ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Basic Counting For Probability
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Basic Counting For ProbabilityQuestion 1 of 20

In a class of 25 students, 15 are taking math and 10 are taking science. If a student is chosen at random, what is the probability that the student is taking math?

12\frac{1}{2}
1510\frac{15}{10}
25\frac{2}{5}
35\frac{3}{5}
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Basic Counting For Probability

Practice Basic Counting For Probability in ACCUPLACER Quantitative Reasoning, Algebra & Statistics 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 Basic Counting For Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

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

In a class of 25 students, 15 are taking math and 10 are taking science. If a student is chosen at random, what is the probability that the student is taking math?

  1. 12\frac{1}{2}
  2. 1510\frac{15}{10}
  3. 25\frac{2}{5}
  4. 35\frac{3}{5} (correct answer)
Explanation: When you encounter a probability question, you're looking for the ratio of favorable outcomes to total possible outcomes. The formula is: Probability = (Number of favorable outcomes) ÷ (Total number of possible outcomes). In this problem, you want to find the probability that a randomly chosen student is taking math. There are 15 students taking math (favorable outcomes) out of 25 total students (total possible outcomes). So the probability is 1525\frac{15}{25}. To simplify this fraction, divide both numerator and denominator by their greatest common factor, which is 5: 15÷525÷5=35\frac{15÷5}{25÷5} = \frac{3}{5}. Looking at the wrong answers: Choice A (12\frac{1}{2}) might tempt you if you mistakenly thought about equal likelihood, but 15 out of 25 is not half. Choice B (1510\frac{15}{10}) incorrectly uses the number of science students (10) in the denominator instead of the total class size—this gives a probability greater than 1, which is impossible. Choice C (25\frac{2}{5}) represents the probability that a student is taking science (10 out of 25 students), not math. The correct answer is D: 35\frac{3}{5}. Study tip: Always check that your probability is between 0 and 1, and make sure you're using the total number of possibilities in the denominator, not just one subset. If your fraction is greater than 1 or uses the wrong total, you've made an error in setting up the problem.

Question 2

From a standard 52-card deck, two cards are drawn without replacement. What is the probability that both cards are face cards (Jack, Queen, or King)?

  1. 1169\frac{1}{169}
  2. 12221\frac{12}{221}
  3. 352\frac{3}{52}
  4. 11221\frac{11}{221} (correct answer)
Explanation: There are 12 face cards in a 52-card deck (3 face cards per suit ×\times 4 suits). The probability of the first card being a face card is 1252\frac{12}{52}. After drawing one face card, there are 11 face cards left and 51 total cards remaining. The probability of the second card also being a face card is 1151\frac{11}{51}. The probability of both events happening is the product of their probabilities: 1252×1151=313×1151=33663\frac{12}{52} \times \frac{11}{51} = \frac{3}{13} \times \frac{11}{51} = \frac{33}{663}. Simplifying this fraction by dividing the numerator and denominator by 3 gives 11221\frac{11}{221}.

Question 3

There are 7 runners in a race. Medals are awarded for first, second, and third place. What is the probability that a specific prediction of the top three runners in the correct order is accurate?

  1. 16\frac{1}{6}
  2. 135\frac{1}{35}
  3. 1210\frac{1}{210} (correct answer)
  4. 1343\frac{1}{343}
Explanation: This is a permutation problem because the order of the runners matters (first, second, third are distinct). The total number of ways the top three places can be filled from 7 runners is P(7,3)=7×6×5=210P(7, 3) = 7 \times 6 \times 5 = 210. A specific prediction of the top three in order (e.g., Runner A first, Runner B second, Runner C third) is exactly one of these possible outcomes. Therefore, the probability of this single outcome occurring is 1210\frac{1}{210}.

Question 4

A company assigns a 4-digit security code using digits 0 through 9, where digits can be repeated. What is the probability that a randomly chosen code contains at least one digit '7'?

  1. 110000\frac{1}{10000}
  2. 410\frac{4}{10}
  3. 343910000\frac{3439}{10000} (correct answer)
  4. 656110000\frac{6561}{10000}
Explanation: The total number of possible 4-digit codes with repetition is 104=10,00010^4 = 10,000. It is easier to calculate the probability of the complement event: that the code contains NO '7's. For each of the 4 digits, there are 9 choices (0-6, 8-9). So, the number of codes with no '7's is 94=65619^4 = 6561. The probability of a code having no '7's is 656110000\frac{6561}{10000}. The probability of a code having at least one '7' is 1−P(no ’7’s)=1−656110000=3439100001 - P(\text{no '7's}) = 1 - \frac{6561}{10000} = \frac{3439}{10000}.

Question 5

A three-digit area code is formed using the digits 0-9. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and no digits may be repeated. What is the probability that a randomly formed code is 203?

  1. 1112\frac{1}{112}
  2. 1160\frac{1}{160}
  3. 1144\frac{1}{144}
  4. 1128\frac{1}{128} (correct answer)
Explanation: First, we determine the total number of possible area codes that fit the rules. First digit: 8 choices (2-9). Second digit: 2 choices (0-1). Third digit: There are 10 total digits. Since two have been used and cannot be repeated, there are 10−2=810 - 2 = 8 choices for the third digit. The total number of possible codes is 8×2×8=1288 \times 2 \times 8 = 128. The code '203' is one specific outcome that fits these rules (2 is not 0/1, 0 is 0/1, and 3 is not 2 or 0). Therefore, the probability of forming this specific code is 1128\frac{1}{128}.

Question 6

A four-character password is created using two letters from the set {A, B, C, D, E} followed by two digits from the set {1, 2, 3, 4}. Repetition of letters is allowed, but repetition of digits is not. What is the probability that a randomly generated password is 'AB12'?

  1. 1100\frac{1}{100}
  2. 1300\frac{1}{300} (correct answer)
  3. 1400\frac{1}{400}
  4. 1600\frac{1}{600}
Explanation: To find the total number of possible passwords, we use the multiplication principle. There are 5 choices for the first letter and 5 for the second (repetition allowed), so 5×5=255 \times 5 = 25 letter combinations. There are 4 choices for the first digit and 3 for the second (no repetition), so 4×3=124 \times 3 = 12 digit combinations. The total number of unique passwords is 25×12=30025 \times 12 = 300. The password 'AB12' is one specific outcome. Therefore, the probability is 1300\frac{1}{300}.

Question 7

A club has 10 members. A president, vice-president, and treasurer are to be selected. If the positions are filled at random, what is the probability that Amy is president, Ben is vice-president, and Carla is treasurer?

  1. 11000\frac{1}{1000}
  2. 1720\frac{1}{720} (correct answer)
  3. 1120\frac{1}{120}
  4. 310\frac{3}{10}
Explanation: The total number of ways to select and assign three ordered positions from 10 members is a permutation: P(10,3)=10×9×8=720P(10, 3) = 10 \times 9 \times 8 = 720. This is the total number of possible outcomes. The specific outcome 'Amy as president, Ben as vice-president, and Carla as treasurer' is only one of these possible outcomes. Therefore, the probability is 1720\frac{1}{720}.

Question 8

In a group of 30 students, 18 take chemistry and 12 take physics. Of these, 5 students take both chemistry and physics. If one student is selected at random, what is the probability that the student takes chemistry but not physics?

  1. 1330\frac{13}{30} (correct answer)
  2. 16\frac{1}{6}
  3. 730\frac{7}{30}
  4. 35\frac{3}{5}
Explanation: We are looking for the number of students who take only chemistry. The total number of students taking chemistry is 18. This total includes the 5 students who also take physics. To find the number of students who take only chemistry, we subtract the students taking both from the total chemistry students: 18−5=1318 - 5 = 13. These 13 students take chemistry but not physics. The total number of students in the group is 30. Therefore, the probability of selecting a student who takes only chemistry is 1330\frac{13}{30}.

Question 9

A cafeteria offers 4 types of sandwiches, 3 types of drinks, and 2 types of chips. If a student randomly selects one item from each category, what is the probability that they select a turkey sandwich (one of the 4 sandwich types), water (one of the 3 drink types), and barbecue chips (one of the 2 chip types)?

  1. 124\frac{1}{24} (correct answer)
  2. 19\frac{1}{9}
  3. 324\frac{3}{24}
  4. 13\frac{1}{3}
Explanation: The total number of possible meal combinations is 4 × 3 × 2 = 24. There is exactly 1 way to select turkey sandwich, water, and barbecue chips. Therefore, the probability is 124\frac{1}{24}. Choice B incorrectly uses only drinks and chips (3 × 2 = 6, giving 16\frac{1}{6} which isn't listed, but students might confuse this with 19\frac{1}{9}). Choice C incorrectly assumes 3 favorable outcomes. Choice D incorrectly treats this as selecting from just 3 categories without considering the specific items within each.

Question 10

In a class of 20 students, 12 students play basketball, 8 students play soccer, and 5 students play both sports. If a student is selected at random, what is the probability that the student plays exactly one sport?

  1. 34\frac{3}{4}
  2. 12\frac{1}{2} (correct answer)
  3. 25\frac{2}{5}
  4. 35\frac{3}{5}
Explanation: Students playing exactly one sport = (students playing only basketball) + (students playing only soccer). Students playing only basketball = 12 - 5 = 7. Students playing only soccer = 8 - 5 = 3. Total playing exactly one sport = 7 + 3 = 10. Probability = 1020=12\frac{10}{20} = \frac{1}{2}. Choice A (34\frac{3}{4}) might result from including students who play both sports. Choice C (25=820\frac{2}{5} = \frac{8}{20}) might come from using only soccer players. Choice D (35=1220\frac{3}{5} = \frac{12}{20}) might come from using only basketball players.

Question 11

A fair six-sided die is rolled three times. What is the probability that the sum of the three rolls is exactly 7?

  1. 572\frac{5}{72}
  2. 15216\frac{15}{216} (correct answer)
  3. 136\frac{1}{36}
  4. 7108\frac{7}{108}
Explanation: Total possible outcomes = 6³ = 216. To find outcomes that sum to 7, we count ordered triples (a,b,c) where a+b+c=7 and 1≤a,b,c≤6. The valid combinations are: (1,1,5), (1,2,4), (1,3,3), (1,4,2), (1,5,1), (2,1,4), (2,2,3), (2,3,2), (2,4,1), (3,1,3), (3,2,2), (3,3,1), (4,1,2), (4,2,1), (5,1,1). That's 15 favorable outcomes. Probability = 15216\frac{15}{216}. Note that 572=15216\frac{5}{72} = \frac{15}{216}, so choice A is equivalent. Choice C assumes two dice. Choice D represents a miscount of favorable outcomes.

Question 12

A box contains 8 identical balls numbered 1 through 8. Three balls are drawn simultaneously without replacement. What is the probability that the three numbers drawn are consecutive integers?

  1. 956\frac{9}{56}
  2. 656\frac{6}{56}
  3. 18\frac{1}{8}
  4. 328\frac{3}{28} (correct answer)
Explanation: When you encounter probability questions involving "without replacement" and specific patterns, you need to count favorable outcomes and divide by total possible outcomes. First, find the total ways to choose 3 balls from 8: (83)=8!3!(8−3)!=8×7×63×2×1=56\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 Next, count the favorable outcomes (three consecutive integers). The possible consecutive triplets from balls numbered 1-8 are: (1,2,3), (2,3,4), (3,4,5), (4,5,6), (5,6,7), and (6,7,8). That's exactly 6 favorable outcomes. Therefore, the probability is 656=328\frac{6}{56} = \frac{3}{28}, which is answer D. Let's examine why the other answers are incorrect: A) 956\frac{9}{56} suggests there are 9 consecutive triplets, but you can only form 6 consecutive triplets from the numbers 1-8. B) 656\frac{6}{56} correctly identifies 6 favorable outcomes but fails to reduce the fraction to lowest terms. Always simplify your final answer. C) 18\frac{1}{8} equals 756\frac{7}{56}, which doesn't match our calculation and likely comes from incorrectly thinking there are 7 favorable outcomes. For combination probability problems, always use this systematic approach: identify what makes an outcome "favorable," count those outcomes carefully, calculate total possible outcomes using combinations, then form your fraction and reduce it. Don't forget that final simplification step—it's often what distinguishes the correct answer from a trap.

Question 13

A standard six-sided die is rolled once. What is the probability of rolling an even number?

  1. 12\frac{1}{2} (correct answer)
  2. 26\frac{2}{6}
  3. 13\frac{1}{3}
  4. 34\frac{3}{4}
Explanation: The correct answer is A. A standard die has faces numbered 1, 2, 3, 4, 5, 6. The even numbers are 2, 4, and 6, so there are 3 favorable outcomes out of 6 total outcomes. The probability is 36=12\frac{3}{6} = \frac{1}{2}. Choice B (26\frac{2}{6}) counts only two even numbers instead of three. Choice C (13\frac{1}{3}) incorrectly counts the total outcomes. Choice D (34\frac{3}{4}) uses an incorrect denominator of 4.

Question 14

A coin is flipped 3 times. What is the probability of getting exactly 2 heads?

  1. 18\frac{1}{8}
  2. 28\frac{2}{8}
  3. 38\frac{3}{8} (correct answer)
  4. 68\frac{6}{8}
Explanation: When you encounter probability questions involving repeated independent events like coin flips, you need to identify all possible outcomes and count how many satisfy your condition. To find the probability of exactly 2 heads in 3 flips, first determine the total possible outcomes. Each flip has 2 possibilities (heads or tails), so 3 flips give you 23=82^3 = 8 total outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Now count the favorable outcomes with exactly 2 heads: HHT, HTH, and THH. That's 3 outcomes out of 8 total, giving you a probability of 38\frac{3}{8}. Looking at the wrong answers: Choice A (18\frac{1}{8}) represents the probability of one specific outcome, like getting exactly HHH or TTT. Choice B (28\frac{2}{8}) might tempt students who incorrectly think there are only 2 ways to get exactly 2 heads, forgetting that the position of the single tail matters. Choice D (68\frac{6}{8}) is far too high and might result from incorrectly calculating the probability of getting "at least" 2 heads instead of "exactly" 2 heads. For future probability problems with multiple trials, remember to systematically list all outcomes when the numbers are manageable, or use the combination formula (nk)\binom{n}{k} for larger problems. The key is distinguishing between "exactly," "at least," and "at most" in the question wording, as these lead to very different calculations.

Question 15

A bag contains 3 red marbles, 4 blue marbles, and 2 green marbles. If one marble is drawn at random, what is the probability that it is blue?

  1. 49\frac{4}{9} (correct answer)
  2. 14\frac{1}{4}
  3. 45\frac{4}{5}
  4. 13\frac{1}{3}
Explanation: The correct answer is A. There are 4 blue marbles out of a total of 3 + 4 + 2 = 9 marbles. The probability is therefore 49\frac{4}{9}. Choice B (14\frac{1}{4}) incorrectly uses only the blue marbles as the denominator. Choice C (45\frac{4}{5}) incorrectly excludes one color from the total count. Choice D (13\frac{1}{3}) treats each color as equally likely regardless of the actual count.

Question 16

A deck of cards has 52 cards with 4 suits. If one card is drawn at random, what is the probability that it is NOT a heart?

  1. 14\frac{1}{4}
  2. 34\frac{3}{4} (correct answer)
  3. 12\frac{1}{2}
  4. 1339\frac{13}{39}
Explanation: When you encounter probability questions about something NOT happening, you're dealing with complement probability. The complement of an event is everything that isn't that event, and the probabilities of an event and its complement always add up to 1. A standard deck has 52 cards divided equally among 4 suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards. Since you want the probability of drawing a card that is NOT a heart, you need to count all the non-heart cards. Non-heart cards include diamonds, clubs, and spades: 13 + 13 + 13 = 39 cards. So the probability is 3952=34\frac{39}{52} = \frac{3}{4}. You can also solve this using complement probability: P(not heart) = 1 - P(heart) = 1 - 1352\frac{13}{52} = 1 - 14\frac{1}{4} = 34\frac{3}{4}. Choice A (14\frac{1}{4}) gives you the probability of drawing a heart, not avoiding one. This is the complement of what you actually want. Choice C (12\frac{1}{2}) would be correct if the deck had only 2 suits instead of 4, or if you were looking for red vs. black cards. Choice D (1339\frac{13}{39}) incorrectly uses 39 as the denominator, which would mean the deck only had 39 cards total. The correct answer is B. Strategy tip: For "NOT" probability questions, you can either count what you want directly or use the complement formula: P(not A) = 1 - P(A). Both methods should give the same answer, so use whichever feels more natural.

Question 17

A spinner has 8 equal sections: 3 yellow, 3 red, and 2 blue. What is the probability of spinning red or blue?

  1. 68\frac{6}{8}
  2. 38\frac{3}{8}
  3. 28\frac{2}{8}
  4. 58\frac{5}{8} (correct answer)
Explanation: When you see probability questions involving "or" scenarios, you're being tested on your ability to add probabilities of separate events that cannot happen simultaneously. To find the probability of spinning red or blue, you need to add the probability of spinning red to the probability of spinning blue. Start by identifying what you know: the spinner has 8 equal sections total, with 3 red sections and 2 blue sections. The probability of spinning red is 38\frac{3}{8} (3 red sections out of 8 total). The probability of spinning blue is 28\frac{2}{8} (2 blue sections out of 8 total). Since these events cannot happen on the same spin, you add these probabilities: 38+28=58\frac{3}{8} + \frac{2}{8} = \frac{5}{8}. Looking at the wrong answers: Choice A (68\frac{6}{8}) incorrectly adds all the colored sections (3 yellow + 3 red), missing that the question asks specifically for red or blue. Choice B (38\frac{3}{8}) gives only the probability of spinning red, ignoring the blue sections entirely. Choice C (28\frac{2}{8}) gives only the probability of spinning blue, ignoring the red sections. The correct answer is D: 58\frac{5}{8}. Remember this pattern: for "or" probability questions with mutually exclusive events (events that can't happen simultaneously), add the individual probabilities. Always double-check that you're including all the relevant outcomes mentioned in the question while excluding any irrelevant ones.

Question 18

A restaurant offers 4 types of soup, 6 types of sandwiches, and 3 types of drinks. How many different lunch combinations are possible if you choose one item from each category?

  1. 13 different combinations
  2. 72 different combinations (correct answer)
  3. 24 different combinations
  4. 18 different combinations
Explanation: When you encounter a problem asking about combinations where you're choosing one item from each of several categories, you're dealing with the fundamental counting principle (also called the multiplication principle). This principle states that if you have multiple independent choices to make, you multiply the number of options for each choice. In this restaurant problem, you're making three independent decisions: selecting one soup, one sandwich, and one drink. Since these choices don't affect each other, you multiply the number of options: 4×6×3=724 \times 6 \times 3 = 72 different lunch combinations. Looking at the wrong answers: Choice A (13) represents adding the options instead of multiplying them: 4+6+3=134 + 6 + 3 = 13. This is a common error when students confuse counting principles. Choice C (24) might result from multiplying only two categories, like 4×6=244 \times 6 = 24, and forgetting about the drinks. Choice D (18) could come from incorrectly calculating 6×3=186 \times 3 = 18 and omitting the soup options. The correct answer is B: 72 different combinations. Remember this key distinction: when you're choosing one item from each category (like building a complete meal), multiply the number of options. When you're choosing between categories (like deciding whether to order soup OR a sandwich), you would add. Watch for keywords like "one from each" versus "one of the following" to identify which counting method to use.

Question 19

A box contains 12 pencils: 7 are sharpened and 5 are unsharpened. If 2 pencils are drawn at random without replacement, how many different ways can this be done?

  1. 24 different ways
  2. 144 different ways
  3. 66 different ways (correct answer)
  4. 12 different ways
Explanation: This question tests your understanding of combinations, which count the number of ways to select items when order doesn't matter. When you see "how many different ways" to choose or draw items without caring about sequence, think combinations. To find the number of ways to draw 2 pencils from 12 total pencils, you use the combination formula: C(n,r)=n!r!(n−r)!C(n,r) = \frac{n!}{r!(n-r)!} where n is the total number of items and r is the number being selected. Here, C(12,2)=12!2!(12−2)!=12!2!×10!=12×112×1=1322=66C(12,2) = \frac{12!}{2!(12-2)!} = \frac{12!}{2! \times 10!} = \frac{12 \times 11}{2 \times 1} = \frac{132}{2} = 66 The information about sharpened versus unsharpened pencils is irrelevant since the question simply asks for the total number of ways to draw any 2 pencils. Choice A (24) might result from incorrectly calculating 12×212 \times 2, confusing this with a simple multiplication problem. Choice B (144) comes from treating this as a permutation problem where order matters: 12×11=13212 \times 11 = 132, or possibly 122=14412^2 = 144. Choice D (12) represents a fundamental misunderstanding, perhaps thinking you can only choose as many ways as there are pencils. Remember: when a problem asks "how many ways" to select items and order doesn't matter, use combinations. The key signal is that swapping two selected items doesn't create a "different way." If the problem cared about which pencil was drawn first versus second, it would be a permutation problem instead.

Question 20

A drawer contains 6 red socks and 4 blue socks. A person pulls out one sock, does not replace it, and then pulls out a second sock. What is the probability that the two socks match in color?

  1. 245\frac{2}{45}
  2. 715\frac{7}{15} (correct answer)
  3. 815\frac{8}{15}
  4. 1325\frac{13}{25}
Explanation: This problem involves two mutually exclusive events: drawing two red socks OR drawing two blue socks. We calculate the probability of each and add them. The probability of drawing two red socks is P(R1 and R2)=P(R1)×P(R2∣R1)=610×59=3090P(R_1 \text{ and } R_2) = P(R_1) \times P(R_2|R_1) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}. The probability of drawing two blue socks is P(B1 and B2)=P(B1)×P(B2∣B1)=410×39=1290P(B_1 \text{ and } B_2) = P(B_1) \times P(B_2|B_1) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}. The total probability of a matching pair is the sum of these probabilities: 3090+1290=4290=715\frac{30}{90} + \frac{12}{90} = \frac{42}{90} = \frac{7}{15}.