Counting and Probability Alıştırma Soruları — GMAT — Sayfa 4

Soru 61Soru →

An integer $n$ is randomly selected from the set of all integers from $10$ to $59$ , inclusive. What is the probability that $n$ is a prime number whose units digit is $3$ ?

\frac{1}{10}

\frac{4}{49}

\frac{2}{25}

\frac{2}{23}

\frac{3}{50}

Cevabı ve açıklamayı göster

Cevap: $\frac{2}{25}$

Cevap

The probability that the selected integer is a prime number whose units digit is

3

is

\frac{2}{25}

.

The correct answer identifies that there are 50 total integers in the inclusive range from 10 to 59. Among the numbers ending in 3 (

13, 23, 33, 43, 53

), exactly four are prime (

13, 23, 43, 53

), while 33 is composite. Dividing 4 favorable outcomes by 50 total outcomes yields

\frac{4}{50}

, which simplifies to

\frac{2}{25}

.

Adım Adım Çözüm

1

Determine the total number of outcomes in the set.

The set contains

59 - 10 + 1 = 50

integers.

For an inclusive set of integers from

a

to

b

, the total number of integers is

b - a + 1

.

2

Identify all integers in the set with a units digit of

3

.

The candidate integers ending in 3 are

13, 23, 33, 43, 53

.

These are all two-digit numbers in the range

[10, 59]

ending with

3

.

3

Determine which candidate integers are prime numbers.

The prime numbers are

13, 23, 43,

and

53

(a total of 4 favorable outcomes). Note that

33 = 3 \times 11

, so it is composite.

A prime number is an integer greater than 1 that has no positive divisors other than 1 and itself.

4

Calculate the single-event probability and simplify the fraction.

\text{Probability} = \frac{4}{50} = \frac{2}{25}

.

Probability is defined as the number of favorable outcomes divided by the total number of possible outcomes.

Anahtar Kavram

Basic Single-Event Probability

Tahmini Süre:1m 30s

Soru 62Soru →

Eight executive team members—3 vice presidents, 3 directors, and 2 managers—are to stand in a single line for a company photograph. If no two vice presidents can stand next to each other, and the 2 managers must stand next to each other, how many different linear arrangements of the eight team members are possible?

Cevabı ve açıklamayı göster

Cevap: 2880

Cevap

2,880

To satisfy both constraints simultaneously, first treat the 2 managers as a single block, which has

2! = 2

internal arrangements. Next, arrange the non-vice-president units—consisting of the 3 distinct directors and the 1 manager block—giving 4 units total, which can be ordered in

4! = 24

ways. These 4 units create 5 distinct slot locations (at the ends and between adjacent units). To ensure no two vice presidents are adjacent, place the 3 distinct vice presidents into 3 of these 5 slots in

P(5, 3) = 60

ways. Multiplying these independent decisions yields

2 \times 24 \times 60 = 2,880

.

Adım Adım Çözüm

1

Group the elements that must remain adjacent and calculate internal permutations

The 2 managers form 1 block with

2! = 2

internal arrangements.

Since the 2 managers must stand next to each other, treating them as a single block ensures they are never separated.

2

Arrange all non-restricted base units in a line

The 3 directors and 1 manager block yield

4! = 24

linear arrangements.

Establishing the sequence of non-vice-president units creates the fixed slots into which the vice presidents will later be inserted.

3

Calculate the available slot arrangements for the separated elements

4 base units create 5 available slots. Permuting 3 vice presidents into 5 slots yields

P(5, 3) = 60

ways.

Placing at most one vice president per slot guarantees that no two vice presidents are placed adjacently.

4

Apply the Fundamental Counting Principle to determine total arrangements

Total arrangements =

2 \times 24 \times 60 = 2,880

.

The choices for internal block arrangement, base unit ordering, and slot placement are independent sequential events.

Anahtar Kavram

Linear Permutations with Simultaneous Grouping and Non-Adjacency Constraints

Soru 63Soru →

A financial firm's investment committee consists of 8 senior analysts and 4 junior analysts. If a project review panel of 3 members is chosen at random without replacement from this committee, what is the probability that the panel contains at least one junior analyst?

\frac{14}{55}

\frac{28}{55}

\frac{19}{27}

\frac{41}{55}

\frac{47}{55}

Cevabı ve açıklamayı göster

Cevap: $\frac{41}{55}$

Cevap

The probability that the panel contains at least one junior analyst is

\frac{41}{55}

.

To find the probability of selecting at least one junior analyst, it is most efficient to use the complementary probability rule:

P(\text{at least one junior}) = 1 - P(\text{no junior})

. The total number of ways to choose 3 panel members from 12 committee members is

\binom{12}{3} = 220

. The number of ways to choose 3 senior analysts from 8 is

\binom{8}{3} = 56

. Thus, the probability of selecting no junior analysts is

\frac{56}{220} = \frac{14}{55}

. Subtracting this from 1 gives

1 - \frac{14}{55} = \frac{41}{55}

.

Adım Adım Çözüm

1

Calculate total possible outcomes for choosing 3 members out of 12.

\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220

total combinations.

Selection order does not matter, so combinations

nCr

are used.

2

Calculate the number of unfavorable outcomes where zero junior analysts are chosen (all 3 selected are senior analysts).

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

ways.

All 3 members must be selected exclusively from the 8 senior analysts.

3

Find the probability of selecting zero junior analysts.

P(\text{no junior analysts}) = \frac{56}{220} = \frac{14}{55}

.

Divide the unfavorable outcomes by the total outcomes.

4

Apply the complementary probability principle:

P(\text{at least 1 junior}) = 1 - P(\text{no junior})

.

P(\text{at least 1 junior}) = 1 - \frac{14}{55} = \frac{41}{55}

.

The event 'at least one junior analyst' is the logical complement of 'no junior analysts'.

Anahtar Kavram

Complementary Probability

Tahmini Süre:2m 0s

Soru 64Soru →

A glass bowl contains 6 red candies and $N$ green candies, where $N$ is a positive integer. If two candies are selected at random one after another without replacement, the probability that both candies selected are red is $\frac{1}{3}$ . What is the value of $N$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4

The correct value of

N

is 4. Because the selection is performed without replacement, the probability of drawing the first red candy is

\frac{6}{N+6}

and the second red candy is

\frac{5}{N+5}

. Setting their product equal to

\frac{1}{3}

yields

\frac{30}{(N+6)(N+5)} = \frac{1}{3}

, which expands to

N^2 + 11N - 60 = 0

. Factoring gives

(N-4)(N+15) = 0

, yielding

N = 4

as the only positive solution.

Adım Adım Çözüm

1

Define total outcomes and event probabilities for dependent selection

Initial total candies =

N + 6

. Probability of first red =

\frac{6}{N+6}

. Probability of second red given first is red =

\frac{5}{N+5}

.

Since selection is without replacement, the total count and red count both decrease by 1 after the first draw, making the events dependent.

2

Set up the probability multiplication equation

P(\text{both red}) = \frac{6}{N+6} \times \frac{5}{N+5} = \frac{30}{(N+6)(N+5)} = \frac{1}{3}

.

The probability of joint dependent events

A

and

B

occurring is

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

.

3

Solve the equation for

N

(N+6)(N+5) = 90 \Rightarrow N^2 + 11N - 60 = 0 \Rightarrow (N-4)(N+15) = 0

.

Cross-multiplying gives a quadratic equation in terms of

N

.

4

Select the valid positive integer root

N = 4

.

The count of candies must be a positive number, so

N = -15

is discarded.

Anahtar Kavram

Probability of Dependent Events (Sampling without Replacement)

Soru 65Soru →

Seven distinct paintings—3 landscapes and 4 portraits, one of which is a portrait of the founder—are to be displayed side-by-side in a single row along a gallery wall. If all 3 landscapes must be displayed together as a contiguous block, and the portrait of the founder cannot be placed at either end of the 7-painting row, in how many different linear arrangements can the 7 paintings be displayed?

72

288

432

576

720

Cevabı ve açıklamayı göster

Cevap: 432

Cevap

432

The correct answer is 432. Treating the 3 landscape paintings as a single block leaves 5 items (the landscape block, the founder's portrait, and 3 other portraits) to arrange. There are 5! = 120 total arrangements of these 5 items. The founder's portrait occupies an end slot of the row if it is placed in either the 1st position (4! = 24 ways) or the 5th position (4! = 24 ways) among the 5 items. Subtracting these 48 invalid arrangements gives 120 - 48 = 72 valid block placements. Finally, accounting for the 3! = 6 internal arrangements of the landscapes inside their block yields 72 × 6 = 432 total linear arrangements.

Adım Adım Çözüm

1

Group the 3 landscape paintings into a single block unit.

We now have 5 items to arrange linearly: the 1 landscape block, the founder's portrait, and the 3 other portraits.

Grouping elements that must remain together simplifies the arrangement problem into smaller independent choices.

2

Calculate the total number of linear arrangements of these 5 items without restrictions on the founder's portrait.

5! = 120 arrangements.

5 distinct objects can be ordered in 5 factorial ways.

3

Determine the number of arrangements where the founder's portrait is placed at either end of the row.

2 × 4! = 48 arrangements.

The founder's portrait is at an end of the 7-painting row if and only if it is in position 1 (first item) or position 5 (last item) among the 5 items. Fixing it at position 1 leaves 4! = 24 ways for the remaining items, and fixing it at position 5 gives another 4! = 24 ways.

4

Subtract the invalid end-position arrangements from the total 5-item arrangements.

120 - 48 = 72 valid arrangements of the 5 items.

Using complementary counting isolates the cases where the founder's portrait is not at either end.

5

Multiply by the number of internal arrangements of the 3 landscape paintings within their block.

72 × 3! = 72 × 6 = 432.

The 3 distinct landscapes within the single block can be ordered internally in 3! = 6 ways for each overall arrangement.

Anahtar Kavram

Permutations with Block Constraints and Complementary Restriction Counting

Soru 66Soru →

If 3 components are randomly selected without replacement from a batch of 10 components containing exactly 2 defective components, the probability that at least one selected component is defective is equal to $\frac{8}{15}$ .

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True. The probability of selecting at least one defective component is indeed

\frac{8}{15}

.

The statement correctly computes the probability of selecting at least one defective component using

1 - P(\text{no defective components}) = 1 - \frac{\binom{8}{3}}{\binom{10}{3}} = 1 - \frac{7}{15} = \frac{8}{15}

.

Adım Adım Çözüm

1

Identify the complementary event

The complement of 'at least one defective component' is 'zero defective components' (all 3 selected components are non-defective).

Calculating

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

avoids calculating multiple dependent individual outcomes.

2

Calculate the total combinations for choosing 3 components from 10

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

.

This establishes the sample space size.

3

Calculate the combinations for choosing 3 non-defective components from 8

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

.

There are 8 non-defective components in the batch.

4

Compute

P(\text{0 defective})

and subtract from 1

P(\text{0 defective}) = \frac{56}{120} = \frac{7}{15}

. Therefore,

P(\text{at least 1 defective}) = 1 - \frac{7}{15} = \frac{8}{15}

.

Completing the complementary calculation confirms the statement is correct.

Anahtar Kavram

Complementary Probability and At-Least-One Scenarios without Replacement

Soru 67Soru →

A charity organization's steering board consists of $6$ donors and $4$ volunteers. Two board members are selected at random, one after another without replacement, to attend a national conference. What is the probability that the first member selected is a donor and the second member selected is a volunteer?

\frac{6}{25}

\frac{4}{15}

\frac{1}{3}

\frac{8}{15}

\frac{12}{25}

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{15}$

Cevap

The probability that the first member selected is a donor and the second member selected is a volunteer is

\frac{4}{15}

.

The correct probability is calculated by multiplying the probability of the first event by the conditional probability of the second event given that the first event occurred. The probability of choosing a donor first is

\frac{6}{10}

. Since the selection is made without replacement, there are

9

total members remaining for the second draw,

4

of whom are volunteers. Therefore, the probability of selecting a volunteer second is

\frac{4}{9}

. The overall probability is

\frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}

.

Adım Adım Çözüm

1

Calculate the probability of selecting a donor on the first choice.

Since there are

6

donors out of

10

total board members,

P(\text{1st Donor}) = \frac{6}{10} = \frac{3}{5}

.

The sample space initially contains

10

members,

6

of whom are donors.

2

Calculate the conditional probability of selecting a volunteer on the second choice given that a donor was chosen first.

After one donor is selected,

9

members remain,

4

of whom are volunteers. Thus,

P(\text{2nd Volunteer} \mid \text{1st Donor}) = \frac{4}{9}

.

The selection is made without replacement, reducing both the total number of members in the pool and the sample space size.

3

Multiply the sequential probabilities for dependent events.

P(\text{1st Donor and 2nd Volunteer}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}

.

For dependent events

A

and

B

, the joint probability is

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

.

Anahtar Kavram

Probability of Dependent Events (Sampling Without Replacement)

Soru 68Soru →

A security analyst is designing a 4-character access code consisting of a digit, followed by two letters, followed by another digit. The code must be constructed according to the following rules:

- The first character must be a prime digit chosen from the set of single-digit integers $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ .
- The second and third characters must be distinct uppercase letters selected from the set $\{A, B, C, D, E\}$ .
- The fourth character must be an odd digit chosen from the set of single-digit integers $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ , such that it is not equal to the first digit.

How many different 4-character access codes can be created following these rules?

300

340

400

420

500

Cevabı ve açıklamayı göster

Cevap: 340

Cevap

340 access codes

The correct count of 340 is obtained by separating the problem into two distinct cases based on whether the prime first digit is even or odd. For the even prime digit (2), there are 5 possible odd digits for the fourth slot, yielding 100 codes. For the three odd prime digits (3, 5, 7), the fourth slot has 4 available choices to avoid duplicating the first digit, yielding 240 codes. Summing both cases gives 340.

Adım Adım Çözüm

1

Identify the set of prime single-digit integers.

The prime digits among

0

through

9

are

2, 3, 5,

and

7

(a total of

4

options). Note that

2

is even, while

3, 5,

and

7

are odd.

1

is not prime by definition, and

0, 4, 6, 8, 9

are composite or neither.

2

Calculate the number of ways to choose the two distinct letter characters.

5 \times 4 = 20

possible two-letter arrangements.

There are

5

choices for the second character and

4

remaining choices for the third character since they must be distinct.

3

Evaluate Case 1: The first digit is the even prime digit (

2

).

1 \times 20 \times 5 = 100

codes.

There is

1

choice for the first digit (

2

). The odd digits available for the fourth character are

1, 3, 5, 7, 9

(

5

choices). Since

2

is even, it never matches any odd digit.

4

Evaluate Case 2: The first digit is an odd prime digit (

3, 5,

or

7

).

3 \times 20 \times 4 = 240

codes.

There are

3

choices for the first digit. The odd digits available for the fourth character are

1, 3, 5, 7, 9

(

5

total), but the fourth digit cannot equal the first digit, leaving

5 - 1 = 4

choices.

5

Sum the total codes across both mutually exclusive cases.

100 + 240 = 340

total codes.

By the Fundamental Counting Principle and Addition Principle for disjoint sets.

Anahtar Kavram

Fundamental Counting Principle with Conditional Restrictions and Case Analysis

Tahmini Süre:2m 0s

Soru 69Soru →

An airline creates a $5$ -character identification code for each of its cargo routes. The characters in the code must satisfy the following sequential rules from left to right:

1. The first character must be a non-zero even digit.
2. The second and third characters must be distinct uppercase vowels chosen from the set $\{A, E, I, O, U\}$ .
3. The fourth character must be a single-digit prime number.
4. The fifth character must be a single digit ( $0$ through $9$ ) such that the sum of the first digit and the fifth digit is odd.

How many such unique identification codes can the airline create?

Cevabı ve açıklamayı göster

Cevap: 1600

Cevap

1,600 unique identification codes

To find the total number of unique identification codes, we evaluate the number of available options for each character position and multiply them together using the Fundamental Counting Principle:

1. First position (non-zero even digit): The even digits from 0 to 9 are

0, 2, 4, 6, 8

. Excluding 0 leaves

4

options (

2, 4, 6, 8

).
2. Second and third positions (distinct vowels): There are

5

vowels (

A, E, I, O, U

). The second position has

5

options, and the third position has

4

remaining options, yielding

5 \times 4 = 20

combinations.
3. Fourth position (prime digit): The single-digit primes are

2, 3, 5, 7

, giving

4

options.
4. Fifth position (parity condition): The sum of the first digit and fifth digit must be odd. Since the first digit is an even number, the fifth digit must be an odd number to make the sum odd. The single-digit odd numbers are

1, 3, 5, 7, 9

, giving

5

options.

Multiplying the choices for all positions gives:

4 \times 20 \times 4 \times 5 = 1,600

Adım Adım Çözüm

1

Determine the number of valid choices for the 1st position (non-zero even digit).

4 choices (

2, 4, 6, 8

).

The digit 0 is excluded because the first character must be non-zero.

2

Determine the number of valid choices for the 2nd and 3rd positions (distinct vowels).

5 \times 4 = 20

choices.

There are 5 vowels available for the 2nd position, and 4 remaining vowels for the 3rd position since they must be distinct.

3

Determine the number of valid choices for the 4th position (prime digit).

4 choices (

2, 3, 5, 7

).

The single-digit prime numbers are 2, 3, 5, and 7.

4

Determine the number of valid choices for the 5th position (parity constraint).

5 choices (

1, 3, 5, 7, 9

).

The sum of an even integer and another integer is odd if and only if the second integer is odd. Since the 1st digit is guaranteed to be even, the 5th digit must be odd.

5

Apply the Fundamental Counting Principle across all 5 sequential stages.

4 \times 20 \times 4 \times 5 = 1,600

unique codes.

Each stage represents an independent decision stage with a fixed number of valid choices.

Anahtar Kavram

Fundamental Counting Principle with Restricted Positions and Parity Rules

Tahmini Süre:2m 0s

Soru 70Soru →

A tech company generates a 5-digit security badge number, represented as $d_1 d_2 d_3 d_4 d_5$ , using digits from $0$ to $9$ , inclusive. The badge numbers must satisfy the following constraints:
- The first digit, $d_1$ , cannot be $0$ or $1$ .
- The second and third digits, $d_2$ and $d_3$ , must be distinct even digits.
- The fourth digit, $d_4$ , must be an odd digit strictly greater than $3$ .
- The fifth digit, $d_5$ , can be any digit except that it cannot be equal to $d_1$ .

How many different 5-digit badge numbers can be created under these rules?

4,320

4,860

5,400

5,760

6,000

Cevabı ve açıklamayı göster

Cevap: 4,320

Cevap

4,320

According to the Fundamental Counting Principle, the total number of configurations is the product of the number of options available at each stage. For the first position, excluding 0 and 1 leaves 8 possible digits. For the second position, any of the 5 even digits can be chosen. For the third position, one of the remaining 4 even digits must be selected to preserve distinctness. For the fourth position, the odd digits strictly greater than 3 are 5, 7, and 9, providing 3 options. Finally, for the fifth position, 9 digits remain available after excluding the specific digit chosen for the first position. Multiplying these independent counts gives 8 × 5 × 4 × 3 × 9 = 4,320.

Adım Adım Çözüm

1

Determine the number of available choices for the first digit,

d_1

.

8 choices

The available digits are

0

through

9

(10 total). Excluding

0

and

1

leaves

\{2, 3, 4, 5, 6, 7, 8, 9\}

, which gives

10 - 2 = 8

choices.

2

Determine the number of available choices for the second and third digits,

d_2

and

d_3

.

20 choices total for

(d_2, d_3)

The set of even digits is

\{0, 2, 4, 6, 8\}

, containing 5 digits. There are 5 choices for

d_2

. Since

d_3

must be distinct from

d_2

, there are

5 - 1 = 4

choices for

d_3

. Thus, there are

5 \times 4 = 20

choices.

3

Determine the number of available choices for the fourth digit,

d_4

.

3 choices

The set of odd digits is

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

. Odd digits strictly greater than

3

are

\{5, 7, 9\}

, which gives 3 choices.

4

Determine the number of available choices for the fifth digit,

d_5

.

9 choices

d_5

can be any digit from

0

through

9

except

d_1

. Out of 10 total digits, excluding 1 digit leaves

10 - 1 = 9

choices.

5

Apply the Fundamental Counting Principle to compute total combinations.

4,320

Multiply the number of independent choices across all 5 stages:

8 \times 5 \times 4 \times 3 \times 9 = 4,320

.

Anahtar Kavram

Fundamental Counting Principle with Multi-Stage Positional Restrictions

Soru 71Soru →

A medical research laboratory creates a 4-digit identification code, $d_1 d_2 d_3 d_4$ , for sample trays using digits selected from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ . The digits in each code must satisfy the following conditions:
- The first digit, $d_1$ , must be an odd prime number.
- The second digit, $d_2$ , must be an even integer.
- All four digits in the code must be distinct.

How many different identification codes can be created under these rules?

18

360

480

504

768

Cevabı ve açıklamayı göster

Cevap: 360

Cevap

360

To form the 4-digit code, we calculate the number of options at each sequential stage. First,

d_1

must be an odd prime from

\{1, 2, 3, 4, 5, 6, 7, 8\}

, which gives

3

options (

3, 5, 7

). Second,

d_2

must be an even integer from the set, giving

4

options (

2, 4, 6, 8

). Third, because all digits must be distinct, picking

d_1

and

d_2

uses

2

digits, leaving

8 - 2 = 6

choices for

d_3

and

5

choices for

d_4

. By the Fundamental Counting Principle, the total number of codes is

3 \times 4 \times 6 \times 5 = 360

.

Adım Adım Çözüm

1

Determine the number of valid choices for the first digit (

d_1

)

3 choices (

3, 5, 7

)

The odd prime numbers in the given set

\{1, 2, 3, 4, 5, 6, 7, 8\}

are

3, 5,

and

7

. Note that

1

is not prime and

2

is an even prime.

2

Determine the number of valid choices for the second digit (

d_2

)

4 choices (

2, 4, 6, 8

)

The even integers in the given set are

2, 4, 6,

and

8

. Notice that the set of choices for

d_1

and

d_2

are disjoint.

3

Determine the number of valid choices for the third digit (

d_3

) and fourth digit (

d_4

)

6 choices for

d_3

and 5 choices for

d_4

Since all 4 digits must be distinct, 2 distinct digits have already been used for

d_1

and

d_2

. From the original set of 8 digits,

8 - 2 = 6

digits remain for

d_3

, and then

5

digits remain for

d_4

.

4

Apply the Fundamental Counting Principle to calculate the total number of codes

3 \times 4 \times 6 \times 5 = 360

The overall number of distinct multi-stage decisions is found by multiplying the number of choices available at each stage.

Anahtar Kavram

Fundamental Counting Principle with Restricted and Non-Repeating Selections

Tahmini Süre:2m 0s

Soru 72Soru →

A corporate compliance archive contains $10$ confidential files and $5$ public files. An auditor randomly selects two files from the archive, one after another without replacement. What is the probability that at least one of the selected files is public?

\frac{2}{21}

\frac{10}{21}

\frac{11}{21}

\frac{5}{9}

\frac{4}{7}

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{7}$

Cevap

The probability that at least one of the selected files is public is

\frac{4}{7}

.

To find the probability that at least one file is public, subtract the probability that no public files are selected (meaning both files are confidential) from 1. The probability of choosing a confidential file on the first draw is

\frac{10}{15} = \frac{2}{3}

. Since the selection is without replacement,

9

confidential files remain out of

14

total files. The probability of choosing a confidential file on the second draw is

\frac{9}{14}

. The joint probability of drawing two confidential files is

\frac{2}{3} \times \frac{9}{14} = \frac{3}{7}

. Subtracting this from 1 yields

1 - \frac{3}{7} = \frac{4}{7}

.

Adım Adım Çözüm

1

Identify the complementary event

The complementary event to 'at least one selected file is public' is 'both selected files are confidential'.

Calculating the probability of the complement is simpler than summing the probabilities of all positive individual outcomes.

2

Calculate the probability of drawing two confidential files sequentially without replacement

The probability of drawing a confidential file on the first selection is

\frac{10}{15} = \frac{2}{3}

. After removing one confidential file, 9 confidential files remain out of 14 total files. The probability on the second selection is

\frac{9}{14}

. Thus,

P(\text{both confidential}) = \frac{10}{15} \times \frac{9}{14} = \frac{2}{3} \times \frac{9}{14} = \frac{3}{7}

.

Because the drawing is done without replacement, the outcome of the first draw affects the total sample space and event space for the second draw.

3

Subtract the complementary probability from 1

P(\text{at least one public}) = 1 - P(\text{both confidential}) = 1 - \frac{3}{7} = \frac{4}{7}

.

The sum of the probabilities of an event and its complement equals 1.

Anahtar Kavram

Probability of Dependent Events and Complementary Probability

Tahmini Süre:1m 30s

Soru 73Soru →

Six distinct letters— $A, B, C, D, E,$ and $F$ —are to be arranged in a single line. How many different linear arrangements are possible such that letter $A$ appears somewhere to the left of letter $B$ , and letters $C$ and $D$ are not adjacent to each other?

Cevabı ve açıklamayı göster

Cevap: 240

Cevap

240

To find the number of valid linear arrangements, apply symmetry and the complement rule. First, in half of all

6! = 720

arrangements (360 arrangements), letter A appears to the left of letter B. Next, find the number of arrangements where A is to the left of B AND letters C and D are adjacent. Treating C and D as one combined block gives

5! \times 2! = 240

total arrangements where C and D are adjacent. By symmetry, letter A appears to the left of letter B in half of these cases (

\frac{240}{2} = 120

). Subtracting these 120 restricted arrangements from the 360 total arrangements where A precedes B gives

360 - 120 = 240

.

Adım Adım Çözüm

1

Determine the number of linear arrangements in which letter A appears somewhere to the left of letter B.

360 arrangements

For 6 distinct letters, there are

6! = 720

total linear arrangements. By symmetry, letter A appears before letter B in exactly half of all arrangements:

\frac{720}{2} = 360

.

2

Determine the number of arrangements where letter A is to the left of letter B AND letters C and D are adjacent.

120 arrangements

Treating C and D as a single block yields 5 items to arrange, which can be done in

5! = 120

ways. The block itself has

2! = 2

internal orderings, giving

120 \times 2 = 240

arrangements where C and D are adjacent. By symmetry, letter A appears before letter B in half of these arrangements:

\frac{240}{2} = 120

.

3

Subtract the arrangements where C and D are adjacent from the total arrangements where A is to the left of B.

240 arrangements

The number of arrangements where letter A is to the left of letter B and letters C and D are not adjacent is

360 - 120 = 240

.

Anahtar Kavram

Permutations with Relative Position and Non-Adjacency Restrictions

Soru 74Soru →

A security system generates unique 7-digit access codes using each of the digits $1, 2, 3, 4, 5, 6,$ and $7$ exactly once. How many such codes can be formed in which all odd digits appear in strictly ascending order from left to right and the digit $2$ appears somewhere to the left of the digit $4$ ?

Cevabı ve açıklamayı göster

Cevap: 105

Cevap

The total number of valid 7-digit security access codes is 105.

Out of the total

7! = 5,040

unrestricted linear arrangements of the 7 distinct digits, the 4 odd digits can be ordered in

4! = 24

ways, but only 1 of these orderings satisfies the strictly ascending condition. Furthermore, the digits 2 and 4 can be ordered in

2! = 2

ways, with digit 2 appearing to the left of digit 4 in exactly 1 of those orderings. Therefore, the number of valid arrangements is given by

\frac{7!}{4! \times 2!} = \frac{5,040}{24 \times 2} = \frac{5,040}{48} = 105

.

Adım Adım Çözüm

1

Calculate slot selections for the odd digits

35 ways

From 7 positions, selecting 4 positions for the odd digits {1, 3, 5, 7} can be done in

\binom{7}{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

ways. Since the odd digits must be ordered in strictly ascending order (1, then 3, then 5, then 7), there is only 1 valid relative order for any selection of 4 slots.

2

Determine valid arrangements for the even digits

3 ways

The remaining 3 slots are occupied by the even digits {2, 4, 6}. The total number of linear arrangements of 3 distinct digits is

3! = 6

. By symmetry, in exactly half of these arrangements, digit 2 appears to the left of digit 4, yielding

\frac{6}{2} = 3

valid ways.

3

Multiply independent choices to find the total arrangements

105 codes

Applying the Fundamental Counting Principle,

35 \text{ (slot choices for odd digits)} \times 1 \text{ (ordering of odd digits)} \times 3 \text{ (arrangements of even digits)} = 105

.

Anahtar Kavram

Permutations with Relative Order Restrictions

Tahmini Süre:2m 0s

Soru 75Soru →

In a plane, there are $n$ points such that no three points are collinear. If the number of distinct triangles that can be formed using these points as vertices is exactly $5$ times the number of distinct line segments that can be formed by joining pairs of these points, what is the value of $n$ ?

7

15

16

17

32

Cevabı ve açıklamayı göster

Cevap: $17$

Cevap

The correct value of

n

is

17

.

The number of distinct triangles formed by

n

points (no three collinear) is

\binom{n}{3} = \frac{n(n-1)(n-2)}{6}

, and the number of line segments is

\binom{n}{2} = \frac{n(n-1)}{2}

. Setting

\binom{n}{3} = 5 \binom{n}{2}

gives

\frac{n(n-1)(n-2)}{6} = \frac{5n(n-1)}{2}

. Dividing both sides by

\frac{n(n-1)}{2}

simplifies the equation to

\frac{n-2}{3} = 5

, yielding

n - 2 = 15

, so

n = 17

.

Adım Adım Çözüm

1

Express the number of triangles and line segments using combinations.

The number of triangles formed by selecting

3

non-collinear points from

n

points is

\binom{n}{3} = \frac{n(n-1)(n-2)}{6}

. The number of line segments formed by selecting

2

points from

n

points is

\binom{n}{2} = \frac{n(n-1)}{2}

.

Order does not matter when selecting vertices for triangles or endpoints for line segments.

2

Set up the given equation relating the two quantities.

\frac{n(n-1)(n-2)}{6} = 5 \times \frac{n(n-1)}{2}

The problem states that the number of triangles is

5

times the number of line segments.

3

Simplify the equation for

n \ge 3

.

Divide both sides by

\frac{n(n-1)}{2}

to get

\frac{n-2}{3} = 5

, which simplifies to

n - 2 = 15

.

Since

n \ge 3

,

n(n-1) \ne 0

, so dividing by common terms is valid.

4

Solve for

n

.

n = 15 + 2 = 17

Adding

2

to both sides isolates

n

.

Anahtar Kavram

Combinations in Geometry

Soru 76Soru →

A software company audited $120$ applications for compliance with two protocols: Accessibility (Protocol A) and Security (Protocol B). The audit revealed that $75$ applications complied with Protocol A, $60$ applications complied with Protocol B, and $35$ applications complied with neither protocol. If an application selected at random from the audited group is known to comply with Protocol A, what is the probability that it also complies with Protocol B?

\frac{5}{12}

\frac{2}{3}

\frac{5}{6}

\frac{2}{15}

\frac{17}{24}

Cevabı ve açıklamayı göster

Cevap: $\frac{2}{3}$

Cevap

\frac{2}{3}

Out of the 120 total applications, 35 comply with neither protocol, leaving 85 that comply with at least one. Using the inclusion-exclusion principle (

75 + 60 - 85 = 50

), exactly 50 applications comply with both protocols. Since the application is already known to comply with Protocol A, the denominator is restricted to the 75 applications in Protocol A. The conditional probability is therefore

50 / 75 = 2/3

.

Adım Adım Çözüm

1

Determine the number of applications that comply with at least one protocol.

Total applications minus those complying with neither:

120 - 35 = 85

.

The total sample space is partitioned into applications complying with at least one protocol and those complying with neither.

2

Find the number of applications complying with both Protocol A and Protocol B using the Principle of Inclusion-Exclusion.

|A \cap B| = |A| + |B| - |A \cup B| = 75 + 60 - 85 = 50

.

Summing the counts of Protocol A and Protocol B counts applications in both protocols twice.

3

Calculate the conditional probability

P(B|A)

.

P(B|A) = \frac{|A \cap B|}{|A|} = \frac{50}{75} = \frac{2}{3}

.

Given that the selected application complies with Protocol A, the sample space is restricted to

|A| = 75

.

Anahtar Kavram

Conditional Probability with Overlapping Sets

Tahmini Süre:1m 30s

Soru 77Soru →

An executive board consisting of 8 distinct members needs to form a subcommittee of 4 members. However, board members Alex and Blair refuse to serve on the subcommittee together unless board member Morgan is also selected. How many different 4-member subcommittees can be formed under these conditions?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

60

The solution uses complementary counting. First, calculate the total possible 4-member subcommittees from 8 members without restrictions, which is

\binom{8}{4} = 70

. Second, identify the restricted scenario that is not allowed: Alex and Blair are both selected, but Morgan is excluded. In this invalid scenario, 2 spots are taken by Alex and Blair, Morgan is excluded from consideration, leaving 2 spots to be filled from the remaining 5 board members, which equals

\binom{5}{2} = 10

invalid subcommittees. Subtracting the invalid subcommittees from the total gives

70 - 10 = 60

valid subcommittees.

Adım Adım Çözüm

1

Calculate the total number of ways to choose a 4-member subcommittee from 8 members without restrictions.

The total unrestricted combinations is

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

.

Since the order of selecting members into a subcommittee does not matter, use the combination formula

\binom{n}{k}

.

2

Determine the condition under which a subcommittee selection is invalid.

A subcommittee is invalid if and only if both Alex and Blair are selected AND Morgan is excluded.

Alex and Blair agree to serve together only if Morgan is also present. Thus, having Alex and Blair together without Morgan violates the condition.

3

Calculate the number of invalid subcommittees.

The number of invalid subcommittees is

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

.

Alex and Blair take 2 of the 4 spots, and Morgan cannot take any spot. The remaining 2 spots must be filled from the remaining

8 - 3 = 5

members.

4

Subtract the invalid subcommittees from the total unrestricted subcommittees using complementary counting.

70 - 10 = 60

valid subcommittees.

Complementary counting yields the total number of subcommittees that satisfy the restriction.

Anahtar Kavram

Combinations with Restrictions and Complementary Counting

Tahmini Süre:2m 0s

Soru 78Soru →

A corporate committee needs to form a 5-member project team selected from a department consisting of 4 data analysts, 5 software developers, and 4 cybersecurity experts. The team must include exactly 1 cybersecurity expert, at least 1 data analyst, and at least 1 software developer. How many different 5-member teams can be formed under these conditions?

124

240

480

504

1,120

Cevabı ve açıklamayı göster

Cevap: 480

Cevap

480

To form a valid 5-member team, we first choose 1 cybersecurity expert from 4 available options in

\binom{4}{1} = 4

ways. The remaining 4 members must be selected from 4 data analysts and 5 software developers such that at least 1 of each role is included. The three mutually exclusive ways to fill these 4 spots are: 1 data analyst and 3 developers (40 ways), 2 data analysts and 2 developers (60 ways), or 3 data analysts and 1 developer (20 ways). Summing these gives 120 valid combinations for the remaining members. Multiplying by the 4 choices for the cybersecurity expert yields

4 \times 120 = 480

total distinct teams.

Adım Adım Çözüm

1

Determine the number of ways to select the required cybersecurity expert

4 ways

Selecting exactly 1 cybersecurity expert from a total of 4 available experts is given by the combination

\binom{4}{1} = 4

.

2

Identify the possible valid distributions for the remaining 4 team members

Three valid cases: (1 DA, 3 SD), (2 DA, 2 SD), and (3 DA, 1 SD)

The team requires 5 members in total. With 1 cybersecurity expert selected, 4 positions remain. Since the team must contain at least 1 data analyst (DA) and at least 1 software developer (SD), we list all valid splits of 4 members.

3

Calculate the combinations for each valid distribution of data analysts and software developers

Case 1: 40 ways; Case 2: 60 ways; Case 3: 20 ways

Case 1 (1 DA, 3 SD):

\binom{4}{1} \times \binom{5}{3} = 4 \times 10 = 40

.
Case 2 (2 DA, 2 SD):

\binom{4}{2} \times \binom{5}{2} = 6 \times 10 = 60

.
Case 3 (3 DA, 1 SD):

\binom{4}{3} \times \binom{5}{1} = 4 \times 5 = 20

.

4

Sum the valid remaining member distributions and apply the Fundamental Counting Principle

480 total teams

Total remaining combinations =

40 + 60 + 20 = 120

. Total team combinations =

\binom{4}{1} \times 120 = 4 \times 120 = 480

.

Anahtar Kavram

Group selections with multiple category constraints and the Fundamental Counting Principle

Tahmini Süre:2m 30s

Soru 79Soru →

A financial firm has a team of 10 auditors consisting of 6 certified public accountants (CPAs) and 4 audit assistants. A special audit task force of 3 auditors is to be selected at random from the team. If the probability that the task force contains at least one CPA is expressed as a fraction $\frac{a}{b}$ in simplest form, what is the value of $a + b$ ?

Cevabı ve açıklamayı göster

Cevap: 59

Cevap

59

To find the probability of selecting at least one CPA, it is most efficient to use the complementary probability formula:

P(\text{at least 1 CPA}) = 1 - P(\text{no CPAs})

. The total number of ways to select any 3 auditors from the team of 10 is given by

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

. The number of ways to select 3 auditors such that none are CPAs (i.e., all 3 are audit assistants) is

\binom{4}{3} = 4

. Therefore, the probability of choosing zero CPAs is

\frac{4}{120} = \frac{1}{30}

. Subtracting from 1 gives

1 - \frac{1}{30} = \frac{29}{30}

. The fraction

\frac{29}{30}

is in simplest form because 29 is a prime number and does not divide 30. Thus,

a = 29

and

b = 30

, making

a + b = 59

.

Adım Adım Çözüm

1

Calculate total possible combinations for selecting 3 members from 10.

\binom{10}{3} = 120

Selection is made without replacement and order of selection does not matter.

2

Calculate combinations of selecting 3 members with no CPAs.

\binom{4}{3} = 4

All 3 selected members must come from the pool of 4 audit assistants.

3

Determine probability of the complement event (no CPAs).

P(\text{no CPA}) = \frac{4}{120} = \frac{1}{30}

Probability is favorable outcomes over total outcomes.

4

Compute probability of at least one CPA.

P(\text{at least 1 CPA}) = 1 - \frac{1}{30} = \frac{29}{30}

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

5

Find the sum of numerator a and denominator b in simplest form.

a + b = 29 + 30 = 59

29/30 cannot be simplified further as 29 is a prime number.

Anahtar Kavram

Complementary Probability and At-Least-One Scenarios

Soru 80Soru →

A furniture manufacturer allows customers to customize a dining set by selecting options from four categories:
- Tabletop shape: rectangular, oval, or round (3 choices)
- Wood finish: oak, walnut, cherry, or maple (4 choices)
- Leg design: tapered, hairpin, or turned (3 choices)
- Number of chairs: 4, 6, or 8 (3 choices)

However, due to space constraints, a round tabletop cannot be paired with a set of 8 chairs. How many distinct dining set configurations can a customer assemble?

Cevabı ve açıklamayı göster

Cevap: 96

Cevap

96 distinct dining set configurations can be assembled.

To find the number of valid dining set configurations, calculate total unrestricted choices using the Fundamental Counting Principle: multiplying 3 tabletop shapes, 4 wood finishes, 3 leg designs, and 3 chair count options gives 108 total combinations. Next, count the prohibited combinations consisting of a round tabletop (1 option) paired with 8 chairs (1 option) across all 4 finishes and 3 leg designs, giving 1 × 4 × 3 × 1 = 12 restricted combinations. Subtracting 12 restricted combinations from 108 total combinations yields 96 valid configurations.

Adım Adım Çözüm

1

Calculate total possible combinations without restrictions

3 × 4 × 3 × 3 = 108 combinations

By the Fundamental Counting Principle, multiplying the number of available options across all independent decision stages gives the total number of unrestricted arrangements.

2

Calculate the number of invalid configurations violating the space constraint

1 × 4 × 3 × 1 = 12 invalid combinations

Restricted configurations consist of 1 tabletop shape (round), 4 wood finishes, 3 leg designs, and 1 chair quantity selection (8 chairs).

3

Subtract the invalid combinations from the total unrestricted combinations

108 - 12 = 96 valid configurations

Subtracting the prohibited configurations from the total possible combinations yields the count of permissible configurations.

Anahtar Kavram

Fundamental Counting Principle with Subtraction of Restricted Cases

Counting and Probability

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