ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Probability And Complements
20 questions · exam conditions
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Probability And ComplementsQuestion 1 of 20

A bag contains 6 red marbles, 4 blue marbles, and 5 green marbles. If one marble is drawn at random, what is the probability that the marble is not blue?

415\frac{4}{15}
14\frac{1}{4}
13\frac{1}{3}
1115\frac{11}{15}
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Probability And Complements

Practice Probability And Complements 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 Probability And Complements, 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

A bag contains 6 red marbles, 4 blue marbles, and 5 green marbles. If one marble is drawn at random, what is the probability that the marble is not blue?

  1. 415\frac{4}{15}
  2. 14\frac{1}{4}
  3. 13\frac{1}{3}
  4. 1115\frac{11}{15} (correct answer)
Explanation: The total number of marbles is 6+4+5=156 + 4 + 5 = 15. The probability of drawing a blue marble is 415\frac{4}{15}. The complement event, drawing a marble that is not blue, has a probability of 1−P(extblue)=1−415=11151 - P( ext{blue}) = 1 - \frac{4}{15} = \frac{11}{15}. Alternatively, the number of non-blue marbles is 6+5=116 + 5 = 11, so the probability is 1115\frac{11}{15}.

Question 2

Two standard six-sided dice are rolled. What is the probability that the sum of the numbers rolled is not equal to 5?

  1. 19\frac{1}{9}
  2. 16\frac{1}{6}
  3. 56\frac{5}{6}
  4. 89\frac{8}{9} (correct answer)
Explanation: There are 6×6=366 \times 6 = 36 possible outcomes. The combinations that sum to 5 are (1, 4), (2, 3), (3, 2), and (4, 1). There are 4 such outcomes. The probability of the sum being 5 is 436=19\frac{4}{36} = \frac{1}{9}. The probability of the sum not being 5 is the complement: 1−19=891 - \frac{1}{9} = \frac{8}{9}.

Question 3

An integer is selected at random from the set {1,2,3,...,25}\{1, 2, 3, ..., 25\}. What is the probability that the selected integer is not a perfect square?

  1. 15\frac{1}{5}
  2. 425\frac{4}{25}
  3. 45\frac{4}{5} (correct answer)
  4. 2125\frac{21}{25}
Explanation: The perfect squares in the set are 12=1,22=4,32=9,42=16,52=251^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25. There are 5 perfect squares. The probability of selecting a perfect square is 525=15\frac{5}{25} = \frac{1}{5}. The probability of not selecting a perfect square is the complement: 1−15=451 - \frac{1}{5} = \frac{4}{5}.

Question 4

The probability that a certain machine part is defective is 0.03. What is the probability that a randomly selected part is not defective?

  1. 0.03
  2. 0.30
  3. 0.97 (correct answer)
  4. 1.03
Explanation: The event that a part is not defective is the complement of the event that it is defective. If P(extdefective)=0.03P( ext{defective}) = 0.03, then P(extnotdefective)=1−P(extdefective)=1−0.03=0.97P( ext{not defective}) = 1 - P( ext{defective}) = 1 - 0.03 = 0.97.

Question 5

The probability of winning a game on a single try is pp. If two independent attempts are made, the probability of winning at least once is 925\frac{9}{25}. What is the probability of not winning on a single try?

  1. 45\frac{4}{5} (correct answer)
  2. 925\frac{9}{25}
  3. 1625\frac{16}{25}
  4. 15\frac{1}{5}
Explanation: Let qq be the probability of not winning on a single try. Then P(extnotwinningoneithertry)=q2P( ext{not winning on either try}) = q^2. The event 'winning at least once' is the complement of 'not winning on either try'. Therefore, P(extatleastonewin)=1−q2P( ext{at least one win}) = 1 - q^2. We are given this probability is 925\frac{9}{25}. So, 925=1−q2\frac{9}{25} = 1 - q^2. Solving for q2q^2 gives q2=1−925=1625q^2 = 1 - \frac{9}{25} = \frac{16}{25}. Taking the square root, q=45q = \frac{4}{5}. The probability of not winning on a single try is 45\frac{4}{5}.

Question 6

A security system has two independent alarms. The probability that the first alarm fails is 0.1, and the probability that the second alarm fails is 0.2. What is the probability that at least one alarm works correctly?

  1. 0.02
  2. 0.30
  3. 0.72
  4. 0.98 (correct answer)
Explanation: The event 'at least one alarm works' is the complement of the event 'both alarms fail'. Since the alarms are independent, the probability that both fail is the product of their individual failure probabilities: P(extbothfail)=0.1×0.2=0.02P( ext{both fail}) = 0.1 \times 0.2 = 0.02. The probability that at least one alarm works is 1−P(extbothfail)=1−0.02=0.981 - P( ext{both fail}) = 1 - 0.02 = 0.98.

Question 7

The probability that a customer orders a drink is 0.8. The probability that a customer orders food is 0.6. The probability a customer orders both is 0.5. What is the probability that a randomly selected customer orders neither a drink nor food?

  1. 0.1 (correct answer)
  2. 0.2
  3. 0.9
  4. 1.0
Explanation: The probability that a customer orders a drink or food is P(D∪F)=P(D)+P(F)−P(D∩F)=0.8+0.6−0.5=0.9P(D \cup F) = P(D) + P(F) - P(D \cap F) = 0.8 + 0.6 - 0.5 = 0.9. The event 'orders neither' is the complement of 'orders a drink or food'. Therefore, the probability of ordering neither is 1−P(D∪F)=1−0.9=0.11 - P(D \cup F) = 1 - 0.9 = 0.1.

Question 8

In a class of 25 students, 15 are taking biology and 12 are taking chemistry. If 8 students are taking both, what is the probability that a randomly selected student is taking neither biology nor chemistry?

  1. 825\frac{8}{25}
  2. 625\frac{6}{25} (correct answer)
  3. 1725\frac{17}{25}
  4. 1925\frac{19}{25}
Explanation: First, find the number of students taking at least one of the subjects using the inclusion-exclusion principle: N(extBiology∪extChemistry)=N(extBiology)+N(extChemistry)−N(extBoth)=15+12−8=19N( ext{Biology} \cup ext{Chemistry}) = N( ext{Biology}) + N( ext{Chemistry}) - N( ext{Both}) = 15 + 12 - 8 = 19. This means 19 students are taking at least one of the courses. The number of students taking neither is the total number of students minus this amount: 25−19=625 - 19 = 6. The probability is therefore 625\frac{6}{25}.

Question 9

A weather forecast states there is a 25% chance of rain on Saturday and a 30% chance of rain on Sunday. If the weather on the two days is independent, what is the probability that it does not rain on either day?

  1. 0.450
  2. 0.525 (correct answer)
  3. 0.550
  4. 0.925
Explanation: The probability of no rain on Saturday is 1−0.25=0.751 - 0.25 = 0.75. The probability of no rain on Sunday is 1−0.30=0.701 - 0.30 = 0.70. Since the events are independent, the probability of no rain on either day is the product of these probabilities: 0.75×0.70=0.5250.75 \times 0.70 = 0.525.

Question 10

A survey shows that 65% of students like pizza, 40% like burgers, and 25% like both pizza and burgers. If a student is selected at random and does NOT like pizza, what is the probability that this student likes burgers?

  1. 47\frac{4}{7}
  2. 25\frac{2}{5}
  3. 1535\frac{15}{35}
  4. 37\frac{3}{7} (correct answer)
Explanation: This is a conditional probability question involving set theory and survey data. When you see problems with overlapping groups (students who like both items), you need to carefully track what happens to each subset when conditions change. First, let's organize the given information. Out of all students: 65% like pizza, 40% like burgers, and 25% like both. This means 35% don't like pizza at all. Among those who don't like pizza, some will like burgers and some won't. To find how many students don't like pizza but do like burgers, subtract the overlap from the total burger lovers: 40% - 25% = 15% of all students like burgers but not pizza. Since 35% don't like pizza total, the probability that a non-pizza lover likes burgers is 15%35%=1535=37\frac{15\%}{35\%} = \frac{15}{35} = \frac{3}{7}. Looking at the wrong answers: Choice A (47\frac{4}{7}) incorrectly uses the complement of our answer, perhaps from confusing conditional probability direction. Choice B (25\frac{2}{5}) appears to come from using the wrong denominator—possibly the total percentage who like burgers (40%) instead of those who don't like pizza. Choice C (1535\frac{15}{35}) shows the unreduced fraction, which equals our answer but isn't simplified. For conditional probability problems involving surveys, always identify your new sample space first (in this case, the 35% who don't like pizza), then find what portion of that group meets your condition (the 15% who like burgers). Reduce fractions to match the answer format.

Question 11

In a game, the probability of winning on any single attempt is 0.3. A player continues until they either win or complete 3 unsuccessful attempts. What is the probability that the player does NOT win the game?

  1. 0.700
  2. 0.657
  3. 0.343 (correct answer)
  4. 0.027
Explanation: When you encounter probability questions involving repeated attempts with stopping conditions, you need to identify all the ways the described outcome can occur and calculate their combined probability. The player does NOT win the game if they fail to win in their first attempt, fail again in their second attempt, and fail once more in their third attempt. Since the probability of winning is 0.3, the probability of losing on any single attempt is 1−0.3=0.71 - 0.3 = 0.7. For three consecutive losses, you multiply the individual probabilities: 0.7×0.7×0.7=0.73=0.3430.7 \times 0.7 \times 0.7 = 0.7^3 = 0.343. Looking at the wrong answers: Choice A (0.700) represents the probability of losing just one attempt, not three consecutive attempts. Choice B (0.657) might come from incorrectly trying to add probabilities rather than multiply them, or from a flawed calculation involving the complement. Choice D (0.027) represents the probability of winning three times in a row (0.330.3^3), which isn't relevant to this question. The correct answer is C (0.343). Remember that when dealing with independent events that must all occur (like losing three attempts in a row), you multiply their individual probabilities. Also, pay careful attention to stopping conditions in probability problems—the player stops after either winning once or losing three times, so "not winning the game" specifically means losing all three allowed attempts.

Question 12

Events X and Y are independent with P(X) = 0.6 and P(Y) = 0.4. Consider the statement: 'Either X occurs or Y does not occur (or both).' What is the probability that this statement is true?

  1. 0.76
  2. 0.84 (correct answer)
  3. 0.60
  4. 0.64
Explanation: When you encounter probability questions involving "either...or" statements, you're dealing with union probability. The key is translating the English into mathematical notation and applying the correct probability rules. The statement "Either X occurs or Y does not occur (or both)" translates to P(X∪Yc)P(X \cup Y^c), where YcY^c represents "Y does not occur." Since these events can overlap (both X can occur AND Y can fail to occur), you need the union formula: P(X∪Yc)=P(X)+P(Yc)−P(X∩Yc)P(X \cup Y^c) = P(X) + P(Y^c) - P(X \cap Y^c). First, find P(Yc)=1−P(Y)=1−0.4=0.6P(Y^c) = 1 - P(Y) = 1 - 0.4 = 0.6. Since X and Y are independent, X and YcY^c are also independent, so P(X∩Yc)=P(X)×P(Yc)=0.6×0.6=0.36P(X \cap Y^c) = P(X) \times P(Y^c) = 0.6 \times 0.6 = 0.36. Therefore: P(X∪Yc)=0.6+0.6−0.36=0.84P(X \cup Y^c) = 0.6 + 0.6 - 0.36 = 0.84. The answer is B. Choice A (0.76) likely comes from incorrectly calculating P(X)+P(Yc)−P(Y)=0.6+0.6−0.4P(X) + P(Y^c) - P(Y) = 0.6 + 0.6 - 0.4. Choice C (0.60) represents just P(X)P(X), ignoring the "or Y does not occur" part entirely. Choice D (0.64) might result from adding P(X)+P(Yc)−P(X)P(Y)P(X) + P(Y^c) - P(X)P(Y) instead of the correct intersection. Remember: "Either A or B" means union (A∪BA \cup B), and always subtract the intersection to avoid double-counting when events can occur simultaneously. Practice translating English probability statements into mathematical notation first.

Question 13

In a quality control process, the probability that a randomly selected item passes inspection is 0.85. If an item fails inspection, it undergoes a repair process that has a 0.70 probability of success. What is the probability that a randomly selected item either passes initial inspection OR is successfully repaired after failing?

  1. 0.955 (correct answer)
  2. 0.595
  3. 0.745
  4. 0.850
Explanation: This requires finding P(passes initial inspection OR successfully repaired). P(passes) = 0.85. P(fails) = 1 - 0.85 = 0.15. P(successfully repaired | fails) = 0.70, so P(fails AND successfully repaired) = 0.15 × 0.70 = 0.105. Therefore, P(passes OR successfully repaired) = 0.85 + 0.105 = 0.955. Choice B incorrectly multiplies 0.85 × 0.70. Choice C uses the average of the two probabilities. Choice D ignores the repair process entirely.

Question 14

The odds in favor of a team winning a championship are 3 to 7. What is the probability that the team does not win the championship?

  1. 310\frac{3}{10}
  2. 37\frac{3}{7}
  3. 47\frac{4}{7}
  4. 710\frac{7}{10} (correct answer)
Explanation: Odds of 3 to 7 mean there are 3 favorable outcomes and 7 unfavorable outcomes, for a total of 3+7=103+7=10 possible outcomes. The probability of the team not winning corresponds to the unfavorable outcomes. Therefore, the probability is 710\frac{7}{10}.

Question 15

A factory produces light bulbs, and the probability that a bulb is functional is 0.995. A quality check involves testing 2 bulbs. What is the probability that not all bulbs tested are functional?

  1. 0.000025
  2. 0.009975 (correct answer)
  3. 0.990025
  4. 0.995000
Explanation: The event 'not all bulbs are functional' is the complement of 'all bulbs are functional'. The probability that both bulbs are functional is 0.995×0.995=0.9900250.995 \times 0.995 = 0.990025. The probability that not all are functional is 1−0.990025=0.0099751 - 0.990025 = 0.009975.

Question 16

A student takes a three-question true-or-false quiz and guesses randomly on every question. What is the probability that the student answers at least one question correctly?

  1. 18\frac{1}{8}
  2. 38\frac{3}{8}
  3. 12\frac{1}{2}
  4. 78\frac{7}{8} (correct answer)
Explanation: The event 'at least one correct' is the complement of the event 'no questions correct' (all wrong). The probability of guessing a single question incorrectly is 12\frac{1}{2}. Since the guesses are independent, the probability of guessing all three incorrectly is (12)×(12)×(12)=18(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{8}. Therefore, the probability of getting at least one correct is 1−18=781 - \frac{1}{8} = \frac{7}{8}.

Question 17

A traffic light is green for 40 seconds, yellow for 5 seconds, and red for 55 seconds. At a random moment, what is the probability that the light is not green?

  1. 60100\frac{60}{100} (correct answer)
  2. 40100\frac{40}{100}
  3. 55100\frac{55}{100}
  4. 5100\frac{5}{100}
Explanation: The total cycle time for the traffic light is 40+5+55=10040 + 5 + 55 = 100 seconds. The probability that the light is green is 40100\frac{40}{100}. The probability that the light is not green is the complement, which is 1−40100=601001 - \frac{40}{100} = \frac{60}{100}. This can also be found by adding the yellow and red times: 5+55=605 + 55 = 60 seconds, so the probability is 60100\frac{60}{100}.

Question 18

A box contains 50 tickets. A student wins a prize if they draw a ticket numbered with a multiple of 6. What is the probability that a student does not win a prize on a single draw?

  1. 425\frac{4}{25}
  2. 850\frac{8}{50}
  3. 2125\frac{21}{25} (correct answer)
  4. 4350\frac{43}{50}
Explanation: The multiples of 6 up to 50 are 6, 12, 18, 24, 30, 36, 42, 48. There are 8 such numbers. The probability of winning is 850=425\frac{8}{50} = \frac{4}{25}. The probability of not winning is the complement: 1−425=21251 - \frac{4}{25} = \frac{21}{25}.

Question 19

If the probability that an event will occur is pp, which expression represents the probability that the event will not occur?

  1. 1−p1 - p (correct answer)
  2. p−1p - 1
  3. 1p\frac{1}{p}
  4. −p-p
Explanation: By definition, the complement of an event is the set of all outcomes that are not in the event. The sum of the probability of an event and its complement is always 1. If the probability of an event is pp, the probability of its complement (the event not occurring) is 1−p1 - p.

Question 20

A spinner is divided into 10 equal sectors, numbered 1 through 10. What is the probability that the spinner does not land on a number that is a multiple of 3?

  1. 310\frac{3}{10}
  2. 13\frac{1}{3}
  3. 710\frac{7}{10} (correct answer)
  4. 23\frac{2}{3}
Explanation: The multiples of 3 between 1 and 10 are 3, 6, and 9. There are 3 such numbers. The probability of landing on a multiple of 3 is 310\frac{3}{10}. The probability of not landing on a multiple of 3 is the complement: 1−310=7101 - \frac{3}{10} = \frac{7}{10}.