Basic Probability and Counting Methods Practice Questions — ACT — Page 2

Question 21Question →

A high school cafeteria offers a lunch meal combination consisting of 1 main dish, 1 side dish, and 1 drink. There are 4 main dish options, 5 side dish options, and 3 drink options available. However, due to dietary restrictions, 1 specific main dish option cannot be paired with 2 specific side dish options. How many different valid lunch meal combinations of 1 main dish, 1 side dish, and 1 drink can a student choose?

Show answer & explanation

Answer: 54

Answer

The total number of valid lunch meal combinations is 54.

The total number of unrestricted meal options is

4 \times 5 \times 3 = 60

. The restricted main dish cannot be paired with 2 specific side dishes across any of the 3 drink options, creating

1 \times 2 \times 3 = 6

invalid combinations. Subtracting these from the total gives

60 - 6 = 54

valid combinations.

Step-by-Step Solution

1

Calculate the total number of unrestricted meal choices

4 \times 5 \times 3 = 60

total combinations

By the Fundamental Counting Principle, multiplying the available choices for each dish component yields the total possible combinations without restrictions.

2

Determine the number of invalid combinations caused by dietary restrictions

1 \times 2 \times 3 = 6

invalid combinations

The 1 restricted main dish cannot be paired with 2 specific side dishes, and each of those 2 pairs can still be matched with any of the 3 drinks.

3

Subtract invalid combinations from the unrestricted total

60 - 6 = 54

valid combinations

Subtracting the disallowed combinations leaves only the combinations that satisfy all dietary restrictions.

Key Concept

Fundamental Counting Principle with Restrictions

Question 22Question →

A music ensemble director must randomly select 3 students from a group consisting of 5 vocalists and 4 instrumentalists to perform at a ceremony. What is the probability that the chosen group contains at least 1 vocalist and at least 1 instrumentalist?

\frac{1}{6}

\frac{10}{21}

\frac{5}{21}

\frac{20}{81}

\frac{5}{6}

Show answer & explanation

Answer: $\frac{5}{6}$

Answer

\frac{5}{6}

To find the probability that the selected trio includes at least 1 vocalist and at least 1 instrumentalist, calculate the total possible combinations of picking 3 students from 9:

\binom{9}{3} = 84

. Next, identify the unfavorable combinations where all 3 students are of the same type:

\binom{5}{3} = 10

ways for all vocalists and

\binom{4}{3} = 4

ways for all instrumentalists, totaling

10 + 4 = 14

unfavorable outcomes. Subtracting these from 84 leaves 70 favorable combinations. Thus, the probability is

\frac{70}{84} = \frac{5}{6}

.

Step-by-Step Solution

1

Calculate the total number of ways to choose any 3 students from the total group of 9 students.

\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

total outcomes.

Establishes the size of the sample space for selecting 3 students without replacement out of 9 (5 vocalists + 4 instrumentalists).

2

Determine the number of unfavorable outcomes (groups consisting entirely of one type of student).

Ways to choose 3 vocalists:

\binom{5}{3} = 10

.
Ways to choose 3 instrumentalists:

\binom{4}{3} = 4

.
Total unfavorable outcomes =

10 + 4 = 14

.

The complement of having 'at least 1 vocalist and at least 1 instrumentalist' is selecting a group of 3 composed entirely of vocalists or entirely of instrumentalists.

3

Subtract the unfavorable outcomes from the total outcomes to get the favorable outcomes.

84 - 14 = 70

favorable combinations.

Finding the complementary set count is direct and prevents missing valid composition cases.

4

Calculate the probability by dividing favorable outcomes by total outcomes.

Probability =

\frac{70}{84} = \frac{5}{6}

.

Dividing favorable combinations by the total size of the sample space gives the required probability.

Key Concept

Complementary Probability and Combinations

Estimated Time:1m 30s

Question 23Question →

A security system requires a 4-digit passcode created using the digits $1, 2, 3, 4, 5, 6,$ and $7$ . No digit may be repeated in a passcode. If the first digit and the last digit must both be odd numbers, how many different passcodes can be formed?

Show answer & explanation

Answer: 240

Answer

240 passcodes can be formed under the given conditions.

To form a 4-digit passcode with distinct digits where the first and last digits are odd: first choose the first digit from the 4 available odd digits (

1, 3, 5, 7

). Next, choose the fourth digit from the remaining 3 odd digits. Then, choose the second digit from the remaining 5 available digits in the full set, and the third digit from the remaining 4 available digits. Multiplying these choices gives

4 \times 5 \times 4 \times 3 = 240

.

Step-by-Step Solution

1

Categorize available digits into odd and even sets

4 odd digits ({1, 3, 5, 7}) and 3 even digits ({2, 4, 6}) out of 7 total digits

The first and last positions require odd digits, so we need to track odd digits separately.

2

Determine options for the first and last positions

4 options for position 1; 3 options for position 4

Position 1 must be odd (4 choices). Because digits cannot repeat, position 4 has 3 remaining odd choices.

3

Determine options for the second and third positions

5 options for position 2; 4 options for position 3

Two digits have already been assigned. Out of 7 total digits, 5 remain for position 2, leaving 4 for position 3.

4

Multiply choices across all positions

4 * 5 * 4 * 3 = 240

By the Fundamental Counting Principle, the total number of combinations is the product of the number of choices at each step.

Key Concept

Fundamental Counting Principle with Position Restrictions and Non-Replacement

Question 24Question →

A bookstore received a shipment of $50$ new books. Among these books, $24$ are fiction, $20$ are hardcover, and $8$ are hardcover fiction books. If $1$ book is selected at random from the shipment, what is the probability that the selected book is either a fiction book or a hardcover book?

\frac{4}{25}

\frac{14}{25}

\frac{18}{25}

\frac{22}{25}

\frac{9}{11}

Show answer & explanation

Answer: $\frac{18}{25}$

Answer

\frac{18}{25}

To find the probability that a randomly chosen book is either fiction or hardcover, calculate the total number of books satisfying at least one condition. By the Principle of Inclusion-Exclusion, the number of favorable books is

24 + 20 - 8 = 36

. Dividing this by the total of

50

books yields

\frac{36}{50}

, which simplifies to

\frac{18}{25}

.

Step-by-Step Solution

1

Identify the total number of items in the sample space and the counts for each event.

Total books

N = 50

, Fiction

N(F) = 24

, Hardcover

N(H) = 20

, Both

N(F \cap H) = 8

.

The sample space consists of all 50 shipment books.

2

Apply the Inclusion-Exclusion Principle to find the number of favorable outcomes.

N(F \cup H) = N(F) + N(H) - N(F \cap H) = 24 + 20 - 8 = 36

.

Hardcover fiction books are counted in both individual totals, so their intersection must be subtracted once to avoid double-counting.

3

Calculate the probability and simplify the fraction.

P(F \cup H) = \frac{36}{50} = \frac{18}{25}

.

Probability is the ratio of favorable outcomes to total outcomes.

Key Concept

Probability of non-mutually exclusive events using the Addition Rule:

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

.

Estimated Time:1m 0s

Question 25Question →

A box contains red, blue, and yellow game tokens. The probability of randomly selecting a red token is $\frac{3}{8}$ , and the probability of randomly selecting a blue token is $\frac{1}{3}$ . If the box contains exactly $14$ yellow tokens, how many total tokens are in the box?

48

24

22

16

72

Show answer & explanation

Answer: $48$

Answer

There are 48 total tokens in the box.

The total probability of all outcomes in the box must equal

1

. Combining the given probabilities for red and blue tokens yields

\frac{3}{8} + \frac{1}{3} = \frac{9}{24} + \frac{8}{24} = \frac{17}{24}

. The remaining probability representing the yellow tokens is

1 - \frac{17}{24} = \frac{7}{24}

. Since the box contains

14

yellow tokens, setting

\frac{7}{24}

of the total number of tokens equal to

14

gives

\frac{7}{24} N = 14

, which simplifies to

N = 48

.

Step-by-Step Solution

1

Find the probability of selecting a red or blue token.

\frac{3}{8} + \frac{1}{3} = \frac{9}{24} + \frac{8}{24} = \frac{17}{24}

Since selecting a red token and a blue token are mutually exclusive events, add their probabilities using a common denominator of

24

.

2

Calculate the probability of selecting a yellow token.

1 - \frac{17}{24} = \frac{7}{24}

The sum of the probabilities of all possible outcomes (red, blue, yellow) must equal

1

.

3

Set up and solve the equation for the total number of tokens

N

.

\frac{7}{24} N = 14 \implies N = 14 \times \frac{24}{7} = 2 \times 24 = 48

Multiply the count of yellow tokens (

14

) by the reciprocal of the yellow token probability (

\frac{24}{7}

).

Key Concept

Determining total sample size using complementary probabilities of mutually exclusive events

Estimated Time:1m 15s

Question 26Question →

A high school planning committee needs to select a subcommittee of $3$ people from a pool of $5$ teachers and $4$ students. How many different subcommittees can be formed consisting of exactly $2$ teachers and $1$ student?

Show answer & explanation

Answer: 40

Answer

The total number of different subcommittees that can be formed is 40.

To find the number of unique subcommittees with 2 teachers and 1 student, calculate the combination for choosing 2 teachers from 5, which equals 10, and 1 student from 4, which equals 4. Multiplying these independent choices together results in 40 possible subcommittees.

Step-by-Step Solution

1

Calculate the number of ways to select 2 teachers from the pool of 5 teachers.

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

ways

Selection order does not matter for committee members, so combinations are used.

2

Calculate the number of ways to select 1 student from the pool of 4 students.

\binom{4}{1} = 4

ways

Choosing 1 item from a set of 4 offers 4 distinct possibilities.

3

Apply the Fundamental Counting Principle to determine total combinations.

10 \times 4 = 40

subcommittees

Multiply the number of choices for teachers by the number of choices for students.

Key Concept

Combinations and Fundamental Counting Principle

Estimated Time:1m 0s

Question 27Question →

A high school photography club has $12$ active members, consisting of $7$ seniors and $5$ juniors. The club needs to elect an executive board composed of a President, a Vice President, and a Secretary, where no member can hold more than one position. If the President must be a senior, how many different executive board arrangements are possible?

Show answer & explanation

Answer: 770

Answer

The total number of different executive board arrangements possible is

770

.

To calculate the total number of distinct executive board arrangements, use the Fundamental Counting Principle by calculating the number of options for each position sequentially. The President must be a senior, so there are 7 choices for President. Once the President is chosen, any of the remaining 11 members can serve as Vice President. After filling the Vice President role, 10 members remain for Secretary. Multiplying these independent choices gives

7 \times 11 \times 10 = 770

.

Step-by-Step Solution

1

Determine the available choices for the President position

7 possibilities

The President position is restricted to seniors only, and there are 7 seniors in the club.

2

Determine the available choices for the Vice President position

11 possibilities

After 1 member is chosen as President, 11 of the original 12 members remain available for the Vice President role.

3

Determine the available choices for the Secretary position

10 possibilities

After 2 members are chosen for President and Vice President, 10 members remain available for Secretary.

4

Calculate the total number of distinct outcomes using the Fundamental Counting Principle

770 total arrangements

Multiplying the choices for each position gives

7 \times 11 \times 10 = 770

.

Key Concept

Fundamental Counting Principle with Position Restrictions

Estimated Time:1m 15s

Question 28Question →

A specialty bakery allows customers to create a custom dessert box. Each box must contain $1$ base, $1$ flavor, and $2$ different toppings. Customers can choose from $3$ base options, $4$ flavor options, and $6$ topping options. How many different custom dessert boxes can a customer create?

72

144

180

216

360

Show answer & explanation

Answer: $180$

Answer

180

different custom dessert boxes can be created.

The total number of outcomes is found by multiplying the number of choices for each component. For the base, there are

3

choices. For the flavor, there are

4

choices. For the

2

different toppings chosen from

6

, order does not matter, so we calculate the combination

\binom{6}{2} = 15

. Multiplying these choices together gives

3 \times 4 \times 15 = 180

.

Step-by-Step Solution

1

Determine the number of ways to choose the base and flavor.

There are

3

base options and

4

flavor options.

Each selection is an independent choice.

2

Calculate the number of ways to choose

2

different toppings out of

6

available toppings using combinations.

\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 ways.

The order in which the toppings are selected does not matter, so we use the combination formula

\binom{n}{k} = \frac{n!}{k!(n-k)!}

.

3

Apply the Fundamental Counting Principle to find the total number of unique dessert boxes.

Total combinations =

3 \times 4 \times 15 = 180

.

Multiply the number of ways to make each independent choice.

Key Concept

Fundamental Counting Principle and Combinations

Estimated Time:1m 15s

Question 29Question →

A box contains $10$ tiles labeled with the integers from $1$ through $10$ . If $2$ tiles are drawn at random without replacement, how many distinct pairs of tiles (where the order of selection does not matter) have a sum that is an odd number?

Show answer & explanation

Answer: 25

Answer

The total number of distinct pairs with an odd sum is 25.

To obtain an odd sum when adding two integers, one integer must be odd and the other must be even. In the range

1

through

10

, there are

5

odd integers (

1, 3, 5, 7, 9

) and

5

even integers (

2, 4, 6, 8, 10

). To form a pair with an odd sum, one tile must be selected from the

5

odd tiles and one tile must be selected from the

5

even tiles. By the Fundamental Counting Principle, the number of such distinct pairs is

5 \times 5 = 25

.

Step-by-Step Solution

1

Determine the parity condition for an odd sum

One tile must be odd and the other must be even

The sum of two integers is odd if and only if one addend is odd and the other addend is even.

2

Count the number of odd and even options

5 odd tiles and 5 even tiles

Among the integers

1

through

10

, the odd numbers are

1, 3, 5, 7, 9

(

5

total) and the even numbers are

2, 4, 6, 8, 10

(

5

total).

3

Apply the Fundamental Counting Principle

5 × 5 = 25 distinct pairs

Selecting one odd tile out of 5 possibilities and one even tile out of 5 possibilities gives

5 \times 5 = 25

distinct unordered pairs.

Key Concept

Fundamental Counting Principle and Parity of Integers

Estimated Time:1m 0s

Question 30Question →

A community art center offers $40$ different workshops during a summer session. Among these workshops, $22$ are scheduled in the evening, $18$ are scheduled on weekends, and $8$ are scheduled in both the evening and on weekends. If one workshop is selected at random from the $40$ workshops, what is the probability that it is scheduled in the evening, on a weekend, or both?

\frac{1}{5}

\frac{3}{10}

\frac{3}{5}

\frac{4}{5}

1

Show answer & explanation

Answer: $\frac{4}{5}$

Answer

The correct probability is

\frac{4}{5}

.

To find the probability that a randomly chosen workshop is in the evening, on a weekend, or both, calculate the total number of distinct workshops meeting at least one criterion using the formula

N(\text{Evening} \cup \text{Weekend}) = N(\text{Evening}) + N(\text{Weekend}) - N(\text{Evening} \cap \text{Weekend})

. Substituting the given values gives

22 + 18 - 8 = 32

workshops. Dividing by the total

40

workshops gives

\frac{32}{40}

, which reduces to

\frac{4}{5}

.

Step-by-Step Solution

1

Identify the given counts for each category

Total workshops =

40

, Evening workshops =

22

, Weekend workshops =

18

, Both =

8

.

Extract the necessary components to apply the inclusion-exclusion principle.

2

Calculate the number of workshops in the evening, on a weekend, or both

Number of favorable workshops =

22 + 18 - 8 = 32

.

Workshops scheduled in both categories are counted twice if evening and weekend counts are added directly, so the intersection must be subtracted once.

3

Compute the probability and simplify the fraction

Probability =

\frac{32}{40} = \frac{4}{5}

.

Divide the number of favorable outcomes by the total sample space size.

Key Concept

Probability of Combined Events (Principle of Inclusion-Exclusion)

Question 31Question →

A student council committee consists of $15$ members: $8$ juniors and $7$ seniors. Among the juniors, $3$ are on the debate team. Among the seniors, $4$ are on the debate team. If one committee member is selected at random, what is the probability that the selected member is a senior or is on the debate team?

\frac{4}{15}

\frac{7}{15}

\frac{2}{3}

\frac{11}{15}

\frac{14}{15}

Show answer & explanation

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

Answer

The probability that the selected member is a senior or on the debate team is

\frac{2}{3}

.

The total number of committee members is

15

. The event consists of selecting someone who is either a senior or on the debate team. There are

7

total seniors and

7

total debate members (

3

juniors and

4

seniors). Using the principle of inclusion-exclusion, the number of favorable outcomes is

7 + 7 - 4 = 10

. The probability is

\frac{10}{15}

, which simplifies to

\frac{2}{3}

.

Step-by-Step Solution

1

Identify the total number of members in the sample space.

The total number of members is

15

.

This is given as the total committee size.

2

Determine the number of favorable outcomes for the event.

Number of seniors =

7

. Number of juniors on the debate team =

3

. Total favorable outcomes =

7 + 3 = 10

.

To find members who are a senior OR on the debate team, count all seniors (

7

) plus non-senior debate members (

3

) to avoid double-counting.

3

Calculate and simplify the probability fraction.

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

.

Divide the favorable outcomes (

10

) by total outcomes (

15

) and simplify by dividing numerator and denominator by

5

.

Key Concept

Probability of Compound Events (Inclusion-Exclusion Principle)

Question 32Question →

A container holds $6$ red tiles, $4$ blue tiles, and $5$ green tiles. If $2$ tiles are selected at random from the container one after another without replacement, what is the probability that both selected tiles are blue?

\frac{7}{29}

\frac{2}{35}

\frac{6}{55}

\frac{16}{225}

\frac{101}{210}

Show answer & explanation

Answer: $\frac{2}{35}$

Answer

The probability that both selected tiles are blue is

\frac{2}{35}

.

The correct answer is determined by calculating the probability of each draw sequentially. The probability of drawing a blue tile first is

\frac{4}{15}

. Because the tile is not replaced,

3

blue tiles and

14

total tiles remain for the second draw, giving a probability of

\frac{3}{14}

. Multiplying these probabilities yields

\frac{4}{15} \times \frac{3}{14} = \frac{12}{210}

, which simplifies to

\frac{2}{35}

.

Step-by-Step Solution

1

Find the total number of tiles in the container.

Total tiles =

6 + 4 + 5 = 15

.

The sample space for the first selection consists of all available tiles.

2

Calculate the probability that the first tile selected is blue.

Probability of first blue tile =

\frac{4}{15}

.

There are

4

blue tiles out of a total of

15

tiles.

3

Determine the remaining number of blue tiles and total tiles for the second selection.

Remaining blue tiles =

3

, remaining total tiles =

14

.

Since the first tile was selected without replacement, both the count of blue tiles and total tiles decrease by

1

.

4

Calculate the probability that the second tile selected is blue given the first was blue.

Probability of second blue tile =

\frac{3}{14}

.

There are now

3

blue tiles remaining out of

14

total remaining tiles.

5

Multiply the probabilities of the sequential dependent events.

\frac{4}{15} \times \frac{3}{14} = \frac{12}{210} = \frac{2}{35}

.

By the multiplication rule for probability,

P(\text{both blue}) = P(\text{1st blue}) \times P(\text{2nd blue | 1st blue})

.

Key Concept

Basic Probability of Dependent Sequential Events (Without Replacement)

Practice More

Try solving a similar question where 3 tiles are selected sequentially without replacement.

Alternative Method

Calculate using combinations: The total ways to choose any 2 tiles out of 15 is

\binom{15}{2} = \frac{15 \times 14}{2} = 105

. The total ways to choose 2 blue tiles out of 4 is

\binom{4}{2} = \frac{4 \times 3}{2} = 6

. The probability is

\frac{6}{105} = \frac{2}{35}

.

Estimated Time:1m 0s

Question 33Question →

A regional sports club assigns identification codes to all of its members. Each code consists of $1$ letter chosen from the set $\{K, L, M, N\}$ , followed by $2$ digits chosen from $\{1, 2, 3, 4, 5\}$ such that no digit is repeated within a code, followed by $1$ symbol chosen from $\{*, \#\}$ . How many unique identification codes can be created using this system?

Show answer & explanation

Answer: 160

Answer

The total number of unique identification codes that can be created is

160

.

According to the Fundamental Counting Principle, to find the total number of multi-stage outcomes, multiply the number of choices at each stage. For the letter slot, there are

4

choices. For the two digit slots without repetition, there are

5 \times 4 = 20

choices. For the symbol slot, there are

2

choices. Multiplying these gives

4 \times 20 \times 2 = 160

unique identification codes.

Step-by-Step Solution

1

Determine the number of available letter choices for the first slot.

4 options

The set of allowed letters

\{K, L, M, N\}

contains 4 distinct elements.

2

Calculate the number of permutations for the two-digit section without repetition.

20 options

Choosing 2 distinct digits from 5 options gives

5 \times 4 = 20

possible outcomes.

3

Determine the number of available symbol choices for the last slot.

2 options

The set of allowed symbols

\{*, \#\}

contains 2 elements.

4

Multiply the number of choices for each slot using the Fundamental Counting Principle.

160 unique codes

Total codes =

4 \times 20 \times 2 = 160

.

Key Concept

Fundamental Counting Principle and Permutations without Repetition

Estimated Time:1m 0s

Question 34Question →

A catering service allows customers to build a custom buffet meal by selecting $1$ appetizer from $4$ available options, $1$ main course from $5$ available options, and $2$ distinct side dishes from $6$ available options. However, due to preparation constraints, $2$ specific side dishes cannot be selected together. How many different valid buffet meal combinations can a customer choose?

23

120

280

300

600

Show answer & explanation

Answer: $280$

Answer

The total number of valid buffet meal combinations is

280

.

There are

4

appetizer choices and

5

main course choices. Choosing

2

side dishes out of

6

options gives

\binom{6}{2} = 15

total pairs. Subtracting the

1

disallowed pair leaves

14

valid side dish pairs. Multiplying these independent choices gives

4 \times 5 \times 14 = 280

valid buffet menu combinations.

Step-by-Step Solution

1

Calculate the total number of ways to select 2 distinct side dishes from 6 options.

\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15

combinations.

Since the order in which the 2 side dishes are picked does not matter, combinations are used.

2

Subtract the restricted pair of side dishes from the total side dish combinations.

15 - 1 = 14

valid side dish combinations.

The problem specifies that 2 specific side dishes cannot be selected together, eliminating 1 specific pair.

3

Apply the Fundamental Counting Principle to find the overall number of valid buffet menu choices.

4 \text{ (appetizers)} \times 5 \text{ (main courses)} \times 14 \text{ (valid side pairs)} = 280

.

The choices for appetizer, main course, and side dish pair are independent, so their individual number of possibilities are multiplied.

Key Concept

Fundamental Counting Principle and Combinations with Restrictions

Estimated Time:1m 15s

Question 35Question →

A fair spinner is divided into $8$ congruent sectors numbered $1$ through $8$ . A player spins the spinner twice in succession. How many of the $64$ possible outcomes result in a sum of the two spins that is strictly greater than $12$ ?

Show answer & explanation

Answer: 10

Answer

There are 10 outcomes that yield a sum strictly greater than 12.

Systematically listing all ordered pairs

(x, y)

from

\{1, 2, \dots, 8\} \times \{1, 2, \dots, 8\}

such that

x + y > 12

yields 4 outcomes for sum 13, 3 outcomes for sum 14, 2 outcomes for sum 15, and 1 outcome for sum 16, totaling 10 valid outcomes.

Step-by-Step Solution

1

Determine the acceptable sums for the two spins

The possible sums strictly greater than 12 are 13, 14, 15, and 16.

Since each spin has a maximum value of 8, the maximum possible sum is 8 + 8 = 16.

2

Count the ordered pairs (spin 1, spin 2) for each valid sum

4 outcomes for sum 13, 3 outcomes for sum 14, 2 outcomes for sum 15, and 1 outcome for sum 16.

First and second spins are ordered, so (5,8) and (8,5) represent distinct outcomes.

3

Sum the outcome counts across all valid cases

4 + 3 + 2 + 1 = 10 outcomes.

The sets of outcomes for distinct sums are mutually exclusive.

Key Concept

Basic Probability and Counting Sample Space Outcomes

Estimated Time:1m 15s

Question 36Question →

A school library receives a shipment of $12$ new books, consisting of $5$ science fiction novels, $4$ historical fiction novels, and $3$ biography books. A librarian wants to choose $3$ books to feature on a display shelf. How many different combinations of $3$ books can be formed that contain exactly $2$ science fiction novels and $1$ biography book?

Show answer & explanation

Answer: 30

Answer

The total number of combinations containing exactly 2 science fiction novels and 1 biography book is 30.

To find the total number of combinations, first determine the number of ways to choose

2

science fiction novels from

5

available:

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

. Next, determine the number of ways to choose

1

biography book from

3

available:

\binom{3}{1} = 3

. Finally, apply the Fundamental Counting Principle by multiplying the counts together:

10 \times 3 = 30

.

Step-by-Step Solution

1

Calculate the combinations for selecting the science fiction novels.

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

Select

2

novels out of the

5

available science fiction novels without regard to order.

2

Calculate the combinations for selecting the biography book.

\binom{3}{1} = 3

Select

1

book out of the

3

available biography books.

3

Apply the Fundamental Counting Principle to find total combinations.

10 \times 3 = 30

Multiply the combinations calculated for each independent selection.

Key Concept

Combinations and the Fundamental Counting Principle

Question 37Question →

A project team of $3$ software developers is to be randomly selected from a pool of $4$ frontend developers and $6$ backend developers. What is the probability that the selected team will consist of exactly $1$ frontend developer and $2$ backend developers?

\frac{1}{6}

\frac{3}{10}

\frac{19}{120}

\frac{1}{2}

\frac{3}{25}

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The probability that the team consists of exactly 1 frontend developer and 2 backend developers is

\frac{1}{2}

.

To find the probability, divide the number of ways to select 1 frontend developer and 2 backend developers by the total number of ways to choose any 3 developers from 10. There are

\binom{4}{1} = 4

ways to select 1 frontend developer and

\binom{6}{2} = 15

ways to select 2 backend developers. Multiplying these yields

4 \times 15 = 60

favorable team combinations. The total number of ways to choose 3 developers from 10 is

\binom{10}{3} = 120

. Thus, the probability is

\frac{60}{120} = \frac{1}{2}

.

Step-by-Step Solution

1

Calculate the total number of possible 3-developer teams from the pool of 10 developers.

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

total outcomes.

The total number of ways to choose 3 people out of 10 without regard to order is given by combinations.

2

Calculate the number of ways to select 1 frontend developer from 4 and 2 backend developers from 6.

\binom{4}{1} \times \binom{6}{2} = 4 \times \frac{6 \times 5}{2 \times 1} = 4 \times 15 = 60

favorable outcomes.

By the Fundamental Counting Principle, multiply the number of ways to choose the frontend developer by the number of ways to choose the backend developers.

3

Divide the number of favorable outcomes by the total possible outcomes to find the probability.

\text{Probability} = \frac{60}{120} = \frac{1}{2}

.

Probability is defined as the ratio of favorable outcomes to total outcomes in a uniform sample space.

Key Concept

Probability of Compound Events using Combinations

Question 38Question →

A bakery allows customers to customize a fruit tart by selecting $1$ crust type from $3$ available choices, $1$ filling type from $4$ available choices, and $2$ different fruit toppings from $5$ available choices. How many different fruit tart combinations are possible?

17

60

120

240

300

Show answer & explanation

Answer: $120$

Answer

The total number of different fruit tart combinations possible is 120.

To find the total number of unique fruit tart combinations, apply the Fundamental Counting Principle across all decision stages. Selecting 1 crust out of 3 yields 3 options. Selecting 1 filling out of 4 yields 4 options. Selecting 2 different fruit toppings out of 5 requires choosing combinations without regard to order, which is calculated as 5 choose 2 = 10. Multiplying the options from each stage gives 3 * 4 * 10 = 120 total combinations.

Step-by-Step Solution

1

Determine the number of ways to choose the crust and filling.

There are

3

crust choices and

4

filling choices.

Each selection stage is independent.

2

Calculate the number of ways to choose

2

different fruit toppings from

5

available options.

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

choices.

The order in which the two toppings are selected does not matter, so combinations must be used.

3

Apply the Fundamental Counting Principle to find the total combinations.

3 \times 4 \times 10 = 120

.

Multiply the number of choices for each independent step.

Key Concept

Fundamental Counting Principle and Combinations

Estimated Time:1m 0s

Question 39Question →

An art gallery manager is arranging 6 unique paintings side by side in a single line on a display wall. Two specific paintings created by the same artist must be placed directly next to each other. How many different line arrangements of the 6 paintings are possible?

Show answer & explanation

Answer: 240

Answer

The total number of different line arrangements is 240.

To satisfy the constraint that two specific paintings must be adjacent, treat them as a single combined block. This reduces the problem to arranging 5 items (the single block plus the other 4 paintings), which can be ordered in 5! = 120 ways. Within the block, the two paintings can be ordered in 2! = 2 ways. Applying the Fundamental Counting Principle gives 120 * 2 = 240 distinct arrangements.

Step-by-Step Solution

1

Group the two specific paintings together as a single block.

There are 5 units to arrange (1 block of 2 paintings plus 4 individual paintings).

Since the two paintings must remain adjacent, treating them as a single item guarantees they will stay together.

2

Calculate the permutations of the 5 units.

5! = 5 * 4 * 3 * 2 * 1 = 120 ways.

There are 5 distinct units being arranged in a straight line.

3

Determine the internal arrangements of the grouped pair.

2! = 2 * 1 = 2 ways.

The two specific paintings within the block can switch positions (Painting A then B, or Painting B then A).

4

Multiply the arrangements together.

120 * 2 = 240 total arrangements.

By the Fundamental Counting Principle, total arrangements equal the number of ways to place the units multiplied by the internal arrangements of the restricted pair.

Key Concept

Permutations with Restrictions (Grouped Elements)

Estimated Time:1m 15s

Question 40Question →

A box contains $30$ raffle tickets numbered $1$ through $30$ . If one ticket is selected at random from the box, what is the probability that the number on the selected ticket is a multiple of $4$ or a multiple of $6$ ?

\frac{1}{15}

\frac{1}{5}

\frac{1}{3}

\frac{2}{5}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{1}{3}$

Answer

The probability that the number on the selected ticket is a multiple of

4

or a multiple of

6

is

\frac{1}{3}

.

The correct answer is determined by applying the addition rule for non-mutually exclusive events:

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

. Out of

30

tickets,

7

are multiples of

4

and

5

are multiples of

6

. The numbers

12

and

24

are multiples of both

4

and

6

, meaning they belong to the intersection. Calculating the total number of unique favorable tickets gives

7 + 5 - 2 = 10

. Dividing

10

favorable outcomes by the

30

total tickets yields a probability of

\frac{10}{30} = \frac{1}{3}

.

Step-by-Step Solution

1

Identify the total number of possible outcomes.

There are

30

total tickets in the box, so the sample space size is

30

.

Each ticket numbered 1 through 30 is equally likely to be selected.

2

Count the outcomes for each individual event.

Multiples of

4

:

\{4, 8, 12, 16, 20, 24, 28\} \rightarrow 7

outcomes.
Multiples of

6

:

\{6, 12, 18, 24, 30\} \rightarrow 5

outcomes.

Listing or dividing

30

by

4

and

6

determines how many tickets satisfy each separate condition.

3

Identify and count the overlapping outcomes (multiples of both

4

and

6

).

Multiples of both

4

and

6

are multiples of

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

, which are

\{12, 24\} \rightarrow 2

outcomes.

These numbers are contained in both individual lists and will be double-counted if not accounted for.

4

Apply the Principle of Inclusion-Exclusion to find the total number of favorable outcomes.

\text{Favorable outcomes} = 7 + 5 - 2 = 10.

Subtracting the intersection prevents double-counting the tickets that satisfy both conditions.

5

Calculate the probability.

P(\text{multiple of } 4 \text{ or } 6) = \frac{10}{30} = \frac{1}{3}.

Probability is the ratio of favorable outcomes to total possible outcomes.

Key Concept

Probability of Non-Mutually Exclusive Events (Inclusion-Exclusion Principle)

Estimated Time:1m 0s

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