ISEE Upper Level Mathematics Achievement Quiz: Single And Compound Probability

What this quiz covers

This quiz focuses on Single And Compound Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Upper Level Mathematics Achievement.

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.

Question 1

1. $\frac{5}{12}$
2. $\frac{7}{12}$
3. $\frac{2}{3}$ (correct answer)
4. $\frac{3}{4}$

Explanation: When you encounter probability questions with "OR" conditions, you're dealing with the union of events. The key insight is recognizing when events overlap and using the inclusion-exclusion principle: P(A OR B) = P(A) + P(B) - P(A AND B). Let's define our events: A = sum greater than 8, and B = at least one die shows 6. First, find P(A). The favorable outcomes for sums greater than 8 are: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6) — that's 10 outcomes out of 36 possible, so P(A) = $\frac{10}{36}$. Next, find P(B). It's easier to calculate the complement: P(no 6's) = $\frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$, so P(at least one 6) = $1 - \frac{25}{36} = \frac{11}{36}$. Now find P(A AND B) — outcomes where sum > 8 AND at least one die shows 6: (3,6), (4,6), (5,6), (6,3), (6,4), (6,5), (6,6). That's 7 outcomes, so P(A AND B) = $\frac{7}{36}$. Therefore: P(A OR B) = $\frac{10}{36} + \frac{11}{36} - \frac{7}{36} = \frac{14}{36} = \frac{2}{3}$, which is answer C. Answer A ($\frac{5}{12}$) likely comes from forgetting to subtract the overlap. Answer B ($\frac{7}{12}$) might result from miscounting favorable outcomes. Answer D ($\frac{3}{4}$) could come from incorrectly adding probabilities without considering overlap. Remember: always subtract the intersection when calculating "OR" probabilities to avoid double-counting overlapping outcomes.

Question 2

1. 0.32
2. 0.48 (correct answer)
3. 0.52
4. 0.64

Explanation: When you encounter probability questions involving "exactly" a certain number of outcomes over multiple trials, you're dealing with binomial probability. The key insight is that "exactly once" means the event happens in one trial but not the other. For red to appear exactly once in two spins, there are two possible scenarios: red on the first spin and not red on the second, OR not red on the first spin and red on the second. Since these scenarios are mutually exclusive, you add their probabilities. First scenario: Red first (probability 0.4), then not red (probability 0.6, since blue and green together total 0.3 + 0.3 = 0.6). This gives us $0.4 \times 0.6 = 0.24$. Second scenario: Not red first (probability 0.6), then red (probability 0.4). This gives us $0.6 \times 0.4 = 0.24$. Total probability: $0.24 + 0.24 = 0.48$, which is answer choice B. Looking at the wrong answers: A (0.32) likely comes from calculating just one scenario instead of both. C (0.52) might result from incorrectly adding 0.4 + 0.3 + 0.3 - 0.4 or similar confusion about complement probabilities. D (0.64) appears to be the probability of getting red at least once, calculated as $1 - (0.6)^2 = 1 - 0.36 = 0.64$. Remember: For "exactly" problems in binomial situations, always consider all the ways the specified outcome can occur, calculate each scenario's probability, then add them together.

Question 3

1. 0.24
2. 0.52 (correct answer)
3. 0.76
4. 1.00

Explanation: When you encounter probability questions involving "A or B but not both," you're dealing with the exclusive or (XOR) - events that can happen individually but not simultaneously. Since events A and B are independent, you can find P(A or B but not both) by calculating P(A and not B) + P(not A and B). For independent events, P(A and not B) = P(A) × P(not B) = 0.6 × (1 - 0.4) = 0.6 × 0.6 = 0.36. Similarly, P(not A and B) = P(not A) × P(B) = (1 - 0.6) × 0.4 = 0.4 × 0.4 = 0.16. Therefore, P(A or B but not both) = 0.36 + 0.16 = 0.52. Choice A (0.24) represents P(A and B) = 0.6 × 0.4 = 0.24, which is the probability that both events occur together - the opposite of what we want. Choice C (0.76) equals P(A or B) = P(A) + P(B) - P(A and B) = 0.6 + 0.4 - 0.24 = 1.00 - 0.24 = 0.76, which includes cases where both events happen. Choice D (1.00) would mean certainty, which only occurs if the events were mutually exclusive and collectively exhaustive. Remember this pattern: "A or B but not both" always equals P(A or B) - P(A and B). This formula works whether events are independent or not, making it a reliable approach for exclusive or problems.

Question 4

1. $\frac{3}{8}$ (correct answer)
2. $\frac{15}{40}$
3. $\frac{1}{4}$
4. $\frac{15}{60}$

Explanation: When you see "given that" in a probability question, you're dealing with conditional probability. This means you're finding the probability of one event happening when you already know another event has occurred. The question asks for P(plays sports | plays music), which reads as "the probability a student plays sports given that the student plays music." To find this, you use the conditional probability formula: P(A|B) = P(A and B) ÷ P(B). Here, you need P(sports and music) ÷ P(music) = 15% ÷ 40% = $\frac{15}{40} = \frac{3}{8}$. Think of it this way: among the 40% of students who play music, 15% play both sports and music. So $\frac{15}{40}$ of music students also play sports. Looking at the wrong answers: Choice B shows $\frac{15}{40}$, which is the correct calculation before simplifying to lowest terms. While mathematically equivalent to $\frac{3}{8}$, choice A is in simplest form. Choice C gives $\frac{1}{4}$, which you might get by incorrectly using $\frac{15}{60}$ (the percentage who play both divided by those who play sports). Choice D shows $\frac{15}{60}$ exactly, representing the common error of finding P(music | sports) instead of P(sports | music). The key strategy is to carefully identify what's given and what you're looking for. "Given that" tells you the denominator in conditional probability - it's your restricted sample space. Always double-check that you have the conditional relationship in the right direction.

Question 5

1. $\frac{1}{4}$ (correct answer)
2. $\frac{4}{16}$
3. $\frac{12}{64}$
4. $\frac{27}{64}$

Explanation: This is a binomial probability problem where you need to find the probability of getting exactly three successes (even numbers) in four independent trials (die rolls). First, identify the probability of success on each roll. A six-sided die has three even numbers (2, 4, 6) out of six possible outcomes, so $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$. The probability of failure is $P(\text{odd}) = \frac{1}{2}$. Use the binomial probability formula: $P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$, where $n = 4$ rolls, $k = 3$ successes, and $p = \frac{1}{2}$. Calculate: $P(X = 3) = \binom{4}{3} \cdot \left(\frac{1}{2}\right)^3 \cdot \left(\frac{1}{2}\right)^1 = 4 \cdot \frac{1}{8} \cdot \frac{1}{2} = 4 \cdot \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$ Choice A gives $\frac{1}{4}$, which is correct. Choice B shows $\frac{4}{16}$, which equals $\frac{1}{4}$ but wasn't simplified—this represents the same correct answer in unreduced form. Choice C shows $\frac{12}{64}$, which simplifies to $\frac{3}{16}$—this likely comes from miscalculating the combinations or using wrong probability values. Choice D shows $\frac{27}{64}$, which doesn't match any reasonable calculation for this scenario and may result from confusion with other probability formulas. When solving binomial probability problems, always identify your success probability first, then carefully apply the formula. Remember that $\binom{4}{3} = 4$, representing the four different ways to arrange three successes in four trials.

Question 6

1. $\frac{74}{105}$ (correct answer)
2. $\frac{31}{105}$
3. $\frac{62}{105}$
4. $\frac{43}{105}$

Explanation: When you encounter probability questions involving "without replacement," you need to consider how each draw affects the subsequent outcomes. This question asks for the probability that two marbles are different colors, which means avoiding the case where both marbles are the same color. The bag contains 15 total marbles (5 red + 4 blue + 6 yellow). To find the probability of different colors, calculate 1 minus the probability of same colors. The probability of drawing two marbles of the same color equals:

• P(both red) = $\frac{5}{15} \times \frac{4}{14} = \frac{20}{210}$
• P(both blue) = $\frac{4}{15} \times \frac{3}{14} = \frac{12}{210}$
• P(both yellow) = $\frac{6}{15} \times \frac{5}{14} = \frac{30}{210}$

Question 7

1. 0.441
2. 0.784 (correct answer)
3. 0.832
4. 0.973

Explanation: When you encounter probability questions asking for "at least" a certain number of successes, you're dealing with binomial probability. The key insight is that "at least two rounds" means exactly two rounds OR exactly three rounds. To find the probability of winning exactly k rounds out of 3, use the binomial formula: $P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$, where n = 3 rounds and p = 0.7. For exactly 2 wins: $P(X = 2) = \binom{3}{2} \cdot (0.7)^2 \cdot (0.3)^1 = 3 \cdot 0.49 \cdot 0.3 = 0.441$ For exactly 3 wins: $P(X = 3) = \binom{3}{3} \cdot (0.7)^3 \cdot (0.3)^0 = 1 \cdot 0.343 \cdot 1 = 0.343$ Therefore, P(at least 2 wins) = 0.441 + 0.343 = 0.784. Choice A (0.441) represents only the probability of winning exactly two rounds—this is a common trap where students forget to include the "exactly three" case. Choice C (0.832) likely comes from calculation errors in the binomial coefficients or powers. Choice D (0.973) is far too high and might result from misunderstanding what "at least two" means or incorrectly calculating 1 minus the probability of winning zero or one round. Remember: when you see "at least" in probability questions, always break it down into separate "exactly" cases and add them together. This systematic approach prevents missing scenarios.

Question 8

1. $\frac{7}{128}$
2. $\frac{11}{128}$
3. $\frac{56}{1024}$ (correct answer)
4. $\frac{176}{1024}$

Explanation: When you encounter a question about repeated independent trials with only two outcomes (like true/false), you're dealing with binomial probability. Each guess has a $\frac{1}{2}$ chance of being correct, and you need to find the probability of getting "more than 7" correct, which means exactly 8, 9, or 10 correct answers. The binomial probability formula is $P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$, where $n = 10$, $p = \frac{1}{2}$, and $k$ is the number of correct answers. For 8 correct: $P(X = 8) = \binom{10}{8} \cdot \left(\frac{1}{2}\right)^{10} = 45 \cdot \frac{1}{1024} = \frac{45}{1024}$ For 9 correct: $P(X = 9) = \binom{10}{9} \cdot \left(\frac{1}{2}\right)^{10} = 10 \cdot \frac{1}{1024} = \frac{10}{1024}$ For 10 correct: $P(X = 10) = \binom{10}{10} \cdot \left(\frac{1}{2}\right)^{10} = 1 \cdot \frac{1}{1024} = \frac{1}{1024}$ Total probability: $\frac{45 + 10 + 1}{1024} = \frac{56}{1024}$, which is answer C. Answer A ($\frac{7}{128}$) equals $\frac{56}{1024}$ when converted, but uses the wrong denominator initially. Answer B ($\frac{11}{128}$) likely comes from miscalculating the binomial coefficients. Answer D ($\frac{176}{1024}$) suggests including cases with 6 or 7 correct answers, misinterpreting "more than 7." Remember: "more than 7" means 8, 9, or 10 — not 7 or more. Always identify exactly which outcomes satisfy the condition before calculating.

Question 9

1. $\frac{1}{3}$ (correct answer)
2. $\frac{10}{30}$
3. $\frac{18}{30}$
4. $\frac{8}{30}$

Explanation: When you encounter problems about students studying multiple subjects, you're dealing with set theory and overlapping groups. The key is organizing the information to avoid double-counting students who belong to both categories. Let's break down what we know: 18 students study French, 16 study Spanish, and 8 study both languages. To find students who study "French but not Spanish," you need to subtract the overlap from the French total: 18 - 8 = 10 students study only French. The probability is therefore $\frac{10}{30} = \frac{1}{3}$, making answer A correct. Now let's examine why the other answers are wrong. Answer B gives $\frac{10}{30}$, which represents the correct number of students (10) but fails to simplify the fraction. While mathematically equivalent to $\frac{1}{3}$, it's not in simplest form. Answer C shows $\frac{18}{30}$, which would be the probability of studying French at all (including those who also study Spanish) – this ignores the "but not Spanish" requirement. Answer D gives $\frac{8}{30}$, which represents students studying both languages, not French exclusively. Study tip: Always draw a Venn diagram for overlap problems. Put the intersection number (8) in the middle first, then work outward to find the exclusive regions. This visual approach prevents the common mistake of forgetting to subtract the overlap when finding "but not" probabilities.

Question 10

1. $\frac{1}{4}$
2. $\frac{3}{7}$ (correct answer)
3. $\frac{3}{8}$
4. $\frac{1}{2}$

Explanation: This is a conditional probability problem, which asks for the probability of one event happening given that another event has already occurred. When you see "given that" in a probability question, you need to use the conditional probability formula or adjust your sample space. First, let's find all possible outcomes when flipping a coin three times. There are $2^3 = 8$ total outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Since we're given that at least one head occurred, we can eliminate TTT from our sample space. This leaves us with 7 favorable outcomes for the condition: HHH, HHT, HTH, HTT, THH, THT, TTH. Now we need exactly two heads from these 7 remaining outcomes. Looking at our list: HHT, HTH, and THH each contain exactly two heads. That's 3 outcomes with exactly two heads. Therefore, the probability is $\frac{3}{7}$. Choice A ($\frac{1}{4}$) incorrectly uses the original sample space of 8 outcomes instead of the reduced space of 7. Choice C ($\frac{3}{8}$) correctly identifies 3 favorable outcomes but fails to account for the conditional restriction, using all 8 original outcomes as the denominator. Choice D ($\frac{1}{2}$) likely comes from incorrectly thinking there are only 6 remaining outcomes or miscounting the favorable cases. Remember: in conditional probability problems, always adjust your sample space to include only the outcomes that satisfy the given condition, then find your target event within that restricted space.

Question 11

1. $\frac{7}{40}$ (correct answer)
2. $\frac{1}{120}$
3. $\frac{3}{10}$
4. $\frac{1}{8}$

Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, 3 numbers are drawn from 1 to 10, and a player chooses 3, finding probability of exactly 2 matches (order irrelevant). Choice A is correct because there are C(3,2) × C(7,1) = 21 favorable combinations out of C(10,3) = 120, yielding 7/40. Choice B is incorrect due to using full match instead, demonstrating a common error where students confuse exact with full matches. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 12

1. $\frac{1}{4}$
2. $\frac{18}{120}$
3. $\frac{3}{10}$ (correct answer)
4. $\frac{54}{120}$

Explanation: When you encounter probability questions involving drawing items without replacement, you need to use combinations to count favorable outcomes and total possible outcomes. To find the probability of exactly 2 red balls in 3 draws, you need to calculate: (ways to choose 2 red from 4) × (ways to choose 1 non-red from 6) ÷ (total ways to choose 3 from 10). The favorable outcomes are: $\binom{4}{2} \times \binom{6}{1} = 6 \times 6 = 36$ The total possible outcomes are: $\binom{10}{3} = 120$ So the probability is $\frac{36}{120} = \frac{3}{10}$, which is answer choice C. Looking at the wrong answers: Choice A ($\frac{1}{4}$) likely comes from incorrectly thinking about this as a simple ratio problem without considering combinations. Choice B ($\frac{18}{120}$) represents a calculation error—perhaps only counting one way to arrange the draws rather than all possible combinations. Choice D ($\frac{54}{120}$) is too large and might result from miscounting the non-red balls or making an error in the combination formula. Notice that choices B and D both have 120 in the denominator, which is correct for the total outcomes, but their numerators reflect different calculation mistakes. For combination probability problems, always identify what you're selecting from each group, calculate the combinations separately, multiply them together for favorable outcomes, then divide by total possible combinations. Double-check your arithmetic, especially when working with factorials.

Question 13

1. $\frac{2}{7}$
2. $\frac{4}{7}$ (correct answer)
3. $\frac{3}{7}$
4. $\frac{5}{7}$

Explanation: When you encounter probability questions involving mutually exclusive events and conditional probability, start by identifying what information you have and what relationships you can use. Since events A and B are mutually exclusive, they cannot occur simultaneously, so $P(A \cap B) = 0$. This means $P(A \cup B) = P(A) + P(B)$. Given $P(A) = 0.3$ and $P(A \cup B) = 0.7$, you can find $P(B) = 0.7 - 0.3 = 0.4$. Now you need $P(B|A \cup B)$, which represents the probability that event B occurs given that either A or B has occurred. Using the conditional probability formula: $P(B|A \cup B) = \frac{P(B \cap (A \cup B))}{P(A \cup B)}$. Since B is a subset of $A \cup B$, we have $P(B \cap (A \cup B)) = P(B) = 0.4$. Therefore: $P(B|A \cup B) = \frac{0.4}{0.7} = \frac{4}{7}$, which is answer choice B. Let's examine the wrong answers: Choice A ($\frac{2}{7}$) might result from incorrectly using $P(A \cap B) = 0$ in the numerator. Choice C ($\frac{3}{7}$) could come from using $P(A)$ instead of $P(B)$ in the numerator. Choice D ($\frac{5}{7}$) might arise from adding probabilities incorrectly or misapplying the complement rule. Remember: with mutually exclusive events, use $P(A \cup B) = P(A) + P(B)$ to find missing probabilities, and always check that your conditional probability makes intuitive sense given the constraint.

Question 14

1. $\frac{1}{4}$
2. $\frac{12}{51}$ (correct answer)
3. $\frac{13}{52}$
4. $\frac{12}{52}$

Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, after drawing a heart first without replacement, students must find the conditional probability of drawing another heart. Choice B is correct because 12 hearts remain out of 51 cards, yielding 12/51. Choice A is incorrect due to using the unconditional probability, demonstrating a common error where students ignore the given condition. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 15

1. $\frac{12}{100}$
2. $\frac{12}{30}$ (correct answer)
3. $\frac{30}{40}$
4. $\frac{40}{100}$

Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, given weather data over 100 days, students must find the conditional probability of high temperatures given rain. Choice B is correct because it's 12 days both divided by 30 rainy days, yielding 12/30. Choice A is incorrect due to using the joint over total instead of conditional, demonstrating a common error where students confuse conditional with joint probability. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 16

1. $\frac{4}{51}$ (correct answer)
2. $\frac{4}{52}$
3. $\frac{1}{13}$
4. $\frac{3}{51}$

Explanation: When you encounter conditional probability problems involving cards dealt without replacement, focus on how the first card changes the composition of the remaining deck. Since the first card was a King, you now have 51 cards left in the deck. The key insight is that removing a King doesn't affect the number of Aces remaining - there are still 4 Aces among those 51 cards. Therefore, the probability that the second card is an Ace is $\frac{4}{51}$. Let's examine why the other answers represent common misconceptions: Answer B ($\frac{4}{52}$) treats this as if you're drawing from a full deck, ignoring that the first card was already removed. This would be correct if the first card were replaced, but that's not the case here. Answer C ($\frac{1}{13}$) equals $\frac{4}{52}$ in reduced form, so it has the same flaw as option B - failing to account for the reduced deck size. Answer D ($\frac{3}{51}$) makes the error of assuming that since a King was drawn first, somehow one fewer Ace remains. This confuses the suits or misunderstands what "given that the first card was a King" means - it's telling you information about what happened, not asking you to calculate it. The correct answer is A: $\frac{4}{51}$. Study tip: In conditional probability card problems, always adjust your denominator for cards already drawn, but only adjust the numerator if the drawn cards actually affect what you're looking for. A King being drawn doesn't change the Ace count.

Question 17

1. $\frac{1}{12}$
2. $\frac{1}{6}$ (correct answer)
3. $\frac{5}{36}$
4. $\frac{1}{18}$

Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, students must identify all doubles when rolling two dice: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Choice B is correct because there are exactly 6 doubles out of 36 total outcomes, giving us 6/36 = 1/6. Choice A incorrectly suggests 3 doubles (1/12 = 3/36), while choice C overcounts with 5/36. To help students: Create a visual representation of all 36 outcomes and highlight the diagonal where both dice show the same number. Emphasize that doubles form a pattern along the main diagonal of the outcome grid.

Question 18

1. $12\%$ (correct answer)
2. $18\%$
3. $48\%$
4. $28\%$

Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, weather data over 100 days is given, and students must find the probability of both rain and high temperatures. Choice A is correct because 12 days have both out of 100, yielding 12%. Choice B is incorrect due to adding instead of finding intersection, demonstrating a common error where students confuse union with intersection. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 19

1. $\frac{1}{2}$ (correct answer)
2. $\frac{28}{56}$
3. $\frac{14}{28}$
4. $\frac{7}{14}$

Explanation: When you encounter probability questions involving sums being even or odd, think about the fundamental rule: a sum is even when you add an even number of odd integers, and odd when you add an odd number of odd integers. In this jar, there are 4 odd-numbered balls (1, 3, 5, 7) and 4 even-numbered balls (2, 4, 6, 8). For the sum of three drawn balls to be even, you need either:

• All three balls to be even numbers, OR
• Exactly two odd numbers and one even number

• Three even balls: $\binom{4}{3} = 4$
• Two odd, one even: $\binom{4}{2} \times \binom{4}{1} = 6 \times 4 = 24$
• Total favorable outcomes: $4 + 24 = 28$

Question 20

1. $\frac{1}{16}$
2. $\frac{13}{204}$ (correct answer)
3. $\frac{1}{4}$
4. $\frac{1}{8}$

Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, two cards are drawn without replacement from a 52-card deck, and students must find the probability of a heart followed by a spade. Choice B is correct because the probability is (13/52) × (13/51), simplifying to 13/204 after accounting for the dependent events. Choice A is incorrect due to treating the draws as independent, demonstrating a common error where students ignore the without-replacement condition. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.