Finite Mathematics Quiz: Counting For Probability

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Counting For ProbabilityQuestion 1 of 17

A committee of 5 people is to be selected from a group of 7 men and 8 women. The committee must include at least 2 men and at least 2 women. What is the probability that the committee will contain exactly 3 women?

\frac{C(8,3) \times C(7,2)}{C(15,5) - C(7,5) - C(8,5) - C(7,1) \times C(8,4) - C(7,4) \times C(8,1)}

\frac{C(8,3) \times C(7,2)}{C(15,5)}

\frac{C(8,3) \times C(7,2)}{C(7,2) \times C(8,3) + C(7,3) \times C(8,2)}

\frac{C(8,3) \times C(7,2)}{C(7,2) \times C(8,3) + C(7,3) \times C(8,2) + C(7,4) \times C(8,1) + C(7,1) \times C(8,4)}

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Finite Mathematics Quiz

Finite Mathematics Quiz: Counting For Probability

Practice Counting For Probability in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Counting For Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\frac{C(8,3) \times C(7,2)}{C(15,5) - C(7,5) - C(8,5) - C(7,1) \times C(8,4) - C(7,4) \times C(8,1)}$
$\frac{C(8,3) \times C(7,2)}{C(15,5)}$
$\frac{C(8,3) \times C(7,2)}{C(7,2) \times C(8,3) + C(7,3) \times C(8,2)}$ (correct answer)
$\frac{C(8,3) \times C(7,2)}{C(7,2) \times C(8,3) + C(7,3) \times C(8,2) + C(7,4) \times C(8,1) + C(7,1) \times C(8,4)}$

Explanation: This is conditional probability. We want P(exactly 3 women | at least 2 men and at least 2 women). With exactly 3 women, we have 2 men, satisfying the constraints. Ways to choose exactly 3 women and 2 men:

C(8,3) \times C(7,2)

. Valid committees satisfying the constraints: exactly 2 women + 3 men:

C(8,2) \times C(7,3)

, or exactly 3 women + 2 men:

C(8,3) \times C(7,2)

. (Can't have 4+ women as that leaves ≤1 man, violating the constraint; similarly for 4+ men). Probability:

\frac{C(8,3) \times C(7,2)}{C(7,2) \times C(8,3) + C(7,3) \times C(8,2)}

. Choice A overcomplicates the denominator. Choice B ignores the constraints. Choice D includes impossible cases.

Question 2

A box contains 8 red balls, 6 blue balls, and 4 green balls. Three balls are drawn simultaneously without replacement. What is the probability that at least one ball of each color is drawn?

$\frac{288}{816}$
$\frac{192}{1224}$
$\frac{384}{816}$
$\frac{192}{816}$ (correct answer)

Explanation: This problem tests your understanding of combinations and the probability of compound events. When you see "at least one of each" in a probability question, you need to count the favorable outcomes where every specified condition is met. To have at least one ball of each color when drawing 3 balls, you must draw exactly one red, one blue, and one green ball (since you're only drawing 3 total). The number of ways to do this is:

\binom{8}{1} \times \binom{6}{1} \times \binom{4}{1} = 8 \times 6 \times 4 = 192

The total number of ways to draw 3 balls from 18 total balls is:

\binom{18}{3} = \frac{18!}{3! \times 15!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816

Therefore, the probability is

\frac{192}{816}

, which is answer D. Answer A (

\frac{288}{816}

) likely comes from incorrectly calculating the favorable outcomes, perhaps by double-counting some combinations. Answer B (

\frac{192}{1224}

) uses the correct numerator but calculates the total outcomes incorrectly—1224 would be the result if you mistakenly used permutations instead of combinations or made an arithmetic error. Answer C (

\frac{384}{816}

) doubles the correct numerator, possibly from misunderstanding the constraint that you need exactly one of each color. Remember: when dealing with "at least one of each" problems with limited draws, first check if it's even possible, then calculate systematically using combinations. Always verify your total number of outcomes matches the selection method described.

Question 3

A student council consists of 12 members. They need to select a president, vice president, and secretary, where no person can hold more than one position. After these positions are filled, they will form a 4-person committee from the remaining members to plan the spring dance. What is the total number of ways this selection process can be completed?

$1320 \times 126 = 166,320$ (correct answer)
$1320 \times 84 = 110,880$
$220 \times 126 = 27,720$
$495 \times 84 = 41,580$

Explanation: First, select the 3 officers from 12 members:

P(12,3) = 12 \times 11 \times 10 = 1320

ways. Then, from the remaining 9 members, choose 4 for the committee:

C(9,4) = \frac{9!}{4!5!} = 126

ways. Total:

1320 \times 126 = 166,320

. Choice B uses

C(9,3) = 84

instead of

C(9,4)

. Choice C incorrectly uses

C(12,3) = 220

for officer selection. Choice D combines both errors.

Question 4

A bag contains 12 marbles: 5 red, 4 blue, and 3 yellow. Three marbles are drawn without replacement. Given that at least one marble drawn is red, what is the probability that exactly two of the drawn marbles are red?

$\frac{C(5,2) \times C(7,1)}{C(12,3) - C(7,3)}$
$\frac{C(5,2) \times C(7,1)}{C(12,3)}$
$\frac{C(5,2) \times C(7,1)}{C(5,1) \times C(7,2) + C(5,2) \times C(7,1) + C(5,3)}$ (correct answer)
$\frac{C(5,2) \times C(7,1)}{C(5,1) \times C(7,2) + C(5,2) \times C(7,1)}$

Explanation: This is conditional probability: P(exactly 2 red | at least 1 red). Ways to get exactly 2 red:

C(5,2) \times C(7,1) = 10 \times 7 = 70

. Ways to get at least 1 red: Total ways minus no red =

C(12,3) - C(7,3)

OR directly: 1 red:

C(5,1) \times C(7,2) = 5 \times 21 = 105

, 2 red:

C(5,2) \times C(7,1) = 70

, 3 red:

C(5,3) = 10

. Total with at least 1 red:

105 + 70 + 10 = 185

. Probability:

\frac{70}{185} = \frac{C(5,2) \times C(7,1)}{C(5,1) \times C(7,2) + C(5,2) \times C(7,1) + C(5,3)}

. Choice A uses complement form (equivalent but different expression). Choice B ignores the condition. Choice D omits the 3-red case.

Question 5

A bookshelf has 8 different mathematics books and 6 different science books. A student wants to select and arrange 5 books in a row such that no two science books are adjacent. In how many ways can this be done if at least one science book must be selected?

$P(8,5) + P(8,4) \times 5 \times P(6,1) + P(8,3) \times C(4,2) \times P(6,2)$
$P(8,4) \times 5 \times 6 + P(8,3) \times C(4,2) \times P(6,2)$ (correct answer)
$P(8,3) \times C(4,2) \times P(6,2) + P(8,4) \times C(5,1) \times 6$
$P(8,3) \times 6 \times 4 \times 5 + P(8,4) \times 6 \times 5$

Explanation: We need exactly 1 or 2 science books (can't have 3+ since they can't be adjacent in 5 positions). Case 1: Exactly 1 science book, 4 math books. Arrange 4 math books:

P(8,4)

ways. This creates 5 gaps for the science book. Choose 1 gap and 1 science book:

5 \times 6 = 30

ways. Total:

P(8,4) \times 5 \times 6

. Case 2: Exactly 2 science books, 3 math books. Arrange 3 math books:

P(8,3)

ways. This creates 4 gaps; choose 2 non-adjacent gaps:

C(4,2) = 6

ways. Arrange 2 science books in chosen gaps:

P(6,2)

ways. Total:

P(8,3) \times C(4,2) \times P(6,2)

. Final answer:

P(8,4) \times 5 \times 6 + P(8,3) \times C(4,2) \times P(6,2)

. Choice A includes the impossible all-math case. Choice C uses wrong gap selection. Choice D miscalculates gap and book selections.

Question 6

A project team of 4 people is to be formed from a group of 6 software engineers and 4 data analysts. If the team is selected at random, what is the probability that it includes at least 3 software engineers?

$\frac{8}{21}$
$\frac{1}{21}$
$\frac{19}{42}$ (correct answer)
$\frac{11}{42}$

Explanation: The total number of ways to form a 4-person team from 10 people is given by the combination formula

\binom{10}{4}

S = \binom{10}{4} = \frac{10!}{4!6!} = 210

The event of having 'at least 3 software engineers' means the team can have either exactly 3 software engineers or exactly 4 software engineers. Case 1: Exactly 3 engineers and 1 analyst. The number of ways is

\binom{6}{3} \times \binom{4}{1} = 20 \times 4 = 80

. Case 2: Exactly 4 engineers and 0 analysts. The number of ways is

\binom{6}{4} \times \binom{4}{0} = 15 \times 1 = 15

. The total number of favorable outcomes is the sum of these cases:

80 + 15 = 95

. The probability is the ratio of favorable outcomes to the total number of outcomes:

P(E) = \frac{95}{210} = \frac{19}{42}

Question 7

A manager has 7 distinct tasks to assign to three employees: Alice, Bob, and Carol. If each task is randomly assigned to one of the three employees, what is the probability that Alice is assigned exactly 3 tasks?

$\frac{\binom{7}{3}}{3^7}$
$\frac{\binom{7}{3} \cdot 3^4}{3^7}$
$\frac{P(7,3) \cdot 2^4}{3^7}$
$\frac{\binom{7}{3} \cdot 2^4}{3^7}$ (correct answer)

Explanation: The total number of ways to assign the 7 distinct tasks to 3 employees is

3^7

, since each of the 7 tasks has 3 possible assignees. To find the number of ways for Alice to get exactly 3 tasks, we first choose which 3 of the 7 tasks are assigned to her. This can be done in

\binom{7}{3}

ways. The remaining

7-3=4

tasks must be assigned to the other two employees, Bob and Carol. For each of these 4 tasks, there are 2 choices (either Bob or Carol). So, there are

2^4

ways to assign the remaining tasks. The total number of favorable outcomes is the product:

\binom{7}{3} \times 2^4

. The probability is the ratio of favorable outcomes to total outcomes:

\frac{\binom{7}{3} \cdot 2^4}{3^7}

Question 8

A security code is formed by a random arrangement of the six distinct letters A, B, C, D, E, F. What is the probability that in the chosen arrangement, the letters A and B are not next to each other?

$\frac{1}{3}$
$\frac{5}{6}$
$\frac{1}{6}$
$\frac{2}{3}$ (correct answer)

Explanation: When you encounter problems asking about arrangements where certain items are NOT next to each other, the complement approach is often the most efficient strategy. Instead of counting arrangements where A and B are separated directly, count the arrangements where they ARE together, then subtract from the total. First, find the total possible arrangements of six distinct letters:

6! = 720

. Next, count arrangements where A and B are adjacent. Treat A and B as a single unit, giving you 5 objects to arrange: (AB), C, D, E, F. These can be arranged in

5! = 120

ways. However, within their unit, A and B can be ordered as AB or BA, so multiply by 2:

120 \times 2 = 240

arrangements where A and B are together. Therefore, arrangements where A and B are NOT together:

720 - 240 = 480

. The probability is

\frac{480}{720} = \frac{2}{3}

, confirming answer D. Looking at the wrong answers: A)

\frac{1}{3}

represents the probability that A and B ARE next to each other (

\frac{240}{720}

) - this is the complement of what we want. B)

\frac{5}{6}

might result from incorrectly calculating the adjacent arrangements as

\frac{1}{6}

of the total and subtracting. C)

\frac{1}{6}

could come from dividing the number of objects incorrectly or misapplying basic probability rules. Remember: for "not adjacent" problems, use the complement rule. Calculate what you don't want, then subtract from the total - it's usually much simpler than direct counting.

Question 9

A bag contains 10 tiles, each labeled with a unique integer from the set $\{1, 2, ..., 10\}$ . If three tiles are drawn from the bag at random without replacement, what is the probability that the sum of the numbers on the three tiles is an even number?

$\frac{1}{4}$
$\frac{2}{5}$
$\frac{3}{5}$
$\frac{1}{2}$ (correct answer)

Explanation: When you encounter probability questions involving the sum of numbers being even or odd, focus on the fundamental rule: a sum is even when you have an even number of odd addends. From the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, there are 5 odd numbers (1, 3, 5, 7, 9) and 5 even numbers (2, 4, 6, 8, 10). For three numbers to sum to an even total, you need either:

All three numbers even (0 odd numbers)
Exactly two numbers odd and one even (2 odd numbers)

Let's calculate these probabilities. The total ways to choose 3 tiles from 10 is

\binom{10}{3} = 120

. For all even numbers:

\binom{5}{3} = 10

ways For exactly 2 odd, 1 even:

\binom{5}{2} \times \binom{5}{1} = 10 \times 5 = 50

ways Total favorable outcomes:

10 + 50 = 60

Probability =

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

Choice A (

\frac{1}{4}

) might result from only considering the all-even case and miscalculating. Choice B (

\frac{2}{5}

) could come from incorrectly computing

\frac{50}{120}

and forgetting the all-even case. Choice C (

\frac{3}{5}

) represents the probability of an odd sum (the complement), which occurs when you select 1 or 3 odd numbers. Study tip: For even/odd sum problems, always identify how many ways you can get an even number of odd addends, then use combinations to count systematically.

Question 10

A robot starts at the origin $(0,0)$ of a coordinate grid and must travel to the point $(5,3)$ by only moving one unit right (R) or one unit up (U) at each step. If all such paths are equally likely, what is the probability that a randomly chosen path passes through the point $(2,2)$ ?

$\frac{3}{7}$ (correct answer)
$\frac{1}{2}$
$\frac{\binom{4}{2} + \binom{4}{1}}{\binom{8}{5}}$
$\frac{5}{14}$

Explanation: A path from

(0,0)

(5,3)

requires 5 right moves and 3 up moves, for a total of 8 moves. The total number of distinct paths is the number of ways to arrange these moves:

\binom{8}{5} = \frac{8!}{5!3!} = 56

. A path that goes through

(2,2)

must first go from

(0,0)

(2,2)

and then from

(2,2)

(5,3)

. Number of paths from

(0,0)

(2,2)

: This requires 2 right and 2 up moves (4 total). Number of paths is

\binom{4}{2} = \frac{4!}{2!2!} = 6

. Number of paths from

(2,2)

(5,3)

: This is equivalent to a path from

(0,0)

(5-2, 3-2) = (3,1)

. This requires 3 right and 1 up move (4 total). Number of paths is

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

. The total number of favorable paths is the product of the paths for each segment:

6 \times 4 = 24

. The probability is the ratio of favorable paths to total paths:

P(E) = \frac{24}{56} = \frac{3}{7}

Question 11

Let $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$ . A non-empty subset of $S$ is chosen at random, with each non-empty subset being equally likely. What is the probability that the chosen subset contains both the smallest and largest elements of $S$ ?

$\frac{1}{4}$
$\frac{64}{255}$ (correct answer)
$\frac{63}{256}$
$\frac{1}{28}$

Explanation: The total number of subsets of a set with 8 elements is

2^8 = 256

. Since the chosen subset must be non-empty, the total number of possible outcomes is

2^8 - 1 = 255

. For a subset to contain both the smallest element (1) and the largest element (8), these two elements must be included. The remaining

8-2=6

elements ({2, 3, 4, 5, 6, 7}) may or may not be in the subset. For each of these 6 elements, there are two choices: either it is in the subset or it is not. Thus, there are

2^6 = 64

subsets of

S

that contain both 1 and 8. The probability is the ratio of the number of favorable subsets to the total number of non-empty subsets:

P(E) = \frac{64}{255}

Question 12

There are 8 guests at a party. Each guest's birthstone is determined by their birth month, with one unique stone for each of the 12 months of the year. Assuming each birth month is equally likely for any guest, what is the probability that at least two guests have the same birthstone?

$\frac{P(12, 8)}{12^8}$
$1 - \frac{C(12, 8)}{12^8}$
$1 - \frac{P(12, 8)}{12^8}$ (correct answer)
$1 - \frac{8!}{12^8}$

Explanation: This is a variation of the birthday problem. It is easier to calculate the probability of the complementary event: that no two guests have the same birthstone (all 8 guests have different birthstones). The total number of ways to assign birthstones to the 8 guests is

12^8

, since each of the 8 guests can have any of the 12 birthstones. The number of ways for all 8 guests to have different birthstones is the number of permutations of choosing 8 distinct items from 12, which is

P(12, 8) = \frac{12!}{(12-8)!}

. The probability of all guests having different birthstones is

P(\text{all different}) = \frac{P(12, 8)}{12^8}

. The probability of at least two guests having the same birthstone is

1 - P(\text{all different})

, which is

1 - \frac{P(12, 8)}{12^8}

Question 13

From a standard 52-card deck, a 5-card hand is dealt. What is the probability of being dealt a hand that contains exactly two pairs (e.g., two kings, two 5s, and one 8)?

$\frac{\binom{13}{2}\binom{4}{2}^2\binom{11}{1}\binom{4}{1}}{\binom{52}{5}}$ (correct answer)
$\frac{\binom{13}{2}\binom{4}{2}^2\binom{44}{1}}{\binom{52}{5}}$
$\frac{\binom{13}{1}\binom{4}{2}\binom{12}{1}\binom{4}{2}\binom{11}{1}\binom{4}{1}}{\binom{52}{5}}$
$\frac{\binom{13}{2}\binom{4}{2}^2}{\binom{52}{5}}$

Explanation: The total number of possible 5-card hands from a 52-card deck is

\binom{52}{5}

. To form a hand with exactly two pairs, we must perform the following steps:

Choose the two ranks for the pairs from the 13 available ranks (Ace through King): $\binom{13}{2}$ ways.
For each of these two ranks, choose 2 cards of the 4 suits available: $\binom{4}{2} \times \binom{4}{2}$ ways.
Choose the rank for the fifth card from the remaining $13 - 2 = 11$ ranks to avoid a full house: $\binom{11}{1}$ ways.
Choose the suit for this fifth card from the 4 available suits: $\binom{4}{1}$ ways. The total number of favorable hands is the product of these steps: $\binom{13}{2}\binom{4}{2}^2\binom{11}{1}\binom{4}{1}$ . The probability is the ratio of favorable hands to the total number of hands.

Question 14

Four married couples (8 people total) are to be seated randomly in a row of 8 chairs. What is the probability that each person is seated next to their spouse?

$\frac{1}{315}$
$\frac{1}{945}$
$\frac{1}{105}$ (correct answer)
$\frac{1}{2520}$

Explanation: The total number of ways to arrange 8 distinct people in 8 chairs is

8! = 40,320

. For the favorable outcomes, we treat each of the 4 couples as a single unit. This means we are arranging 4 units, which can be done in

4!

ways. Within each of the 4 units (couples), the two people (husband and wife) can be arranged in

2!

ways. Since there are 4 couples, this internal arrangement can be done in

(2!)^4 = 2^4 = 16

ways. The total number of favorable arrangements is the product of these two parts:

4! \times (2!)^4 = 24 \times 16 = 384

. The probability is the ratio of favorable arrangements to the total arrangements:

P(E) = \frac{384}{40,320} = \frac{1}{105}

Question 15

In a lottery game, a player selects 6 distinct numbers from 1 to 40. Later, 6 winning numbers are drawn. What is the probability that a player's ticket matches exactly 4 of the 6 winning numbers?

$\frac{\binom{6}{4}}{\binom{40}{6}}$
$\frac{\binom{6}{4}\binom{34}{2}}{\binom{40}{6}}$ (correct answer)
$\frac{\binom{6}{4} + \binom{34}{2}}{\binom{40}{6}}$
$\frac{\binom{34}{4}\binom{6}{2}}{\binom{40}{6}}$

Explanation: The total number of ways to choose 6 numbers from 40 is

\binom{40}{6}

. For the player to match exactly 4 winning numbers, their ticket must consist of 4 numbers from the 6 winning numbers and 2 numbers from the

40-6=34

losing numbers. The number of ways to choose 4 of the 6 winning numbers is

\binom{6}{4}

. The number of ways to choose 2 of the 34 losing numbers is

\binom{34}{2}

. The total number of favorable combinations is the product of these two values:

\binom{6}{4}\binom{34}{2}

. The probability is the ratio of favorable outcomes to the total number of outcomes:

\frac{\binom{6}{4}\binom{34}{2}}{\binom{40}{6}}

Question 16

A string is formed by a random arrangement of the letters in the word STATISTICS. What is the probability that the arrangement begins and ends with the letter S?

$\frac{1}{15}$ (correct answer)
$\frac{1}{12}$
$\frac{3}{50}$
$\frac{9}{100}$

Explanation: The word STATISTICS has 10 letters with repetitions: S (3), T (3), A (1), I (2), C (1). The total number of unique arrangements is:

S = \frac{10!}{3!3!2!1!1!} = \frac{3,628,800}{6 \times 6 \times 2} = 50,400

For an arrangement to begin and end with S, we fix two S's at the ends. We then need to arrange the remaining 8 letters: S (1), T (3), A (1), I (2), C (1). The number of ways to arrange these middle letters is:

E = \frac{8!}{1!3!1!2!1!} = \frac{40,320}{6 \times 2} = 3,360

The probability is the ratio of favorable arrangements to the total arrangements:

P(E) = \frac{3,360}{50,400} = \frac{336}{5040} = \frac{1}{15}

Question 17

A crate contains 20 routers, of which 4 are defective. A quality control inspector randomly selects 5 routers for testing. What is the probability that exactly 2 of the selected routers are defective?

$\frac{\binom{4}{2}}{\binom{20}{5}}$
$\frac{\binom{4}{2}\binom{16}{3}}{\binom{20}{5}}$ (correct answer)
$\binom{5}{2}(0.2)^2(0.8)^3$
$\frac{P(4, 2)P(16, 3)}{P(20, 5)}$

Explanation: This is a hypergeometric probability problem, as the selections are made without replacement. The total number of ways to choose 5 routers from 20 is

\binom{20}{5}

. To find the number of favorable outcomes, we need to select exactly 2 defective routers and 3 non-defective routers. Number of ways to choose 2 defective routers from the 4 available:

\binom{4}{2}

. Number of ways to choose 3 non-defective routers from the

20-4=16

available:

\binom{16}{3}

. The total number of favorable outcomes is the product of these two values:

\binom{4}{2}\binom{16}{3}

. The probability is the ratio of favorable outcomes to the total number of outcomes:

\frac{\binom{4}{2}\binom{16}{3}}{\binom{20}{5}}