers:

\{1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, 91, 96\}

.

These integers take the form

5k + 1

for integers

k \ge 0

where

1 \le 5k + 1 \le 100

.

4

Find the intersection of both sets (numbers that are prime AND leave a remainder of 1 when divided by 5).

The 5 numbers satisfying both conditions are

\{11, 31, 41, 61, 71\}

.

Any integer leaving a remainder of 1 when divided by 5 must end in 1 or 6. No prime ends in 6. The primes ending in 1 in this range are 11, 31, 41, 61, and 71.

5

Calculate the number of favorable outcomes satisfying 'either condition, but not both' (symmetric difference).

Favorable outcomes =

(25 - 5) + (20 - 5) = 20 + 15 = 35

.

We must count numbers that are only prime (20) plus numbers that are only remainder 1 (15), excluding the 5 overlapping numbers.

6

Calculate the single-event probability.

P = \frac{35}{100} = \frac{7}{20}

.

Divide favorable outcomes by total sample space size.

Anahtar Kavram

Basic single-event probability with set operations and number properties (prime numbers, remainders).

Tahmini Süre:2m 0s

Soru 39Soru →

At an international conference of $200$ delegates, every delegate speaks at least one of three languages: French, German, or Spanish. Exactly $115$ delegates speak French, $85$ speak German, and $67$ speak Spanish. Furthermore, exactly $15$ delegates speak all three languages. If the number of delegates who speak only one language is four times the number of delegates who speak exactly two languages, how many delegates speak exactly two of these three languages?

Cevabı ve açıklamayı göster

Cevap: 37

Cevap

The number of delegates who speak exactly two of the three languages is 37.

By dividing the 200 delegates into three disjoint groups—those speaking exactly 1 language (

E_1

), exactly 2 languages (

E_2

), and all 3 languages (

E_3 = 15

)—we establish that

E_1 + E_2 + 15 = 200

, giving

E_1 + E_2 = 185

. Given

E_1 = 4E_2

, substituting yields

5E_2 = 185

, which gives

E_2 = 37

.

Adım Adım Çözüm

1

Define exact-count variables for the overlapping sets

Let

E_1

be the number of delegates speaking exactly one language,

E_2

be the number speaking exactly two languages, and

E_3 = 15

be the number speaking all three languages.

Categorizing elements into mutually exclusive subsets (exactly 1, exactly 2, exactly 3) simplifies 3-set inclusion-exclusion problems.

2

Set up the total population equation

E_1 + E_2 + E_3 = 200 \implies E_1 + E_2 + 15 = 200 \implies E_1 + E_2 = 185

Since every delegate speaks at least one language, the sum of all mutually exclusive regions equals the total number of delegates.

3

Apply the given relationship between

E_1

and

E_2

Substitute

E_1 = 4E_2

into

E_1 + E_2 = 185

to get

4E_2 + E_2 = 185 \implies 5E_2 = 185

The problem establishes that the number of single-language speakers is four times the number of dual-language speakers.

4

Solve for the target quantity

E_2

E_2 = 37

Dividing 185 by 5 yields the exact count of delegates speaking exactly two languages.

Anahtar Kavram

Three-Set Inclusion-Exclusion Principle using Exact-Overlap Subsets

Soru 40Soru →

A security analyst needs to generate a 6-symbol access code using all of the following characters: $A, A, A, B, B,$ and $C$ . How many distinct 6-symbol access codes can be formed such that the two $B$ 's are not adjacent to each other?

20

24

40

60

120

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

40 distinct 6-symbol access codes can be formed.

First arrange the four symbols that have no adjacency restrictions (

A, A, A,

and

C

). The number of unique ways to arrange these 4 symbols with 3 identical

A

's is

\frac{4!}{3!} = 4

. Placing these 4 symbols in a row creates 5 available spaces (one before the first character, three between characters, and one after the last character). To ensure the two

B

's are not adjacent, choose 2 distinct spaces out of the 5 for the

B

's. The number of ways to select 2 spaces out of 5 is

\binom{5}{2} = 10

. Multiplying the arrangement of base characters by the space selections yields

4 \times 10 = 40

total valid codes.

Adım Adım Çözüm

1

Arrange the non-restricted characters

There are 4 ways to arrange the remaining characters

A, A, A,

and

C

.

The number of distinct arrangements of 4 characters where 3 are identical is

\frac{4!}{3! \times 1!} = 4

.

2

Determine the available slots for the restricted character

Arranging 4 characters creates 5 distinct slots (including the two ends): _ X _ X _ X _ X _.

To ensure no two

B

's are adjacent, each

B

must occupy a distinct slot around or between the arranged characters.

3

Calculate the combinations of slot positions

There are

\binom{5}{2} = 10

ways to choose 2 slots for the two identical

B

's.

Since the two

B

's are identical, order of selection within the slots does not matter.

4

Apply the Fundamental Counting Principle

4 \times 10 = 40

distinct codes.

Multiply the number of ways to arrange the base characters by the number of ways to place the

B

's into the available slots.

Anahtar Kavram

Counting with restrictions using the slotting method for non-adjacent elements

Counting and Probability

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm