ISEE Middle Level Quiz: Calculating Probability
20 questions · exam conditions
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Calculating ProbabilityQuestion 1 of 20

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

1126\frac{11}{26}
2552\frac{25}{52}
313\frac{3}{13}
12\frac{1}{2}
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ISEE Middle Level Quiz

ISEE Middle Level Quiz: Calculating Probability

Practice Calculating Probability in ISEE Middle Level 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 Calculating Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Middle Level.

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

  1. 1126\frac{11}{26} (correct answer)
  2. 2552\frac{25}{52}
  3. 313\frac{3}{13}
  4. 12\frac{1}{2}
Explanation: There are 13 hearts in a 52-card deck. There are 12 face cards (3 in each of the 4 suits). However, the Jack, Queen, and King of hearts are both hearts and face cards. To avoid double-counting, we use the formula P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B). The number of favorable outcomes is the number of hearts plus the number of face cards minus the number of cards that are both: 13+12−3=2213 + 12 - 3 = 22. The total number of possible outcomes is 52. Therefore, the probability is 2252\frac{22}{52}, which simplifies to 1126\frac{11}{26}.

Question 2

A number is randomly chosen from the set of all factors of 72. What is the probability that the chosen number is a multiple of 3?

  1. 12\frac{1}{2}
  2. 23\frac{2}{3} (correct answer)
  3. 34\frac{3}{4}
  4. 56\frac{5}{6}
Explanation: First, list all the factors of 72. The factors are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. There are a total of 12 factors. Next, identify which of these factors are multiples of 3. The multiples of 3 in this list are 3, 6, 9, 12, 18, 24, 36, 72. There are 8 such numbers. The probability is the ratio of the number of favorable outcomes to the total number of outcomes: 812=23\frac{8}{12} = \frac{2}{3}.

Question 3

The probability that event A occurs is 0.4, and the probability that event B occurs is 0.5. If the probability that both A and B occur is 0.1, what is the probability that neither A nor B occurs?

  1. 0.1
  2. 0.2 (correct answer)
  3. 0.8
  4. 0.9
Explanation: The probability that either A or B (or both) occurs is given by the formula P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). Using the given values, P(A∪B)=0.4+0.5−0.1=0.8P(A \cup B) = 0.4 + 0.5 - 0.1 = 0.8. The event that 'neither A nor B occurs' is the complement of the event that 'either A or B occurs'. Therefore, the probability is 1−P(A∪B)=1−0.8=0.21 - P(A \cup B) = 1 - 0.8 = 0.2.

Question 4

A fair coin is tossed once before a quiz. Outcomes are heads or tails, each 1/21/2. What is the probability of not heads outcome?

  1. 1/21/2 (correct answer)
  2. $0%$
  3. 1/41/4
  4. 2/32/3
Explanation: This question tests middle school mathematics skills, specifically calculating probability from outcomes (aligned with ISEE standards). Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to total possible outcomes. In this scenario, students must identify and count the not heads outcome from a coin toss, which is tails, with 1 out of 2. The correct answer works by accurately calculating the probability as 1/2, showing a clear understanding of the total outcome space. A common distractor fails by assuming impossible outcomes, leading to 0%. To help students, teach them to list all possible outcomes and practice converting between fractions and percentages. Encourage checking calculations by ensuring the probabilities sum to 1 or 100%.

Question 5

Two distinct numbers are selected at random from the set {1, 2, 3, 4, 5}. What is the probability that their sum is even?

  1. 25\frac{2}{5} (correct answer)
  2. 310\frac{3}{10}
  3. 12\frac{1}{2}
  4. 35\frac{3}{5}
Explanation: The sum of two numbers is even if both numbers are even or both numbers are odd. The set is {1, 2, 3, 4, 5}, which has 3 odd numbers {1, 3, 5} and 2 even numbers {2, 4}. The total number of ways to choose two distinct numbers is C(5,2)=5×42=10C(5,2) = \frac{5 \times 4}{2} = 10. Case 1: Both are odd. The number of ways to choose 2 odd numbers from 3 is C(3,2)=3C(3,2) = 3. Case 2: Both are even. The number of ways to choose 2 even numbers from 2 is C(2,2)=1C(2,2) = 1. The total number of favorable outcomes is 3+1=43 + 1 = 4. The probability is 410=25\frac{4}{10} = \frac{2}{5}.

Question 6

Alex has a bag with 4 red and 6 blue marbles. Beth has a bag with 3 red and 4 blue marbles. Alex draws one marble from his bag, and Beth draws one from hers. What is the probability that both marbles drawn are the same color?

  1. 1235\frac{12}{35}
  2. 1735\frac{17}{35}
  3. 1835\frac{18}{35} (correct answer)
  4. 12\frac{1}{2}
Explanation: There are two ways for the marbles to be the same color: both are red, or both are blue. We calculate the probability of each case and add them. Case 1: Both red. The probability Alex draws red is 410\frac{4}{10}. The probability Beth draws red is 37\frac{3}{7}. The probability of both drawing red is 410×37=1270\frac{4}{10} \times \frac{3}{7} = \frac{12}{70}. Case 2: Both blue. The probability Alex draws blue is 610\frac{6}{10}. The probability Beth draws blue is 47\frac{4}{7}. The probability of both drawing blue is 610×47=2470\frac{6}{10} \times \frac{4}{7} = \frac{24}{70}. The total probability of drawing the same color is the sum of these probabilities: 1270+2470=3670\frac{12}{70} + \frac{24}{70} = \frac{36}{70}, which simplifies to 1835\frac{18}{35}.

Question 7

If a fair coin is tossed 4 times, what is the probability of getting exactly 3 heads?

  1. 116\frac{1}{16}
  2. 316\frac{3}{16}
  3. 14\frac{1}{4} (correct answer)
  4. 38\frac{3}{8}
Explanation: When a coin is tossed 4 times, the total number of possible outcomes is 24=162^4 = 16. We need to find the number of outcomes with exactly 3 heads. The possible arrangements are HHHT, HHTH, HTHH, and THHH. There are 4 favorable outcomes. The probability is the number of favorable outcomes divided by the total number of outcomes: 416=14\frac{4}{16} = \frac{1}{4}.

Question 8

In a group of 30 students, 18 play basketball, 15 play soccer, and 8 play both. If one student is chosen at random, what is the probability that the student plays neither basketball nor soccer?

  1. 16\frac{1}{6} (correct answer)
  2. 15\frac{1}{5}
  3. 415\frac{4}{15}
  4. 730\frac{7}{30}
Explanation: To find the number of students who play at least one sport, we add the number who play each sport and subtract the number who play both (to avoid double-counting): 18+15−8=2518 + 15 - 8 = 25. So, 25 students play either basketball or soccer or both. The total number of students is 30. The number of students who play neither sport is the total number of students minus those who play at least one sport: 30−25=530 - 25 = 5. The probability of choosing a student who plays neither sport is 530=16\frac{5}{30} = \frac{1}{6}.

Question 9

The letters of the word 'MEDIAN' are written on six separate cards. If two cards are drawn at random without replacement, what is the probability that the first card is a vowel and the second card is a consonant?

  1. 14\frac{1}{4}
  2. 310\frac{3}{10} (correct answer)
  3. 12\frac{1}{2}
  4. 35\frac{3}{5}
Explanation: The word MEDIAN has 6 letters. There are 3 vowels (E, I, A) and 3 consonants (M, D, N). The probability that the first card drawn is a vowel is 36=12\frac{3}{6} = \frac{1}{2}. After a vowel is drawn, there are 5 cards left. Of these, 3 are consonants. The probability that the second card drawn is a consonant is 35\frac{3}{5}. The probability of both events happening in this order is the product of their individual probabilities: 12×35=310\frac{1}{2} \times \frac{3}{5} = \frac{3}{10}.

Question 10

A bag contains 20 balls, some of which are white and the rest are black. The probability of drawing a white ball is 25\frac{2}{5}. After 5 white balls are removed, what is the new probability of drawing a white ball?

  1. 15\frac{1}{5} (correct answer)
  2. 320\frac{3}{20}
  3. 14\frac{1}{4}
  4. 13\frac{1}{3}
Explanation: Initially, there are 20 balls. The probability of drawing a white ball is 25\frac{2}{5}. The number of white balls is 25×20=8\frac{2}{5} \times 20 = 8. The number of black balls is 20−8=1220 - 8 = 12. After 5 white balls are removed, there are 8−5=38 - 5 = 3 white balls left. The total number of balls in the bag is now 20−5=1520 - 5 = 15. The new probability of drawing a white ball is the number of remaining white balls divided by the new total number of balls: 315=15\frac{3}{15} = \frac{1}{5}.

Question 11

A box contains red and green apples. The probability of picking a green apple is 38\frac{3}{8}. After 10 green apples are added to the box, the probability of picking a green apple becomes 12\frac{1}{2}. How many red apples are in the box?

  1. 15
  2. 20
  3. 25 (correct answer)
  4. 40
Explanation: Let G be the initial number of green apples and R be the number of red apples. The initial total is G+RG+R. We are given GG+R=38\frac{G}{G+R} = \frac{3}{8}, which implies 8G=3(G+R)8G = 3(G+R) or 5G=3R5G = 3R. After adding 10 green apples, the new number of green apples is G+10G+10 and the new total is G+R+10G+R+10. The new probability is G+10G+R+10=12\frac{G+10}{G+R+10} = \frac{1}{2}, which implies 2(G+10)=G+R+102(G+10) = G+R+10 or 2G+20=G+R+102G+20 = G+R+10, simplifying to R=G+10R = G+10. We now have a system of two equations: 5G=3R5G = 3R and R=G+10R = G+10. Substitute the second equation into the first: 5G=3(G+10)⇒5G=3G+30⇒2G=30⇒G=155G = 3(G+10) \Rightarrow 5G = 3G + 30 \Rightarrow 2G = 30 \Rightarrow G = 15. Now find R using R=G+10R = G+10: R=15+10=25R = 15+10 = 25. There are 25 red apples.

Question 12

A bag contains only red, blue, and yellow chips. The probability of randomly selecting a red chip is 15\frac{1}{5}, and the probability of selecting a blue chip is 12\frac{1}{2}. If there are 12 yellow chips in the bag, what is the total number of chips in the bag?

  1. 24
  2. 36
  3. 40 (correct answer)
  4. 50
Explanation: The probabilities of all possible outcomes must sum to 1. The probability of picking a yellow chip is P(yellow)=1−P(red)−P(blue)P(\text{yellow}) = 1 - P(\text{red}) - P(\text{blue}). P(yellow)=1−15−12=1010−210−510=310P(\text{yellow}) = 1 - \frac{1}{5} - \frac{1}{2} = \frac{10}{10} - \frac{2}{10} - \frac{5}{10} = \frac{3}{10}. We are told that there are 12 yellow chips, and these represent 310\frac{3}{10} of the total. Let T be the total number of chips. Then 310T=12\frac{3}{10}T = 12. To solve for T, multiply both sides by 103\frac{10}{3}: T=12×103=4×10=40T = 12 \times \frac{10}{3} = 4 \times 10 = 40. There are 40 chips in the bag.

Question 13

Two fair six-sided dice are rolled. What is the probability that the sum of the numbers rolled is a prime number?

  1. 512\frac{5}{12} (correct answer)
  2. 718\frac{7}{18}
  3. 29\frac{2}{9}
  4. 511\frac{5}{11}
Explanation: There are 6×6=366 \times 6 = 36 possible outcomes. The possible sums range from 2 to 12. The prime numbers in this range are 2, 3, 5, 7, and 11. We count the ways to get each prime sum: Sum of 2: (1,1) - 1 way Sum of 3: (1,2), (2,1) - 2 ways Sum of 5: (1,4), (4,1), (2,3), (3,2) - 4 ways Sum of 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) - 6 ways Sum of 11: (5,6), (6,5) - 2 ways The total number of favorable outcomes is 1+2+4+6+2=151 + 2 + 4 + 6 + 2 = 15. The probability is 1536\frac{15}{36}, which simplifies to 512\frac{5}{12}.

Question 14

A spinner is divided into 8 equal sectors, numbered 1 through 8. If the spinner is spun twice, what is the probability that the product of the two numbers is odd?

  1. 116\frac{1}{16}
  2. 14\frac{1}{4} (correct answer)
  3. 38\frac{3}{8}
  4. 12\frac{1}{2}
Explanation: For the product of two numbers to be odd, both numbers must be odd. In the set {1, 2, 3, 4, 5, 6, 7, 8}, the odd numbers are {1, 3, 5, 7}. There are 4 odd numbers. The probability of spinning an odd number on the first spin is 48=12\frac{4}{8} = \frac{1}{2}. The probability of spinning an odd number on the second spin is also 48=12\frac{4}{8} = \frac{1}{2}. Since the spins are independent events, the probability that both outcomes are odd is the product of the individual probabilities: 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.

Question 15

A jar contains 5 red marbles, 4 blue marbles, and 3 green marbles. If two marbles are drawn from the jar at random without replacement, what is the probability that both marbles are blue?

  1. 111\frac{1}{11} (correct answer)
  2. 19\frac{1}{9}
  3. 13\frac{1}{3}
  4. 211\frac{2}{11}
Explanation: The total number of marbles in the jar is 5+4+3=125 + 4 + 3 = 12. The probability of the first marble being blue is 412\frac{4}{12}. After one blue marble is drawn, there are 11 marbles left, and 3 of them are blue. So, the probability of the second marble being blue is 311\frac{3}{11}. To find the probability of both events happening, multiply their probabilities: 412×311=13×311=111\frac{4}{12} \times \frac{3}{11} = \frac{1}{3} \times \frac{3}{11} = \frac{1}{11}.

Question 16

An integer is randomly selected from the integers 1 to 40, inclusive. What is the probability that the selected integer is a multiple of 4 or a multiple of 6?

  1. 14\frac{1}{4}
  2. 1340\frac{13}{40} (correct answer)
  3. 25\frac{2}{5}
  4. 12\frac{1}{2}
Explanation: There are 40 integers in total. The number of multiples of 4 is 40÷4=1040 \div 4 = 10. The number of multiples of 6 is 40÷6=640 \div 6 = 6 with a remainder (the multiples are 6, 12, 18, 24, 30, 36). To find the number of integers that are multiples of 4 or 6, we add the counts and subtract the overlap. The overlap consists of multiples of the least common multiple of 4 and 6, which is 12. The multiples of 12 up to 40 are 12, 24, and 36 (3 multiples). The number of favorable outcomes is 10+6−3=1310 + 6 - 3 = 13. The probability is 1340\frac{13}{40}.

Question 17

In a board game, you roll a fair die once. What is the probability of rolling a 6 outcome?

  1. 5/65/6 (83%)
  2. 1/61/6 (17%) (correct answer)
  3. 1/51/5 (20%)
  4. 1/31/3 (33%)
Explanation: This question tests middle school mathematics skills, specifically calculating probability from outcomes (aligned with ISEE standards). Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to total possible outcomes. In this scenario, students must recognize that a fair die has 6 faces numbered 1 through 6, with only one face showing 6 (favorable outcome). The correct answer works by calculating 1/6 or approximately 17%, showing understanding that each face has equal probability. A common distractor like 5/6 might arise from students calculating the probability of NOT rolling a 6 instead. To help students, use visual aids like dice and emphasize that "fair" means each outcome is equally likely.

Question 18

A bag has 2 yellow, 3 purple, and 5 orange marbles. What is the probability of drawing purple outcome?

  1. 3/103/10 (30%) (correct answer)
  2. 7/107/10 (70%)
  3. 3/53/5 (60%)
  4. 1/31/3 (33%)
Explanation: This question tests middle school mathematics skills, specifically calculating probability from outcomes (aligned with ISEE standards). Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to total possible outcomes. In this scenario, students must identify that there are 3 purple marbles (favorable outcomes) out of 2 + 3 + 5 = 10 total marbles. The correct answer works by calculating 3/10 or 30%, showing accurate counting and fraction formation. A common distractor like 7/10 might arise from adding the other colors instead of focusing on purple. To help students, teach them to underline or circle the specific outcome requested in the problem and double-check their counting before calculating.

Question 19

In gym class, you roll a fair die once. What is the probability of an even outcome?

  1. 1/21/2 (50%) (correct answer)
  2. 1/31/3 (33%)
  3. 2/32/3 (67%)
  4. 1/61/6 (17%)
Explanation: This question tests middle school mathematics skills, specifically calculating probability from outcomes (aligned with ISEE standards). Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to total possible outcomes. In this scenario, students must identify the even outcomes when rolling a fair die: 2, 4, and 6 (3 favorable outcomes) out of 6 total possible outcomes (1, 2, 3, 4, 5, 6). The correct answer works by calculating 3/6 = 1/2 or 50%, showing understanding that half the numbers on a die are even. A common distractor might be 1/3, assuming only two outcomes are even, or misunderstanding what constitutes an even number. To help students, teach them to list all possible outcomes systematically and identify which satisfy the given condition.

Question 20

A bag has 3 red, 5 blue, and 2 green marbles. What is the probability of drawing a blue outcome?

  1. 1/21/2 (50%) (correct answer)
  2. 1/101/10 (10%)
  3. 3/103/10 (30%)
  4. 2/102/10 (20%)
Explanation: This question tests middle school mathematics skills, specifically calculating probability from outcomes (aligned with ISEE standards). Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to total possible outcomes. In this scenario, students must identify that there are 5 blue marbles (favorable outcomes) out of 3 + 5 + 2 = 10 total marbles (total possible outcomes). The correct answer should be 5/10 = 1/2 or 50%, but the marked answer A shows 1/2 (50%) which appears correct. However, upon verification, the probability of drawing a blue marble is indeed 5/10 = 1/2, confirming answer A is correct. To help students, teach them to first count all items, then identify favorable outcomes, and simplify fractions when possible.