Basic Probability and Counting Methods Practice Questions — ACT

Question 1Question →

A library display shelf has space for $5$ distinct books aligned in a row. The librarian has $3$ different science fiction books and $2$ different biography books. How many different row arrangements of these $5$ books are possible if the $2$ biography books must not be placed next to each other?

Show answer & explanation

Answer: 72

Answer

72 different row arrangements are possible.

To find the number of arrangements where the

2

biography books are not adjacent, subtract the number of arrangements where they are adjacent (

4! \times 2! = 48

) from the total possible arrangements (

5! = 120

), resulting in

120 - 48 = 72

.

Step-by-Step Solution

1

Calculate the total number of unrestricted arrangements of all 5 distinct books.

5! = 5 \times 4 \times 3 \times 2 \times 1 = 120

There are

5

distinct books, so there are

5!

total ways to arrange them in a line.

2

Calculate the number of arrangements in which the 2 biography books are placed next to each other.

4! \times 2! = 24 \times 2 = 48

Treat the

2

biography books as a single combined unit. This leaves

4

units to arrange (

3

science fiction books plus

1

biography unit), which can be ordered in

4! = 24

ways. Inside the unit, the

2

biography books can be arranged in

2! = 2

ways.

3

Subtract the number of adjacent biography arrangements from the total number of arrangements.

120 - 48 = 72

Subtracting the restricted outcomes (biography books adjacent) from the total possible outcomes gives the number of valid arrangements (biography books separated).

Key Concept

Permutations and Counting Methods with Complementary Counting

Question 2Question →

An art gallery owner wants to display 5 paintings in a row on a wall. The owner selects these 5 paintings from a collection of 4 different landscape paintings and 4 different portrait paintings. The display must meet the following guidelines:

1. No two landscape paintings can be placed next to each other.
2. At least 2 landscape paintings must be displayed.

How many different arrangements of 5 paintings are possible?

Show answer & explanation

Answer: 2016

Answer

2016

The correct answer is 2016. By breaking down the problem into two mutually exclusive cases based on the number of landscape paintings (either 3 landscapes and 2 portraits, or 2 landscapes and 3 portraits), we can find the valid arrangements for each. For 3 landscapes, the only valid layout is

L-P-L-P-L

, yielding

P(4,3) \times P(4,2) = 288

arrangements. For 2 landscapes, there are

\binom{4}{2} = 6

layout configurations, each yielding

P(4,2) \times P(4,3) = 288

arrangements, for a total of 1728. Adding these two cases gives

288 + 1728 = 2016

.

Step-by-Step Solution

1

Determine the possible number of landscape paintings (

k

) that can be displayed.

k = 2

or

k = 3

Since at least 2 landscapes must be displayed,

k \ge 2

. Since no two landscapes can be adjacent in a 5-painting row, we cannot have 4 landscapes (as that would require at least 3 portraits to separate them, making the total count at least 7 paintings). Thus,

k

can only be 2 or 3.

2

Calculate the arrangements for the case with 3 landscape paintings and 2 portrait paintings.

288 arrangements

For 3 landscapes (

L

) and 2 portraits (

P

) to have no adjacent landscapes, the only possible pattern of positions is

L-P-L-P-L

. The number of ways to choose and arrange 3 landscapes from 4 is

P(4, 3) = 4 \times 3 \times 2 = 24

. The number of ways to choose and arrange 2 portraits from 4 is

P(4, 2) = 4 \times 3 = 12

. The total arrangements for this case is

24 \times 12 = 288

.

3

Calculate the arrangements for the case with 2 landscape paintings and 3 portrait paintings.

1728 arrangements

For 2 landscapes (

L

) and 3 portraits (

P

) to have no adjacent landscapes, we place the 3 portraits first:

\_ P \_ P \_ P \_

. We choose 2 of the 4 available spaces for the landscapes in

\binom{4}{2} = 6

ways. For each pattern, the number of ways to choose and arrange 2 landscapes from 4 is

P(4, 2) = 12

, and the number of ways to choose and arrange 3 portraits from 4 is

P(4, 3) = 24

. The total arrangements for this case is

6 \times 12 \times 24 = 1728

.

4

Sum the arrangements from both cases.

2016 arrangements

Since the two cases are mutually exclusive, we add their individual counts:

288 + 1728 = 2016

.

Key Concept

Permutations and Combinations with Constraints

Question 3Question →

A box contains only red, blue, and green marbles. The ratio of red marbles to blue marbles is $2:3$ , and the ratio of blue marbles to green marbles is $3:5$ . If one marble is drawn at random from the box, the probability of drawing a red marble is $P$ . If $6$ green marbles are added to the box and no other marbles are removed, the probability of drawing a red marble becomes $Q$ . Given that $P - Q = \frac{3}{115}$ , how many blue marbles were originally in the box?

3

8

12

15

20

Show answer & explanation

Answer: 12

Answer

12

The correct answer is 12. By expressing the original counts of red, blue, and green marbles as

2k

,

3k

, and

5k

, the total number of marbles is

10k

. The initial probability of drawing a red marble is

P = \frac{2k}{10k} = \frac{1}{5}

. When 6 green marbles are added, the new total becomes

10k + 6

, yielding a new probability

Q = \frac{2k}{10k + 6} = \frac{k}{5k + 3}

. Setting their difference to

\frac{3}{115}

results in

\frac{1}{5} - \frac{k}{5k + 3} = \frac{3}{115}

. Simplifying this rational equation gives

25k + 15 = 115

, which solves to

k = 4

. Therefore, the original number of blue marbles is

3k = 3(4) = 12

.

Step-by-Step Solution

1

Define the number of red, blue, and green marbles in terms of a variable

k

.

Let red marbles =

2k

, blue marbles =

3k

, and green marbles =

5k

. The initial total number of marbles is

2k + 3k + 5k = 10k

.

The given ratios are red:blue = 2:3 and blue:green = 3:5, which combine to a continuous ratio of red:blue:green = 2:3:5.

2

Express the initial probability

P

of drawing a red marble.

P = \frac{2k}{10k} = \frac{1}{5}

.

Probability is the number of favorable outcomes (red marbles) divided by the total number of outcomes.

3

Express the new probability

Q

of drawing a red marble after adding 6 green marbles.

The number of green marbles becomes

5k + 6

, so the new total is

10k + 6

. Thus,

Q = \frac{2k}{10k + 6}

.

Adding 6 green marbles increases the total marble count by 6 while the number of red marbles remains

2k

.

4

Set up the equation using the given difference

P - Q = \frac{3}{115}

and solve for

k

.

\frac{1}{5} - \frac{2k}{10k + 6} = \frac{3}{115}

\frac{1}{5} - \frac{k}{5k + 3} = \frac{3}{115}

\frac{(5k + 3) - 5k}{5(5k + 3)} = \frac{3}{115}

\frac{3}{25k + 15} = \frac{3}{115}

25k + 15 = 115 \implies 25k = 100 \implies k = 4

.

Solving the rational equation gives the multiplier

k

that determines the actual number of marbles.

5

Calculate the original number of blue marbles.

Blue marbles =

3k = 3(4) = 12

.

The question asks for the original number of blue marbles, which is represented by

3k

.

Key Concept

Basic Probability and Counting Methods

Estimated Time:3m 0s

Question 4Question →

A security passcode consists of a sequence of 4 digits chosen from the digits 0 through 9. To be valid, a passcode must contain at least one repeated digit, but no digit can appear more than twice. Additionally, the passcode cannot end with an odd digit. How many different valid security passcodes can be created?

Show answer & explanation

Answer: 2295

Answer

There are 2,295 different valid security passcodes.

The correct answer of 2,295 is found by subtracting all invalid passcodes from the total possible passcodes. The total number of 4-digit passcodes ending in an even digit is 5,000. The invalid passcodes are those with all distinct digits (2,520), those with a digit repeated three times (180), and those with all four digits identical (5). Subtracting these gives 5,000 - 2,520 - 180 - 5 = 2,295.

Step-by-Step Solution

1

Calculate the total number of 4-digit passcodes ending in an even digit.

5,000

The last digit must be even (0, 2, 4, 6, or 8) to not be odd, giving 5 choices. The first three digits can be any of the 10 digits from 0 through 9. By the Fundamental Counting Principle, the total number of passcodes is 10 * 10 * 10 * 5 = 5,000.

2

Calculate the number of invalid passcodes where all 4 digits are distinct.

2,520

For all 4 digits to be distinct, the last digit must be chosen from the 5 even digits. The remaining 3 positions must be filled with 3 distinct digits chosen from the remaining 9 digits. There are 9 * 8 * 7 = 504 ways to choose these. This gives 504 * 5 = 2,520 passcodes.

3

Calculate the number of invalid passcodes where a single digit is repeated three times.

180

If the repeated digit is the last digit, there are 5 choices for that digit, and the single distinct digit (9 choices) can be placed in any of the 3 remaining positions, giving 5 * 9 * 3 = 135 passcodes. If the repeated digit is not the last digit, the three identical digits occupy the first three positions, and the last digit (5 choices) is distinct from them (9 choices), giving 5 * 9 = 45 passcodes. In total, 135 + 45 = 180 passcodes.

4

Calculate the number of invalid passcodes where all four digits are identical.

5

Since the last digit must be even, all four digits must be the same even digit (0000, 2222, 4444, 6666, or 8888), which gives 5 passcodes.

5

Subtract the invalid passcodes from the total number of passcodes.

2,295

Subtracting the passcodes with all distinct digits (2,520) and those with a digit repeated three or four times (180 + 5 = 185) from the total of 5,000 gives 5,000 - 2,520 - 185 = 2,295.

Key Concept

Complementary counting using permutations and partition analysis

Alternative Method

Instead of complementary counting, count the valid cases directly: Case A where the digit frequencies are [2, 1, 1] (which yields 2,160 codes) and Case B where the digit frequencies are [2, 2] (which yields 135 codes). Adding these yields 2,160 + 135 = 2,295 codes.

Estimated Time:3m 0s

Question 5Question →

A high school debate club consists of 15 members: 6 freshmen, 5 sophomores, and 4 juniors. If a committee of 3 members is to be selected at random from the club, what is the probability that the committee will contain at least 1 freshman and at least 1 sophomore?

\frac{24}{91}

\frac{247}{455}

\frac{51}{91}

\frac{251}{455}

\frac{6}{7}

Show answer & explanation

Answer: $\frac{51}{91}$

Answer

The correct probability is

\frac{51}{91}

.

To find the probability that a randomly chosen 3-member committee from a club of 15 members (6 freshmen, 5 sophomores, 4 juniors) has at least 1 freshman and at least 1 sophomore, we can find the total number of possible committees and subtract the number of unfavorable committees. The total number of committees is

\binom{15}{3} = 455

. The unfavorable committees are those with no freshmen (chosen from the 9 sophomores and juniors:

\binom{9}{3} = 84

) or no sophomores (chosen from the 10 freshmen and juniors:

\binom{10}{3} = 120

). Because these two groups overlap when only juniors are chosen (

\binom{4}{3} = 4

), the total number of unfavorable committees is

84 + 120 - 4 = 200

by the Principle of Inclusion-Exclusion. Thus, there are

455 - 200 = 255

favorable committees, yielding a probability of

\frac{255}{455} = \frac{51}{91}

.

Step-by-Step Solution

1

Calculate the total number of possible 3-member committees that can be formed from the 15 club members.

\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455

total committees.

To find the probability, we first need to determine the size of the entire sample space (the total number of possible outcomes).

2

Determine the number of committees that do not meet the requirement of containing at least one freshman and at least one sophomore. This occurs if a committee has no freshmen, no sophomores, or neither.

Let

A

be the event of choosing a committee with no freshmen (only sophomores and juniors):

\binom{9}{3} = 84

ways. Let

B

be the event of choosing a committee with no sophomores (only freshmen and juniors):

\binom{10}{3} = 120

ways. Let

A \cap B

be the event of choosing a committee with neither freshmen nor sophomores (only juniors):

\binom{4}{3} = 4

ways.

It is easier to count the complement (the unfavorable outcomes) and subtract it from the total.

3

Use the Principle of Inclusion-Exclusion to find the total number of unfavorable committees (no freshmen or no sophomores).

|A \cup B| = |A| + |B| - |A \cap B| = 84 + 120 - 4 = 200

unfavorable committees.

Since the events of having no freshmen and having no sophomores overlap when only juniors are selected, we must subtract the intersection to avoid double-counting.

4

Subtract the number of unfavorable committees from the total number of committees to find the number of favorable committees, and then compute the probability.

Favorable committees:

455 - 200 = 255

ways. Probability:

\frac{255}{455} = \frac{51}{91}

.

Subtracting the complement from the total gives the number of valid outcomes, and dividing this by the total outcomes yields the desired probability.

Key Concept

Using combinations and the Principle of Inclusion-Exclusion to calculate probabilities of compound events.

Estimated Time:3m 0s

Question 6Question →

A cafe offers a lunch special where a customer selects $1$ sandwich from $4$ available options, $1$ side dish from $3$ available options, and $1$ drink from $5$ available options. How many different total lunch combinations consisting of one sandwich, one side dish, and one drink can a customer choose?

Show answer & explanation

Answer: 60

Answer

60 combinations

According to the Fundamental Counting Principle, if one event can occur in

m

ways, a second in

n

ways, and a third in

p

ways, the total number of combinations for all three events occurring together is

m \times n \times p

. Multiplying

4

sandwiches by

3

sides by

5

drinks yields

4 \times 3 \times 5 = 60

distinct lunch combinations.

Step-by-Step Solution

1

Identify the number of options available for each independent selection.

4 sandwiches, 3 side dishes, 5 drinks

The total outcomes depend on making one selection from each distinct category.

2

Apply the Fundamental Counting Principle.

4 × 3 × 5 = 60

The total number of outcomes for independent sequential choices is the product of the number of options for each choice.

Key Concept

Fundamental Counting Principle

Estimated Time:45s

Question 7Question →

A box contains $5$ red tickets, $7$ blue tickets, and $8$ green tickets. If one ticket is selected at random from the box, what is the probability that the selected ticket is NOT blue?

\frac{7}{20}

\frac{13}{20}

\frac{7}{13}

\frac{3}{5}

\frac{13}{7}

Show answer & explanation

Answer: $\frac{13}{20}$

Answer

The probability that the selected ticket is not blue is

\frac{13}{20}

.

To find the probability of selecting a ticket that is not blue, divide the number of non-blue tickets by the total number of tickets. There are

5 + 7 + 8 = 20

total tickets, of which

5 + 8 = 13

are not blue. Thus, the probability is

\frac{13}{20}

.

Step-by-Step Solution

1

Find the total number of outcomes (total tickets in the box).

Total tickets =

5 + 7 + 8 = 20

.

Basic probability requires dividing the number of favorable outcomes by the total sample space.

2

Calculate the number of favorable outcomes (tickets that are NOT blue).

Favorable tickets =

5\text{ (red)} + 8\text{ (green)} = 13

(or

20 - 7 = 13

).

The complement of choosing a blue ticket is choosing a red or green ticket.

3

Form the probability ratio of favorable outcomes to total outcomes.

Probability =

\frac{13}{20}

.

Probability is defined as

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

.

Key Concept

Basic Probability of Complementary Events

Estimated Time:45s

Question 8Question →

A jar contains $3$ green marbles, $5$ yellow marbles, and $12$ purple marbles. If one marble is drawn at random from the jar, what is the probability that the drawn marble is NOT purple? Express your answer as a decimal.

Show answer & explanation

Answer: 0.4

Answer

The probability that the drawn marble is NOT purple is 0.4.

To find the probability of drawing a non-purple marble, identify the number of non-purple marbles (3 green + 5 yellow = 8) and divide by the total number of marbles in the jar (3 + 5 + 12 = 20). The resulting fraction 8/20 simplifies to 0.4.

Step-by-Step Solution

1

Calculate the total number of outcomes in the sample space.

Total marbles = 3 + 5 + 12 = 20

The total sample space includes all marbles contained in the jar.

2

Determine the number of favorable outcomes.

Non-purple marbles = 3 + 5 = 8

A marble that is not purple must be either green or yellow.

3

Compute the probability.

Probability = 8 / 20 = 0.4

The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes.

Key Concept

Basic Probability of Complementary Events

Question 9Question →

At a local vehicle dealership, a buyer customizing a new car model can select $1$ exterior paint color from $5$ available colors, $1$ interior seat material from $3$ available materials, and $1$ transmission type from $2$ available options (manual or automatic). According to the Fundamental Counting Principle, how many different unique configurations of the car can a buyer create?

Show answer & explanation

Answer: 30

Answer

There are 30 different unique configurations possible.

According to the Fundamental Counting Principle, when making a series of independent choices, the total number of distinct outcomes is found by multiplying the number of options for each choice. Multiplying 5 exterior colors, 3 interior materials, and 2 transmission types gives

5 \times 3 \times 2 = 30

unique car configurations.

Step-by-Step Solution

1

Count the number of choices for each feature

Paint colors = 5, Interior materials = 3, Transmissions = 2

Each feature selection is an independent choice in the configuration process.

2

Multiply the number of options for all features

5 \times 3 \times 2 = 30

The Fundamental Counting Principle states that if there are

n_1

ways to make one choice,

n_2

ways to make a second, and

n_3

ways to make a third, the total number of combined outcomes is

n_1 \times n_2 \times n_3

.

Key Concept

Fundamental Counting Principle

Estimated Time:45s

Question 10Question →

A bookshelf contains $4$ mystery novels, $6$ science fiction novels, and $5$ biography books. If one book is chosen at random from the bookshelf, what is the probability that the chosen book is NOT a science fiction novel?

\frac{3}{5}

\frac{2}{5}

\frac{2}{3}

\frac{3}{2}

\frac{11}{15}

Show answer & explanation

Answer: $\frac{3}{5}$

Answer

The probability that the chosen book is NOT a science fiction novel is

\frac{3}{5}

.

The correct answer is

\frac{3}{5}

. To find the probability of selecting a book that is not science fiction, divide the number of books that are not science fiction (

4

mystery +

5

biography =

9

) by the total number of books (

4 + 6 + 5 = 15

). This yields

\frac{9}{15}

, which simplifies to

\frac{3}{5}

.

Step-by-Step Solution

1

Find the total number of books on the shelf

Total books =

4 + 6 + 5 = 15

The sample space consists of all mystery, science fiction, and biography books combined.

2

Count the number of books that are NOT science fiction

Non-science fiction books =

4 + 5 = 9

The mystery novels (

4

) and biography books (

5

) are the favorable outcomes.

3

Calculate the probability and simplify the fraction

Probability =

\frac{9}{15} = \frac{3}{5}

Probability is the number of favorable outcomes divided by the total number of outcomes.

Key Concept

Basic Probability of Complementary Events

Estimated Time:45s

Question 11Question →

A theater group offers souvenir t-shirts in $4$ different sizes (Small, Medium, Large, and Extra-Large) and $5$ different colors. In addition, each t-shirt can be printed either with or without a custom emblem on the front. How many distinct t-shirt design options are available?

11

20

40

80

100

Show answer & explanation

Answer: $40$

Answer

There are 40 distinct t-shirt design options available.

According to the Fundamental Counting Principle, to find the total number of distinct outcomes from a series of independent choices, multiply the number of choices available for each feature. Since there are

4

sizes,

5

colors, and

2

emblem choices (with or without an emblem), the total number of distinct t-shirt options is

4 \times 5 \times 2 = 40

.

Step-by-Step Solution

1

Identify the number of independent choices for each attribute.

Size choices = 4; Color choices = 5; Emblem choices = 2 (with emblem or without emblem).

Each attribute represents an independent stage of selection.

2

Apply the Fundamental Counting Principle.

Total options =

4 \times 5 \times 2 = 40

.

When an event consists of multiple sequential independent choices, the total number of outcomes is the product of the number of options for each choice.

Key Concept

Fundamental Counting Principle

Estimated Time:45s

Question 12Question →

A conference schedule allows attendees to select 1 morning workshop out of 6 options, 1 keynote session out of 3 options, and 1 afternoon panel out of 4 options. However, 2 specific morning workshops conflict with 1 specific afternoon panel and cannot be chosen together. How many different valid 3-session schedule combinations can an attendee create?

Show answer & explanation

Answer: 66

Answer

An attendee can create 66 different valid 3-session schedule combinations.

According to the Fundamental Counting Principle, the total unrestricted number of schedule combinations is

6 \times 3 \times 4 = 72

. The restriction specifies that 2 morning workshops cannot be paired with 1 specific afternoon panel. Since there are 3 keynote speaker options available for any schedule, the number of invalid combinations is

2 \times 3 \times 1 = 6

. Subtracting the invalid options from the total gives

72 - 6 = 66

valid schedule combinations.

Step-by-Step Solution

1

Calculate total unrestricted schedule combinations

72 combinations

Multiply the choices for each session:

6 \times 3 \times 4 = 72

.

2

Calculate the number of conflicting schedule combinations

6 invalid combinations

The 2 restricted morning workshops combined with 1 restricted afternoon panel can occur alongside any of the 3 keynote choices (

2 \times 3 \times 1 = 6

).

3

Subtract conflicting combinations from total combinations

66 valid combinations

Subtracting 6 invalid schedules from 72 total schedules leaves 66 allowable options.

Key Concept

Fundamental Counting Principle with Restrictions

Estimated Time:1m 30s

Question 13Question →

A fair game spinner is divided into $8$ equal sectors numbered $1$ through $8$ . If Maya spins the spinner once, what is the probability that the spinner lands on a number strictly greater than $5$ ?

\frac{3}{8}

\frac{5}{8}

\frac{3}{5}

\frac{1}{4}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{3}{8}$

Answer

The probability that the spinner lands on a number strictly greater than 5 is

\frac{3}{8}

.

The sample space consists of 8 equally likely outcomes: {1, 2, 3, 4, 5, 6, 7, 8}. The event 'landing on a number strictly greater than 5' consists of 3 outcomes: {6, 7, 8}. The probability is the ratio of favorable outcomes to total outcomes, which gives 3/8.

Step-by-Step Solution

1

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

The total number of equal sectors is

8

, so total outcomes =

8

.

The spinner is divided into 8 equal sections numbered 1 through 8.

2

Identify the favorable outcomes that satisfy the condition 'strictly greater than 5'.

The numbers strictly greater than

5

are

6

,

7

, and

8

. Thus, there are

3

favorable outcomes.

Strictly greater than 5 excludes 5 itself.

3

Calculate the probability using the basic probability formula.

P(\text{greater than } 5) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8}

.

Theoretical probability is the ratio of favorable outcomes to total sample space outcomes.

Key Concept

Basic Probability of Simple Events

Estimated Time:45s

Question 14Question →

A juice bar allows customers to create a custom smoothie by selecting exactly $1$ fruit base out of $4$ available options and $1$ liquid base out of $3$ available options. How many different custom smoothie combinations of $1$ fruit base and $1$ liquid base are possible?

Show answer & explanation

Answer: 12

Answer

The total number of possible smoothie combinations is 12.

According to the Fundamental Counting Principle, if there are

m

ways to make a first choice and

n

ways to make a second choice, there are

m \times n

total possible pairs. Here, multiplying

4

fruit choices by

3

liquid choices yields

4 \times 3 = 12

total smoothie combinations.

Step-by-Step Solution

1

Determine the number of options for each independent choice

Fruit base options = 4; Liquid base options = 3

Each choice is independent of the other.

2

Multiply the number of options for each choice using the Fundamental Counting Principle

4 x 3 = 12

To find the total number of combinations across multiple independent categories, multiply the number of choices in each category together.

Key Concept

Fundamental Counting Principle

Estimated Time:45s

Question 15Question →

A high school student council must select a 3-member executive committee consisting of a President, a Vice President, and a Treasurer from a group of 6 juniors and 4 seniors. If the positions are filled sequentially at random and no student can hold more than one position, what is the probability that the President is a senior, the Vice President is a junior, and the Treasurer is a senior?

\frac{1}{10}

\frac{3}{10}

\frac{2}{15}

\frac{12}{125}

\frac{17}{120}

Show answer & explanation

Answer: $\frac{1}{10}$

Answer

The probability that the President is a senior, the Vice President is a junior, and the Treasurer is a senior is

\frac{1}{10}

.

To find the probability of dependent sequential events, multiply the probability of each event given that the preceding events have occurred. First, selecting a senior President has a probability of

\frac{4}{10}

. Next, selecting a junior Vice President from the remaining 9 students has a probability of

\frac{6}{9}

. Finally, selecting another senior as Treasurer from the remaining 8 students (which now contains 3 seniors) has a probability of

\frac{3}{8}

. Multiplying these probabilities yields

\frac{4}{10} \times \frac{6}{9} \times \frac{3}{8} = \frac{72}{720} = \frac{1}{10}

.

Step-by-Step Solution

1

Determine the probability of choosing a senior as President.

There are 4 seniors out of 10 total students, so

P(\text{President is Senior}) = \frac{4}{10}

.

The first selection is made from the full group of 10 students.

2

Determine the probability of choosing a junior as Vice President given the first event.

There are 6 juniors remaining out of 9 total remaining students, so

P(\text{VP is Junior} \mid \text{President is Senior}) = \frac{6}{9}

.

One student (a senior) has already been selected, reducing the total count to 9.

3

Determine the probability of choosing a senior as Treasurer given the first two events.

There are 3 seniors remaining out of 8 total remaining students, so

P(\text{Treasurer is Senior} \mid \text{first two selections}) = \frac{3}{8}

.

Two students have now been selected, leaving 8 total students and 3 remaining seniors.

4

Multiply the sequential conditional probabilities together.

\frac{4}{10} \times \frac{6}{9} \times \frac{3}{8} = \frac{72}{720} = \frac{1}{10}

.

According to the Multiplication Rule for Dependent Events, the joint probability is the product of the sequential probabilities.

Key Concept

Probability of Dependent Compound Events

Estimated Time:1m 30s

Question 16Question →

A software development team consists of $4$ front-end developers, $5$ back-end developers, and $3$ quality assurance (QA) engineers. The project manager needs to select a $3$ -member subcommittee to present a new product feature. If the subcommittee must contain at least $1$ front-end developer and at least $1$ back-end developer, how many different $3$ -member subcommittees can be formed?

Show answer & explanation

Answer: 130

Answer

The total number of valid 3-member subcommittees that can be formed is 130.

To form a 3-member committee with at least 1 front-end developer and at least 1 back-end developer from 4 front-end, 5 back-end, and 3 QA engineers, we break the problem into three mutually exclusive valid scenarios: selecting 1 member from each role gives

4 \times 5 \times 3 = 60

ways; selecting 2 front-end and 1 back-end developer gives

C(4,2) \times C(5,1) = 6 \times 5 = 30

ways; and selecting 1 front-end and 2 back-end developers gives

C(4,1) \times C(5,2) = 4 \times 10 = 40

ways. Summing these possibilities gives

60 + 30 + 40 = 130

. Alternatively, subtracting invalid committees (those with no front-end developers:

C(8,3) = 56

, no back-end developers:

C(7,3) = 35

, minus double-counted 3 QA engineers:

1

) from total 3-member committees (

C(12,3) = 220

) yields

220 - (56 + 35 - 1) = 130

.

Step-by-Step Solution

1

Determine the required subcommittee size and role counts

The team has 4 front-end developers, 5 back-end developers, and 3 QA engineers (12 members total). A subcommittee of size 3 is required.

Establishing the total population and category counts is necessary before computing combination constraints.

2

Enumerate the mutually exclusive cases that satisfy all constraints

Case 1: (1 front-end, 1 back-end, 1 QA)
Case 2: (2 front-end, 1 back-end, 0 QA)
Case 3: (1 front-end, 2 back-end, 0 QA)

The committee requires at least 1 front-end and at least 1 back-end member in a 3-person group.

3

Compute combinations for each valid case using the combination formula

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

Case 1:

C(4,1) \times C(5,1) \times C(3,1) = 4 \times 5 \times 3 = 60

Case 2:

C(4,2) \times C(5,1) \times C(3,0) = 6 \times 5 \times 1 = 30

Case 3:

C(4,1) \times C(5,2) \times C(3,0) = 4 \times 10 \times 1 = 40

Order of selection does not matter when forming a committee, so combinations are used.

4

Add the counts from all mutually exclusive cases

Total valid subcommittees =

60 + 30 + 40 = 130

According to the addition rule of counting, the total number of outcomes across disjoint cases is the sum of their individual counts.

Key Concept

Counting combinations across multiple groups with specific distribution constraints

Question 17Question →

An art gallery features a collection of $60$ paintings, each categorized by its medium (oil or watercolor) and its historical period ( $19\text{th}$ -century or $20\text{th}$ -century). Among the collection, $35$ paintings are $19\text{th}$ -century works, $40$ paintings are oil paintings, and $10$ paintings are $20\text{th}$ -century watercolor works. If a single painting is selected at random from the collection, what is the probability that the selected painting is a $19\text{th}$ -century oil painting?

\frac{1}{6}

\frac{1}{4}

\frac{7}{18}

\frac{5}{12}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{5}{12}$

Answer

\frac{5}{12}

To find the probability of choosing a

19\text{th}

-century oil painting, determine the number of paintings that fit both criteria. Out of

60

paintings,

40

are oil, meaning

20

are watercolors. Given that

10

watercolors are

20\text{th}

-century, the remaining

10

watercolors must be

19\text{th}

-century. Since there are

35

total

19\text{th}

-century paintings, subtracting the

10

19\text{th}

-century watercolors yields

25

19\text{th}

-century oil paintings. The probability is therefore

\frac{25}{60}

, which simplifies to

\frac{5}{12}

.

Step-by-Step Solution

1

Find the total number of watercolor paintings in the collection.

Since there are

60

total paintings and

40

oil paintings, the number of watercolor paintings is

60 - 40 = 20

.

Paintings are categorized into oil or watercolor, so the remaining paintings must be watercolors.

2

Find the number of

19\text{th}

-century watercolor paintings.

There are

20

total watercolor paintings and

10

are from the

20\text{th}

century, so the number of

19\text{th}

-century watercolors is

20 - 10 = 10

.

Subtracting the

20\text{th}

-century watercolors from the total watercolors gives the

19\text{th}

-century watercolors.

3

Calculate the number of

19\text{th}

-century oil paintings.

There are

35

total

19\text{th}

-century paintings. Subtracting the

10

19\text{th}

-century watercolors leaves

35 - 10 = 25

19\text{th}

-century oil paintings.

All

19\text{th}

-century paintings are either oil or watercolor.

4

Compute the probability of picking a

19\text{th}

-century oil painting.

The probability is

\frac{25}{60} = \frac{5}{12}

.

Divide the number of favorable outcomes (

25

) by the total number of paintings in the sample space (

60

).

Key Concept

Two-way categorization and conditional counting for compound probabilities

Estimated Time:2m 0s

Question 18Question →

A container holds $40$ marbles, each of which is red, blue, or green. The probability of randomly drawing a red marble from the container is $\frac{2}{5}$ . Among the remaining marbles, the ratio of blue marbles to green marbles is $1:3$ . If one marble is selected at random from the container, what is the probability that it is green?

\frac{3}{20}

\frac{1}{5}

\frac{9}{20}

\frac{3}{5}

\frac{3}{4}

Show answer & explanation

Answer: $\frac{9}{20}$

Answer

The probability that the randomly selected marble is green is

\frac{9}{20}

.

The correct answer is

\frac{9}{20}

. The probability of drawing a non-red marble is

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

. Given that the non-red marbles are split in a

1:3

ratio between blue and green, green marbles represent

\frac{3}{1+3} = \frac{3}{4}

of the non-red group. Multiplying the probability of drawing a non-red marble by the proportion of green marbles within that group gives

\frac{3}{5} \times \frac{3}{4} = \frac{9}{20}

.

Step-by-Step Solution

1

Find the total number of red marbles and the number of remaining marbles.

Since the probability of drawing a red marble is

\frac{2}{5}

, the number of red marbles is

\frac{2}{5} \times 40 = 16

. The remaining number of marbles (blue and green) is

40 - 16 = 24

.

Determining the count of non-red marbles isolates the sample space for blue and green marbles.

2

Use the ratio of blue to green marbles to find the number of green marbles.

The ratio of blue to green is

1:3

, meaning there are

1 + 3 = 4

equal parts. Each part contains

\frac{24}{4} = 6

marbles. Therefore, there are

3 \times 6 = 18

green marbles.

Converting a part-to-part ratio into actual counts allows calculation of the favorable outcome.

3

Calculate the probability of randomly drawing a green marble.

The probability is the number of green marbles divided by the total number of marbles:

\frac{18}{40} = \frac{9}{20}

.

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

Key Concept

Combining complement probability rules with part-to-whole ratio conversions to compute event probabilities.

Estimated Time:1m 30s

Question 19Question →

A community library's summer program requires participants to create a 3-book reading list consisting of exactly $1$ biography, $1$ science fiction novel, and $1$ history book selected from a featured list. The featured list contains $5$ biographies (2 of which have over $400$ pages), $6$ science fiction novels (3 of which have over $400$ pages), and $4$ history books (1 of which has over $400$ pages). If a participant selects $1$ book of each genre at random from the featured list, what is the probability that at least $1$ of the selected books has over $400$ pages? Express your answer as a decimal.

Show answer & explanation

Answer: 0.775

Answer

0.775

To find the probability of selecting at least one book over

400

pages, it is most efficient to use the complement rule:

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

. There are

5 \times 6 \times 4 = 120

total 3-book combinations. The number of books with

400

pages or fewer in each category are

3

biographies,

3

sci-fi novels, and

3

history books. Thus, there are

3 \times 3 \times 3 = 27

combinations with no books over

400

pages. The probability of choosing no books over

400

pages is

\frac{27}{120} = 0.225

. Subtracting this from

1

gives

1 - 0.225 = 0.775

.

Step-by-Step Solution

1

Calculate the total number of possible combinations of selecting 1 book from each genre

Total combinations =

5 \times 6 \times 4 = 120

By the Fundamental Counting Principle, multiplying the number of choices in each independent category yields the total outcomes.

2

Determine the complement event: selecting a reading list where NO book has over 400 pages

Available books of 400 pages or fewer: 3 biographies, 3 sci-fi novels, and 3 history books

Subtracting the number of books over 400 pages from the total in each category gives the count of books with 400 pages or fewer.

3

Calculate the number of combinations consisting entirely of books with 400 pages or fewer

Complement combinations =

3 \times 3 \times 3 = 27

Applying the Fundamental Counting Principle to the non-qualifying choices gives the total outcomes for the complement event.

4

Find the probability of the complement event and subtract from 1 to find the target probability

P(\text{at least one over } 400) = 1 - \frac{27}{120} = 1 - 0.225 = 0.775

The complement rule states that

P(A) = 1 - P(A')

, which is much more efficient than calculating probabilities for 1, 2, or 3 long books separately.

Key Concept

Complementary Probability and Fundamental Counting Principle

Estimated Time:1m 30s

Question 20Question →

A security code consists of four characters generated in order: a letter, a digit, a letter, and a digit.

- The 1st character is a letter chosen from the 5-element set $\{A, B, C, D, E\}$ .
- The 2nd character is a digit chosen from the 9 non-zero digits $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ .
- The 3rd character is a letter chosen from $\{A, B, C, D, E\}$ , but it cannot be the same as the 1st character.
- The 4th character is an even digit chosen from $\{2, 4, 6, 8\}$ , but it cannot be the same as the 2nd character.

How many different security codes can be formed following these rules?

Show answer & explanation

Answer: 640

Answer

640

The problem requires counting combinations with dependent restrictions, which can be solved by splitting into two mutually exclusive cases. If the 2nd character is an odd digit (5 possibilities), the 4th character has all 4 even digits available, giving 5 × 5 × 4 × 4 = 400 valid codes. If the 2nd character is an even digit (4 possibilities), the 4th character cannot match it, leaving 3 options, giving 5 × 4 × 4 × 3 = 240 valid codes. Summing the two distinct cases gives 400 + 240 = 640.

Step-by-Step Solution

1

Analyze the conditions for the digit positions (2nd and 4th characters).

The restrictions require dividing the counting into two cases based on the parity of the 2nd character.

The 4th character must be chosen from the 4 even digits {2, 4, 6, 8}. If the 2nd character is an odd digit, it does not reduce the choices for the 4th character. If the 2nd character is an even digit, it reduces the remaining choices for the 4th character from 4 to 3.

2

Calculate total possibilities for Case 1 (2nd character is odd).

5 × 5 × 4 × 4 = 400 codes.

There are 5 options for the 1st letter, 5 odd digits {1, 3, 5, 7, 9} for the 2nd character, 4 remaining letters for the 3rd character, and 4 even digits for the 4th character.

3

Calculate total possibilities for Case 2 (2nd character is even).

5 × 4 × 4 × 3 = 240 codes.

There are 5 options for the 1st letter, 4 even digits {2, 4, 6, 8} for the 2nd character, 4 remaining letters for the 3rd character, and 3 remaining even digits for the 4th character.

4

Sum the results of the two cases.

400 + 240 = 640 security codes.

According to the Addition Principle of counting, the total number of outcomes for mutually exclusive events is the sum of their individual outcomes.

Key Concept

Fundamental Counting Principle with Case Analysis

Basic Probability and Counting Methods

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