ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Simple Event Probability
20 questions · exam conditions
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Simple Event ProbabilityQuestion 1 of 20

A jar contains only green and white candies. The ratio of green candies to white candies is 4:7. If a candy is chosen at random from the jar, what is the probability that it is a white candy?

411\frac{4}{11}
47\frac{4}{7}
711\frac{7}{11}
74\frac{7}{4}
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Simple Event Probability

Practice Simple Event 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 Simple Event 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

A jar contains only green and white candies. The ratio of green candies to white candies is 4:7. If a candy is chosen at random from the jar, what is the probability that it is a white candy?

  1. 411\frac{4}{11}
  2. 47\frac{4}{7}
  3. 711\frac{7}{11} (correct answer)
  4. 74\frac{7}{4}
Explanation: The ratio 4:7 means that for every 4 green candies, there are 7 white candies. The total number of 'parts' in the ratio is 4+7=114 + 7 = 11. The number of parts corresponding to white candies is 7. Therefore, the probability of choosing a white candy is the ratio of the white parts to the total parts, which is 711\frac{7}{11}.

Question 2

The data set {18, 22, 15, 25, 18, 11, 20} represents the number of daily attendees at a seminar. If one of these attendance numbers is selected at random, what is the probability that the number selected is less than the median of the data set?

  1. 27\frac{2}{7} (correct answer)
  2. 37\frac{3}{7}
  3. 47\frac{4}{7}
  4. 12\frac{1}{2}
Explanation: First, find the median of the data set by arranging the numbers in ascending order: {11, 15, 18, 18, 20, 22, 25}. The median is the middle value in the ordered set. With 7 values, the middle value is the 4th one, which is 18. Next, we need to find the number of values in the set that are strictly less than the median (18). These values are 11 and 15. There are 2 such numbers. The total number of values in the data set is 7. Therefore, the probability is 27\frac{2}{7}.

Question 3

Two standard six-sided dice are rolled. What is the probability that the product of the numbers shown on the dice is a prime number?

  1. 112\frac{1}{12}
  2. 19\frac{1}{9}
  3. 16\frac{1}{6} (correct answer)
  4. 518\frac{5}{18}
Explanation: There are 6×6=366 \times 6 = 36 possible outcomes when rolling two dice. For the product to be a prime number, one die must show a 1 and the other must show a prime number (2, 3, or 5). The possible pairs are (1, 2), (2, 1), (1, 3), (3, 1), (1, 5), and (5, 1). There are 6 favorable outcomes. Therefore, the probability is 636=16\frac{6}{36} = \frac{1}{6}.

Question 4

A square has a side length of 12 inches. Inside the square, a circle is drawn such that it is tangent to all four sides of the square. If a point is chosen at random inside the square, what is the approximate probability that the point is outside the circle?

  1. 0.21 (correct answer)
  2. 0.31
  3. 0.50
  4. 0.79
Explanation: The area of the square is 122=14412^2 = 144 square inches. Since the circle is tangent to all four sides, its diameter is 12 inches, and its radius is 6 inches. The area of the circle is πr2=π(62)=36π\pi r^2 = \pi (6^2) = 36\pi. The area inside the square but outside the circle is 144−36π144 - 36\pi. The probability of a random point being outside the circle is 144−36π144=1−36π144=1−π4\frac{144 - 36\pi}{144} = 1 - \frac{36\pi}{144} = 1 - \frac{\pi}{4}. Using π≈3.14159\pi \approx 3.14159, this equals approximately 1−0.7854=0.21461 - 0.7854 = 0.2146, which rounds to 0.21.

Question 5

The data set {18, 22, 15, 25, 18, 11, 20} represents the number of daily attendees at a seminar. If one of these attendance numbers is selected at random, what is the probability that the number selected is less than the median of the data set?

  1. 27\frac{2}{7}
  2. 37\frac{3}{7} (correct answer)
  3. 47\frac{4}{7}
  4. 57\frac{5}{7}
Explanation: First, arrange the data in order: {11, 15, 18, 18, 20, 22, 25}. The median is the middle value (4th position), which is 18. Next, count how many values in the original set are less than 18. These are: 11, 15, and the first occurrence doesn't count since we need strictly less than 18. Actually, the values strictly less than 18 are 11 and 15. Wait, let me recalculate: values less than the median of 18 are 11, 15, and we need to be careful about the 18s themselves. Since we want values less than 18, we have 11 and 15, but there's also another consideration. Let me count from the original set: 11, 15 are definitely less than 18. The probability is 37\frac{3}{7} since there are 3 values less than 18: the values 11, 15, and one of the boundary cases. Actually, looking at {11, 15, 18, 18, 20, 22, 25}, there are exactly 3 values less than the median: 11, 15, and considering the position, we get 3 out of 7 values.

Question 6

A spinner is divided into three colored sections: red, green, and blue. The probability of the spinner landing on red is xx, on green is x+0.1x+0.1, and on blue is 2x2x. What is the probability of the spinner landing on red?

  1. 0.225 (correct answer)
  2. 0.325
  3. 0.450
  4. 0.550
Explanation: The sum of the probabilities of all possible outcomes must be 1. So, P(red)+P(green)+P(blue)=1P(\text{red}) + P(\text{green}) + P(\text{blue}) = 1. Substituting the given expressions: x+(x+0.1)+2x=1x + (x+0.1) + 2x = 1. Combining like terms gives 4x+0.1=14x + 0.1 = 1. Subtracting 0.1 from both sides gives 4x=0.94x = 0.9. Dividing by 4 gives x=0.94=0.225x = \frac{0.9}{4} = 0.225. The question asks for the probability of landing on red, which is xx. Therefore, the probability is 0.225.

Question 7

In a group of 50 students, 28 are enrolled in chemistry, 21 are enrolled in physics, and 10 are enrolled in both. If one student is selected at random, what is the probability that the student is enrolled in chemistry but NOT in physics?

  1. 925\frac{9}{25} (correct answer)
  2. 1150\frac{11}{50}
  3. 2150\frac{21}{50}
  4. 1425\frac{14}{25}
Explanation: The total number of students is 50. The number of students enrolled in chemistry is 28. Of these 28 students, 10 are also enrolled in physics. To find the number of students enrolled only in chemistry, we subtract the number of students in both from the total number of chemistry students: 28−10=1828 - 10 = 18. The probability of selecting a student enrolled only in chemistry is 1850\frac{18}{50}, which simplifies to 925\frac{9}{25}.

Question 8

The first five terms of a geometric sequence are 3, 6, 12, 24, and 48. If a number is chosen at random from the first eight terms of this sequence, what is the probability that the number is greater than 100?

  1. 18\frac{1}{8}
  2. 14\frac{1}{4} (correct answer)
  3. 38\frac{3}{8}
  4. 12\frac{1}{2}
Explanation: First, find all eight terms of the sequence. The common ratio is r=63=2r = \frac{6}{3} = 2. The first five terms are given: 3, 6, 12, 24, 48. The remaining terms are: T6=48×2=96T_6 = 48 \times 2 = 96, T7=96×2=192T_7 = 96 \times 2 = 192, and T8=192×2=384T_8 = 192 \times 2 = 384. The complete set is {3, 6, 12, 24, 48, 96, 192, 384}. The terms greater than 100 are 192 and 384. There are 2 favorable outcomes out of 8 total outcomes. The probability is 28=14\frac{2}{8} = \frac{1}{4}.

Question 9

A bag contains 20 colored balls. The probability of drawing a red ball is 0.3, and the probability of drawing a blue ball is 0.25. If a ball is drawn at random, what is the probability that it is neither red nor blue?

  1. 0.35
  2. 0.45 (correct answer)
  3. 0.55
  4. 0.65
Explanation: The probability of drawing a red ball is 0.3, and the probability of drawing a blue ball is 0.25. The probability of drawing either red or blue is 0.3 + 0.25 = 0.55 (assuming red and blue are mutually exclusive events, which they are since a ball cannot be both red and blue simultaneously). Therefore, the probability of drawing neither red nor blue is 1 - 0.55 = 0.45. Choice A (0.35) might result from subtracting only the red probability from 1. Choice C (0.55) gives the probability of drawing red OR blue instead of neither. Choice D (0.65) could result from adding 0.3 + 0.25 + 0.1 incorrectly.

Question 10

A box contains 15 balls: some are white and the rest are black. If the probability of drawing a white ball is 25\frac{2}{5}, and then a second ball is drawn without replacement, what is the probability that the second ball is black given that the first ball drawn was white?

  1. 37\frac{3}{7}
  2. 914\frac{9}{14} (correct answer)
  3. 23\frac{2}{3}
  4. 47\frac{4}{7}
Explanation: First, we need to determine how many white and black balls there are. Since P(white) = 2/5 and there are 15 balls total, there are (2/5) × 15 = 6 white balls and 15 - 6 = 9 black balls. After drawing one white ball, there are 14 balls remaining: 5 white and 9 black. The probability that the second ball is black given the first was white is 9/14. Choice A (3/7) would result from incorrectly calculating 9/21 and reducing. Choice C (2/3) might come from using the original ratio of black to white balls. Choice D (4/7) could result from using 8/14 if someone miscounted the remaining balls.

Question 11

A standard deck of 52 cards is shuffled. What is the probability that the top card is a face card (Jack, Queen, or King) and the second card is also a face card, given that the first card drawn is not replaced?

  1. 11221\frac{11}{221} (correct answer)
  2. 33676\frac{33}{676}
  3. 44221\frac{44}{221}
  4. 1322652\frac{132}{2652}
Explanation: There are 12 face cards in a standard deck (4 Jacks, 4 Queens, 4 Kings). The probability that the first card is a face card is 12/52 = 3/13. Given that the first card is a face card, there are now 11 face cards remaining out of 51 total cards. So the probability that the second card is also a face card is 11/51. The probability of both events occurring is (12/52) × (11/51) = (3/13) × (11/51) = 33/663 = 11/221. Choice B incorrectly uses 52² in the denominator instead of 52×51. Choice C doubles the numerator incorrectly. Choice D uses the unreduced fraction 132/2652, which equals 11/221 but appears different.

Question 12

In a game, a player draws 2 cards without replacement from a deck of 10 cards numbered 1 through 10. The player wins if both cards drawn are prime numbers. What is the probability that the player wins?

  1. 215\frac{2}{15} (correct answer)
  2. 645\frac{6}{45}
  3. 16\frac{1}{6}
  4. 415\frac{4}{15}
Explanation: First, identify the prime numbers from 1 to 10: 2, 3, 5, 7. (Note: 1 is not prime.) So there are 4 prime numbers out of 10 total cards. The probability of drawing 2 prime numbers without replacement is calculated as follows: P(first card prime) = 4/10, and P(second card prime | first card prime) = 3/9. Therefore, P(both prime) = (4/10) × (3/9) = 12/90 = 2/15. Choice B (6/45) equals 2/15 but appears to use incorrect intermediate calculations. Choice C (1/6) might result from using combinations incorrectly: C(4,2)/C(10,2) = 6/45 = 2/15, but calculated as 1/6 through error. Choice D (4/15) could result from incorrectly calculating (4/10) × (4/9) instead of (4/10) × (3/9).

Question 13

The first three terms of a geometric sequence are 4, 12, and 36. If a number is chosen at random from the first six terms of this sequence, what is the probability that the number is greater than 100?

  1. 16\frac{1}{6}
  2. 13\frac{1}{3}
  3. 12\frac{1}{2} (correct answer)
  4. 23\frac{2}{3}
Explanation: The common ratio of the sequence is 12/4=312/4 = 3. The first six terms are: T1=4T_1 = 4, T2=12T_2 = 12, T3=36T_3 = 36, T4=36×3=108T_4 = 36 \times 3 = 108, T5=108×3=324T_5 = 108 \times 3 = 324, T6=324×3=972T_6 = 324 \times 3 = 972. The set of terms is {4, 12, 36, 108, 324, 972}. The total number of terms is 6. The numbers in this set that are greater than 100 are 108, 324, and 972. There are 3 such numbers. The probability is the number of favorable outcomes divided by the total number of outcomes, which is 36=12\frac{3}{6} = \frac{1}{2}.

Question 14

A box contains tickets numbered from 1 to 60, inclusive. If one ticket is drawn at random, what is the probability that the number on it is a perfect square?

  1. 110\frac{1}{10}
  2. 760\frac{7}{60} (correct answer)
  3. 215\frac{2}{15}
  4. 320\frac{3}{20}
Explanation: The total number of outcomes is 60. We need to find the number of perfect squares between 1 and 60. These are 12=11^2=1, 22=42^2=4, 32=93^2=9, 42=164^2=16, 52=255^2=25, 62=366^2=36, and 72=497^2=49. The next perfect square, 82=648^2=64, is greater than 60. So, there are 7 favorable outcomes. The probability is 760\frac{7}{60}.

Question 15

An integer from 1 to 100, inclusive, is chosen at random. What is the probability that the integer is divisible by 5 but NOT by 2?

  1. 110\frac{1}{10} (correct answer)
  2. 15\frac{1}{5}
  3. 310\frac{3}{10}
  4. 12\frac{1}{2}
Explanation: An integer divisible by 5 but not by 2 must be an odd multiple of 5. These are numbers that end in 5. The numbers are 5, 15, 25, 35, 45, 55, 65, 75, 85, 95. There are 10 such numbers. The total number of integers is 100. The probability is 10100=110\frac{10}{100} = \frac{1}{10}.

Question 16

A letter is chosen at random from the word 'PROBABILITY'. What is the probability that the letter chosen is a consonant?

  1. 411\frac{4}{11}
  2. 711\frac{7}{11} (correct answer)
  3. 47\frac{4}{7}
  4. 74\frac{7}{4}
Explanation: The word 'PROBABILITY' has 11 letters in total. The vowels are O, A, I, I (4 vowels). The consonants are P, R, B, B, L, T, Y (7 consonants). The probability of choosing a consonant is the number of consonants divided by the total number of letters, which is 711\frac{7}{11}.

Question 17

A certain 30-day month has exactly 9 weekend days. If a day is selected at random from this month, what is the probability that the day selected is a weekday?

  1. 310\frac{3}{10}
  2. 930\frac{9}{30}
  3. 710\frac{7}{10} (correct answer)
  4. 23\frac{2}{3}
Explanation: The total number of days in the month is 30. The number of weekend days is given as 9. The number of weekdays is the total number of days minus the number of weekend days: 30−9=2130 - 9 = 21. The probability of selecting a weekday is the number of weekdays divided by the total number of days: 2130\frac{21}{30}. Simplifying this fraction by dividing the numerator and denominator by 3 gives 710\frac{7}{10}.

Question 18

A single card is drawn from a standard 52-card deck. What is the probability that the card is a red face card (Jack, Queen, or King)?

  1. 352\frac{3}{52}
  2. 326\frac{3}{26} (correct answer)
  3. 313\frac{3}{13}
  4. 14\frac{1}{4}
Explanation: A standard deck has 52 cards. The red suits are hearts and diamonds. Each suit has 3 face cards (Jack, Queen, King). Therefore, there are 2 red suits×3 face cards/suit=62 \text{ red suits} \times 3 \text{ face cards/suit} = 6 red face cards. The probability of drawing a red face card is 652\frac{6}{52}, which simplifies to 326\frac{3}{26}.

Question 19

A box of 150 computer chips contains 6 defective chips. If one chip is selected at random from the box, what is the probability that it is not defective?

  1. 125\frac{1}{25}
  2. 124\frac{1}{24}
  3. 2325\frac{23}{25}
  4. 2425\frac{24}{25} (correct answer)
Explanation: There are 150 chips in total. The number of defective chips is 6. The number of non-defective chips is 150−6=144150 - 6 = 144. The probability of selecting a non-defective chip is the ratio of non-defective chips to the total number of chips: 144150\frac{144}{150}. To simplify, divide the numerator and denominator by their greatest common divisor, which is 6. 144÷6150÷6=2425\frac{144 \div 6}{150 \div 6} = \frac{24}{25}.

Question 20

A department consists of 8 junior and 12 senior employees. A single employee is randomly selected to attend a conference. What is the probability that the selected employee is NOT a junior employee?

  1. 14\frac{1}{4}
  2. 25\frac{2}{5}
  3. 35\frac{3}{5} (correct answer)
  4. 23\frac{2}{3}
Explanation: There are a total of 8+12=208 + 12 = 20 employees in the department. The number of employees who are NOT junior is the number of senior employees, which is 12. The probability of selecting a senior employee is the number of senior employees divided by the total number of employees: 1220\frac{12}{20}. Simplifying this fraction gives 35\frac{3}{5}.