ISEE Upper Level Mathematics Achievement Quiz: Comparing Probabilities

What this quiz covers

This quiz focuses on Comparing Probabilities, 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. The probability of same color is greater than the probability of different colors
2. The probability of different colors is greater than the probability of same color (correct answer)
3. The probabilities are equal for both outcomes occurring
4. The relationship cannot be determined without additional information about the drawing process

Explanation: This problem tests conditional probability with dependent events, where the outcome of the first draw affects the second draw since there's no replacement. To find which probability is greater, you need to calculate both scenarios. The bag has 18 total marbles (8 red + 6 blue + 4 green). For same-color pairs, calculate each possibility:

• Two red: $\frac{8}{18} \times \frac{7}{17} = \frac{56}{306}$
• Two blue: $\frac{6}{18} \times \frac{5}{17} = \frac{30}{306}$
• Two green: $\frac{4}{18} \times \frac{3}{17} = \frac{12}{306}$

Question 2

1. P(both A and B) is greater than P(at least one of A or B)
2. P(at least one of A or B) is greater than P(both A and B) (correct answer)
3. P(both A and B) equals P(at least one of A or B)
4. The comparison depends on whether the events are mutually exclusive

Explanation: When you encounter probability questions involving independent events, you need to distinguish between the probability of both events happening versus at least one event happening. These represent very different scenarios and require different calculations. For independent events, the probability of both A and B occurring is found by multiplying their individual probabilities: $P(A \text{ and } B) = P(A) \times P(B) = \frac{3}{7} \times \frac{4}{9} = \frac{12}{63} = \frac{4}{21}$. The probability of at least one event occurring is easier to calculate using the complement: $P(\text{at least one}) = 1 - P(\text{neither})$. Since the events are independent, $P(\text{neither}) = P(\text{not A}) \times P(\text{not B}) = \frac{4}{7} \times \frac{5}{9} = \frac{20}{63}$. Therefore, $P(\text{at least one}) = 1 - \frac{20}{63} = \frac{43}{63}$. Comparing these: $\frac{4}{21} = \frac{12}{63}$ versus $\frac{43}{63}$. Since $\frac{43}{63} > \frac{12}{63}$, the probability of at least one event occurring is greater than the probability of both occurring. Choice A incorrectly reverses this relationship. Choice C suggests they're equal, which would only occur in very specific circumstances that don't apply here. Choice D mentions mutual exclusivity, but the question already tells us the events are independent, making this consideration irrelevant. Remember: "At least one" is typically much more likely than "both" because it covers more favorable outcomes. When multiplying probabilities less than 1, the result gets smaller, while "at least one" encompasses multiple scenarios.

Question 3

1. P(exactly two sixes) is greater than P(at most one six)
2. P(at most one six) is greater than P(exactly two sixes) (correct answer)
3. P(exactly two sixes) equals P(at most one six)
4. The comparison varies depending on the order of the rolls

Explanation: When you encounter probability questions involving multiple independent trials, you need to use the binomial probability formula or carefully count favorable outcomes. Let's calculate both probabilities. For exactly two sixes in three rolls, you need to find the number of ways to get two sixes and one non-six. There are 3 ways to arrange this: (6,6,not-6), (6,not-6,6), or (not-6,6,6). Each specific outcome has probability $\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{216}$. So P(exactly two sixes) = $3 \times \frac{5}{216} = \frac{15}{216}$. For at most one six, you need P(zero sixes) + P(exactly one six). P(zero sixes) = $(\frac{5}{6})^3 = \frac{125}{216}$. P(exactly one six) = $3 \times \frac{1}{6} \times (\frac{5}{6})^2 = 3 \times \frac{25}{216} = \frac{75}{216}$. Therefore, P(at most one six) = $\frac{125}{216} + \frac{75}{216} = \frac{200}{216}$. Comparing: $\frac{200}{216} > \frac{15}{216}$, so P(at most one six) is greater. Choice A incorrectly reverses the inequality. Choice C suggests they're equal, which contradicts our calculations. Choice D incorrectly implies that order matters for probability calculations—while order affects specific sequences, it doesn't change the overall probabilities we're comparing. Strategy tip: When comparing probabilities involving "at most" or "at least," remember these typically yield larger probabilities than "exactly" scenarios because they include multiple outcomes.

Question 4

1. P(multiple of 8) is greater than P(contains digit 7)
2. P(contains digit 7) is greater than P(multiple of 8) (correct answer)
3. P(multiple of 8) equals P(contains digit 7)
4. The relationship depends on how the lottery drawing is conducted

Explanation: When you encounter probability comparison questions, you need to calculate each probability separately and then compare the results. To find P(multiple of 8), count the multiples of 8 from 1 to 100: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96. That's 12 numbers out of 100, so P(multiple of 8) = $\frac{12}{100} = 0.12$. For P(contains digit 7), systematically count numbers containing the digit 7:

• Single-digit: 7 (1 number)
• Teens: 17 (1 number)
• Seventies: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79 (10 numbers)
• Other decades with 7 in units place: 27, 37, 47, 57, 67, 87, 97 (7 numbers)

Question 5

1. P(exactly one event) is greater than P(neither event) (correct answer)
2. P(neither event) is greater than P(exactly one event)
3. P(exactly one event) equals P(neither event)
4. The comparison requires knowledge of whether the events are mutually exclusive

Explanation: When you encounter probability questions involving independent events, you need to calculate compound probabilities by multiplying the individual probabilities for joint outcomes. Let's find each probability systematically. For exactly one event to occur, either A happens and B doesn't, or B happens and A doesn't. Since the events are independent:

• P(A occurs and B doesn't) = P(A) × P(not B) = 0.6 × 0.6 = 0.36
• P(B occurs and A doesn't) = P(not A) × P(B) = 0.4 × 0.4 = 0.16
• P(exactly one event) = 0.36 + 0.16 = 0.52

Question 6

1. Both cards being the same color is more likely than different colors (correct answer)
2. Cards being different colors is more likely than both being the same color
3. Both outcomes have equal probability due to the replacement process
4. The comparison depends on whether red or black cards are drawn first

Explanation: When you encounter probability questions involving "with replacement," focus on calculating the probability of each specific outcome, then compare them directly. Let's break down this modified deck: 20 red cards + 10 black cards = 30 total cards. The probability of drawing red is $\frac{20}{30} = \frac{2}{3}$, and the probability of drawing black is $\frac{10}{30} = \frac{1}{3}$. For both cards being the same color, you need either both red OR both black:

• P(both red) = $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$
• P(both black) = $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$
• P(same color) = $\frac{4}{9} + \frac{1}{9} = \frac{5}{9}$

• P(red, then black) = $\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$
• P(black, then red) = $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$
• P(different colors) = $\frac{2}{9} + \frac{2}{9} = \frac{4}{9}$

Question 7

1. P(at least one) is greater than P(both events) (correct answer)
2. P(both events) is greater than P(at least one)
3. P(at least one) equals P(both events)
4. The comparison depends on the specific nature of the independence

Explanation: When you encounter independent events in probability, remember that independence means the occurrence of one event doesn't affect the probability of the other. This allows you to use the multiplication rule: $P(A \text{ and } B) = P(A) \times P(B)$. Let's calculate both probabilities. For the probability that both events occur: $P(A \text{ and } B) = 0.4 \times 0.7 = 0.28$. For the probability that at least one event occurs, use the complement rule. It's easier to calculate the probability that neither event occurs, then subtract from 1: $P(\text{at least one}) = 1 - P(\text{neither}) = 1 - P(\text{not A and not B})$. Since the events are independent, $P(\text{not A and not B}) = P(\text{not A}) \times P(\text{not B}) = 0.6 \times 0.3 = 0.18$. Therefore, $P(\text{at least one}) = 1 - 0.18 = 0.82$. Comparing the results: 0.82 > 0.28, so choice A is correct. Choice B incorrectly reverses this relationship. Choice C suggests they're equal, which would only happen in very specific probability combinations that don't apply here. Choice D is wrong because independence actually gives us enough information to make a definitive comparison—the specific nature doesn't matter once we know the events are independent and have the given probabilities. Study tip: For "at least one" probability questions, always consider using the complement rule (1 minus the probability of none). It's usually faster than adding multiple probability combinations.

Question 8

1. P(product is even) is greater than P(sum is even) (correct answer)
2. P(sum is even) is greater than P(product is even)
3. P(product is even) equals P(sum is even)
4. The comparison depends on whether the same die is used for both rolls

Explanation: When comparing probabilities involving products and sums, you need to analyze when each outcome occurs by considering the parity (odd/even nature) of the individual rolls. For the product to be even, at least one roll must be even. On an 8-sided die, there are 4 even numbers (2, 4, 6, 8) and 4 odd numbers (1, 3, 5, 7). The probability that both rolls are odd is $\frac{4}{8} \times \frac{4}{8} = \frac{1}{4}$. Therefore, the probability that the product is even is $1 - \frac{1}{4} = \frac{3}{4}$. For the sum to be even, both rolls must have the same parity (both even or both odd). The probability both are even is $\frac{4}{8} \times \frac{4}{8} = \frac{1}{4}$. The probability both are odd is also $\frac{1}{4}$. So the probability the sum is even is $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$. Since $\frac{3}{4} > \frac{1}{2}$, answer A is correct. Answer B incorrectly reverses the comparison. Answer C suggests they're equal, but $\frac{3}{4} \neq \frac{1}{2}$. Answer D implies the physical die matters, but probability calculations depend only on the number of equally likely outcomes, not whether you use one die twice or two separate dice. Remember this pattern: for any fair die with equal numbers of odd and even faces, the product being even is always more likely than the sum being even, because products need only one even factor while sums require matching parities.

Question 9

1. The probability of drawing two hearts is greater
2. The probability of drawing one heart and one spade is greater (correct answer)
3. Both probabilities are equal since hearts and spades are equally represented
4. The comparison depends on which card is drawn first

Explanation: When you encounter probability questions involving drawing cards without replacement, you need to calculate conditional probabilities since each draw affects the next. Let's compare these two scenarios by finding the exact probabilities. For drawing two hearts: The probability of the first heart is $\frac{13}{52}$. After removing one heart, the probability of the second heart is $\frac{12}{51}$. So the probability of two hearts is $\frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} = \frac{1}{17} \approx 0.059$. For drawing one heart and one spade: This can happen in two ways - heart then spade, or spade then heart. Heart then spade: $\frac{13}{52} \times \frac{13}{51} = \frac{169}{2652}$. Spade then heart: $\frac{13}{52} \times \frac{13}{51} = \frac{169}{2652}$. Total probability: $\frac{169}{2652} + \frac{169}{2652} = \frac{338}{2652} = \frac{2}{17} \approx 0.118$. Since $\frac{2}{17} > \frac{1}{17}$, drawing one heart and one spade is more likely, making B correct. A is wrong because the calculations show two hearts is less likely. C incorrectly assumes equal representation means equal probability - but drawing two cards of the same suit is harder than drawing two of different suits. D is wrong because the total probability doesn't depend on order when we consider all possible arrangements. Remember: when comparing "same suit" versus "different suits" scenarios, different suits typically have higher probability because there are more ways to achieve that outcome.

Question 10

1. P(all face cards) is greater than P(all same suit)
2. P(all same suit) is greater than P(all face cards) (correct answer)
3. P(all face cards) equals P(all same suit)
4. The comparison depends on whether Aces are considered face cards

Explanation: When comparing probabilities involving card combinations, you need to calculate each probability separately and consider how the constraints affect the available choices. For all 3 cards to be face cards: There are 12 face cards (4 Jacks, 4 Queens, 4 Kings) in a standard deck. The probability is $\frac{12}{52} \times \frac{11}{51} \times \frac{10}{50} = \frac{1320}{132600} = \frac{11}{1105}$. For all 3 cards to be from the same suit: You can pick any suit for the first card, then the remaining two cards must match that suit. The probability is $1 \times \frac{12}{51} \times \frac{11}{50} = \frac{132}{2550} = \frac{22}{425}$. Converting to compare: $\frac{11}{1105} ≈ 0.00995$ and $\frac{22}{425} ≈ 0.0518$. The same-suit probability is about 5 times larger. Choice A incorrectly suggests face cards are more likely, but 12 face cards create a much tighter constraint than 13 cards of any suit. Choice C claims they're equal, but the calculations clearly show different values. Choice D brings up Aces as face cards, but this is a red herring—the standard definition uses only Jacks, Queens, and Kings, and even if Aces were included, it wouldn't make the probabilities equal. Therefore, B is correct: P(all same suit) is greater than P(all face cards). Strategy tip: In probability comparisons, always calculate both values explicitly. Don't rely on intuition—constraints that seem similar often have very different mathematical outcomes.

Question 11

1. P(exactly 3 heads) is greater than P(at least 3 heads)
2. P(at least 3 heads) is greater than P(exactly 3 heads) (correct answer)
3. P(exactly 3 heads) equals P(at least 3 heads)
4. The relationship varies depending on which specific flips result in heads

Explanation: When you encounter probability questions involving multiple events, you need to carefully distinguish between "exactly" and "at least" scenarios. Both require counting favorable outcomes, but "at least" encompasses multiple cases. For exactly 3 heads in 4 flips, you need to find the number of ways to arrange 3 heads and 1 tail. Using combinations: $\binom{4}{3} = 4$ ways. Since each outcome has probability $(\frac{1}{2})^4 = \frac{1}{16}$, the probability is $\frac{4}{16} = \frac{1}{4}$. For at least 3 heads, you need either exactly 3 heads OR exactly 4 heads. We already found exactly 3 heads has probability $\frac{1}{4}$. For exactly 4 heads: $\binom{4}{4} = 1$ way, so probability is $\frac{1}{16}$. Therefore, P(at least 3 heads) = $\frac{1}{4} + \frac{1}{16} = \frac{4}{16} + \frac{1}{16} = \frac{5}{16}$. Since $\frac{5}{16} > \frac{4}{16}$, answer B is correct. Answer A incorrectly reverses the relationship. The "exactly" probability cannot exceed the "at least" probability since "exactly 3" is just one component of "at least 3." Answer C suggests they're equal, but this ignores that "at least 3" includes the additional case of exactly 4 heads. Answer D incorrectly suggests the relationship changes based on which specific flips are heads, but probability calculations depend only on the total count, not the sequence. Remember: "At least" probabilities always equal or exceed their corresponding "exactly" probabilities because they include additional favorable cases.

Question 12

1. Drawing exactly 2 red balls is more likely than drawing exactly 2 blue balls (correct answer)
2. Drawing exactly 2 blue balls is more likely than drawing exactly 2 red balls
3. Both outcomes have equal probability since exactly 2 of each color is drawn
4. The comparison depends on the specific order in which balls are drawn

Explanation: When you encounter probability questions involving drawing items without replacement, you need to calculate the exact probability of each outcome and compare them directly. To find the probability of drawing exactly 2 red balls out of 3 draws, you can have one red ball and two blue balls in the remaining draw. There are $\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30$ ways to choose 2 red from 5 red balls and 1 blue from 3 blue balls. The total ways to choose any 3 balls from 8 is $\binom{8}{3} = 56$. So P(exactly 2 red) = $\frac{30}{56} = \frac{15}{28}$. For exactly 2 blue balls, you need 2 blue and 1 red. There are $\binom{3}{2} \times \binom{5}{1} = 3 \times 5 = 15$ ways to do this. So P(exactly 2 blue) = $\frac{15}{56}$. Comparing these probabilities: $\frac{15}{28} = \frac{30}{56}$ versus $\frac{15}{56}$. Since $\frac{30}{56} > \frac{15}{56}$, drawing exactly 2 red balls is more likely, making choice A correct. Choice B reverses the comparison incorrectly. Choice C falls into the trap of thinking that because you're drawing "exactly 2" of each color, the probabilities must be equal—but this ignores that there are different numbers of each color available. Choice D incorrectly suggests order matters, but we're only concerned with the final composition, not sequence. Remember: in probability comparisons, always calculate the actual values rather than making intuitive assumptions about equality.

Question 13

1. P(same color twice) is greater than P(red at least once)
2. P(red at least once) is greater than P(same color twice) (correct answer)
3. P(same color twice) equals P(red at least once)
4. The relationship depends on the specific sequence of spins performed

Explanation: When comparing probabilities in multi-step scenarios, you need to calculate each probability systematically and consider all possible outcomes. First, let's find P(same color twice). This occurs when you get red-red, blue-blue, or yellow-yellow:

• P(red twice) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• P(blue twice) = $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$
• P(yellow twice) = $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

Question 14

1. P(exactly 2 chocolate) is greater than P(exactly 1 chocolate) (correct answer)
2. P(exactly 1 chocolate) is greater than P(exactly 2 chocolate)
3. P(exactly 2 chocolate) equals P(exactly 1 chocolate)
4. The comparison depends on whether the selection process is truly random

Explanation: When you encounter probability questions involving "without replacement," you're dealing with combinations since the order doesn't matter and each item can only be selected once. To find the probability of exactly 2 chocolate candies out of 3 selections, you need the number of ways to choose 2 chocolates from 15 and 1 vanilla from 10, divided by the total ways to choose any 3 candies from 25. $P(\text{exactly 2 chocolate}) = \frac{\binom{15}{2} \times \binom{10}{1}}{\binom{25}{3}} = \frac{105 \times 10}{2300} = \frac{1050}{2300}$ $P(\text{exactly 1 chocolate}) = \frac{\binom{15}{1} \times \binom{10}{2}}{\binom{25}{3}} = \frac{15 \times 45}{2300} = \frac{675}{2300}$ Since $\frac{1050}{2300} > \frac{675}{2300}$, the probability of exactly 2 chocolate candies is greater. Choice A correctly identifies this relationship. Choice B reverses the comparison and is incorrect. Choice C suggests the probabilities are equal, but our calculations show they differ significantly. Choice D implies the comparison depends on randomness, but probability calculations assume random selection by definition—this is a mathematical distractor that sounds sophisticated but misunderstands the nature of probability. Study tip: In combination probability problems, always set up your calculation as (favorable outcomes)/(total outcomes). Use the multiplication principle: if you need multiple conditions met simultaneously, multiply the individual combinations together.