ISEE Upper Level Quiz: Comparing Probabilities
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Comparing ProbabilitiesQuestion 1 of 14

A bag contains 8 red marbles, 6 blue marbles, and 4 green marbles. Two marbles are drawn without replacement. What is the probability that both marbles are the same color compared to the probability that the two marbles are different colors?

The probability of same color is greater than the probability of different colors
The probability of different colors is greater than the probability of same color
The probabilities are equal for both outcomes occurring
The relationship cannot be determined without additional information about the drawing process
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ISEE Upper Level Quiz

ISEE Upper Level Quiz: Comparing Probabilities

Practice Comparing Probabilities in ISEE Upper 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 Comparing Probabilities, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Upper 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 bag contains 8 red marbles, 6 blue marbles, and 4 green marbles. Two marbles are drawn without replacement. What is the probability that both marbles are the same color compared to the probability that the two marbles are different colors?

  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: 818×717=56306\frac{8}{18} \times \frac{7}{17} = \frac{56}{306}
  • Two blue: 618×517=30306\frac{6}{18} \times \frac{5}{17} = \frac{30}{306}
  • Two green: 418×317=12306\frac{4}{18} \times \frac{3}{17} = \frac{12}{306}
Total probability of same color: 56+30+12306=98306\frac{56 + 30 + 12}{306} = \frac{98}{306} Since probabilities must sum to 1, the probability of different colors is: 1−98306=2083061 - \frac{98}{306} = \frac{208}{306} Since 208306>98306\frac{208}{306} > \frac{98}{306}, different colors is more likely than same color. Choice A incorrectly suggests same color is more probable. This might stem from focusing only on the largest group (red marbles) without considering all possibilities. Choice C incorrectly claims the probabilities are equal, which would only occur with very specific marble distributions. Choice D is wrong because the drawing process is clearly defined as "without replacement," giving us all necessary information. Remember: When comparing compound probabilities, calculate systematically rather than making intuitive guesses. The math often reveals counterintuitive results, especially with dependent events.

Question 2

In a certain game, Event A occurs with probability 37\frac{3}{7} and Event B occurs with probability 49\frac{4}{9}. If the events are independent, which statement correctly compares the probability of both events occurring to the probability of at least one event occurring?

  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 and B)=P(A)×P(B)=37×49=1263=421P(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(at least one)=1−P(neither)P(\text{at least one}) = 1 - P(\text{neither}). Since the events are independent, P(neither)=P(not A)×P(not B)=47×59=2063P(\text{neither}) = P(\text{not A}) \times P(\text{not B}) = \frac{4}{7} \times \frac{5}{9} = \frac{20}{63}. Therefore, P(at least one)=1−2063=4363P(\text{at least one}) = 1 - \frac{20}{63} = \frac{43}{63}. Comparing these: 421=1263\frac{4}{21} = \frac{12}{63} versus 4363\frac{43}{63}. Since 4363>1263\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

In a lottery, tickets are numbered 1 through 100. A winning ticket is drawn randomly. Compare the probability that the winning number is a multiple of 8 to the probability that it contains the digit 7.

  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) = 12100=0.12\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)
Total: 1 + 1 + 10 + 7 = 19 numbers, so P(contains digit 7) = 19100=0.19\frac{19}{100} = 0.19. Since 0.19 > 0.12, P(contains digit 7) is greater than P(multiple of 8). Choice A incorrectly reverses the relationship. Choice C suggests they're equal, but 19 ≠ 12. Choice D implies the drawing method affects the comparison, but since we're told tickets are drawn randomly, the probabilities are fixed regardless of the specific drawing procedure. Strategy tip: For "contains digit" problems, organize your counting by digit position (units, tens) to avoid missing numbers. For multiples, use division to check your count: ⌊100÷8⌋=12\lfloor 100 ÷ 8 \rfloor = 12 confirms there are 12 multiples of 8 up to 100.

Question 4

A deck of cards has been modified so that it contains 20 red cards and 10 black cards. Two cards are drawn with replacement. Which probability is greater: both cards being the same color or the cards being different colors?

  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 2030=23\frac{20}{30} = \frac{2}{3}, and the probability of drawing black is 1030=13\frac{10}{30} = \frac{1}{3}. For both cards being the same color, you need either both red OR both black:
  • P(both red) = 23×23=49\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}
  • P(both black) = 13×13=19\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}
  • P(same color) = 49+19=59\frac{4}{9} + \frac{1}{9} = \frac{5}{9}
For different colors, you need red then black OR black then red:
  • P(red, then black) = 23×13=29\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}
  • P(black, then red) = 13×23=29\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}
  • P(different colors) = 29+29=49\frac{2}{9} + \frac{2}{9} = \frac{4}{9}
Since 59>49\frac{5}{9} > \frac{4}{9}, same color is more likely, making A correct. B incorrectly reverses the comparison. C misunderstands replacement—replacement affects whether draws are independent, but doesn't make probabilities equal when the deck composition is uneven. D suggests order matters for the final probability, but we're comparing total probabilities regardless of which color comes first. Remember: when comparing probability scenarios, calculate each total probability separately, then compare the results numerically. Uneven group sizes often create surprising probability outcomes.

Question 5

A standard deck of 52 cards is shuffled. The top 3 cards are revealed. Compare the probability that all 3 cards are face cards (Jack, Queen, King) to the probability that all 3 cards are from the same suit.

  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 1252×1151×1050=1320132600=111105\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 1×1251×1150=1322550=224251 \times \frac{12}{51} \times \frac{11}{50} = \frac{132}{2550} = \frac{22}{425}. Converting to compare: 111105≈0.00995\frac{11}{1105} ≈ 0.00995 and 22425≈0.0518\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 6

A spinner has regions colored red, blue, and yellow with probabilities 12\frac{1}{2}, 13\frac{1}{3}, and 16\frac{1}{6} respectively. The spinner is spun twice. Compare the probability of getting the same color twice to the probability of getting red at least once.

  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) = 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
  • P(blue twice) = 13×13=19\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}
  • P(yellow twice) = 16×16=136\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}
So P(same color twice) = 14+19+136=9+4+136=1436=718\frac{1}{4} + \frac{1}{9} + \frac{1}{36} = \frac{9 + 4 + 1}{36} = \frac{14}{36} = \frac{7}{18} For P(red at least once), it's easier to use the complement: P(red at least once) = 1 - P(no red in either spin). The probability of not getting red on one spin is 1−12=121 - \frac{1}{2} = \frac{1}{2}, so P(no red twice) = 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. Therefore, P(red at least once) = 1−14=341 - \frac{1}{4} = \frac{3}{4} Comparing: 34=2736\frac{3}{4} = \frac{27}{36} versus 718=1436\frac{7}{18} = \frac{14}{36} Since 2736>1436\frac{27}{36} > \frac{14}{36}, P(red at least once) is greater than P(same color twice), making B correct. Choice A reverses this relationship. Choice C incorrectly suggests they're equal. Choice D is wrong because probability calculations are deterministic—the actual spins don't affect the theoretical probabilities. Strategy tip: For "at least once" problems, always consider using the complement rule (1 minus the probability of the opposite event) as it's often simpler than calculating multiple cases directly.

Question 7

A box contains 15 chocolate candies and 10 vanilla candies. Three candies are selected without replacement. Compare the probability of selecting exactly 2 chocolate candies to the probability of selecting exactly 1 chocolate candy.

  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(exactly 2 chocolate)=(152)×(101)(253)=105×102300=10502300P(\text{exactly 2 chocolate}) = \frac{\binom{15}{2} \times \binom{10}{1}}{\binom{25}{3}} = \frac{105 \times 10}{2300} = \frac{1050}{2300} P(exactly 1 chocolate)=(151)×(102)(253)=15×452300=6752300P(\text{exactly 1 chocolate}) = \frac{\binom{15}{1} \times \binom{10}{2}}{\binom{25}{3}} = \frac{15 \times 45}{2300} = \frac{675}{2300} Since 10502300>6752300\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.

Question 8

A fair six-sided die is rolled three times. Compare the probability of getting exactly two sixes to the probability of getting at most one six.

  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 16×16×56=5216\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{216}. So P(exactly two sixes) = 3×5216=152163 \times \frac{5}{216} = \frac{15}{216}. For at most one six, you need P(zero sixes) + P(exactly one six). P(zero sixes) = (56)3=125216(\frac{5}{6})^3 = \frac{125}{216}. P(exactly one six) = 3×16×(56)2=3×25216=752163 \times \frac{1}{6} \times (\frac{5}{6})^2 = 3 \times \frac{25}{216} = \frac{75}{216}. Therefore, P(at most one six) = 125216+75216=200216\frac{125}{216} + \frac{75}{216} = \frac{200}{216}. Comparing: 200216>15216\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 9

Two independent events A and B have probabilities P(A) = 0.6 and P(B) = 0.4. Compare the probability of exactly one of these events occurring to the probability of neither event occurring.

  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
For neither event to occur: P(neither A nor B) = P(not A) × P(not B) = 0.4 × 0.6 = 0.24 Since 0.52 > 0.24, the probability of exactly one event occurring is greater than the probability of neither event occurring. Choice A is correct because our calculation shows P(exactly one event) = 0.52 is indeed greater than P(neither event) = 0.24. Choice B incorrectly reverses this relationship. Choice C suggests they're equal, which contradicts our calculations (0.52 ≠ 0.24). Choice D incorrectly implies we need additional information about mutual exclusivity, but independence is sufficient information to solve this problem completely. Study tip: For independent events, always remember that P(not A) = 1 - P(A), and joint probabilities multiply. Practice breaking "exactly one" scenarios into their component parts: (A and not B) or (not A and B).

Question 10

Two events A and B are independent with P(A) = 0.4 and P(B) = 0.7. Compare the probability that at least one event occurs to the probability that both events occur.

  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 and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B). Let's calculate both probabilities. For the probability that both events occur: P(A and B)=0.4×0.7=0.28P(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(at least one)=1−P(neither)=1−P(not A and not B)P(\text{at least one}) = 1 - P(\text{neither}) = 1 - P(\text{not A and not B}). Since the events are independent, P(not A and not B)=P(not A)×P(not B)=0.6×0.3=0.18P(\text{not A and not B}) = P(\text{not A}) \times P(\text{not B}) = 0.6 \times 0.3 = 0.18. Therefore, P(at least one)=1−0.18=0.82P(\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 11

A fair octahedral die (8 faces numbered 1-8) is rolled twice. Compare the probability that the product of the two rolls is even to the probability that the sum of the two rolls is even.

  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 48×48=14\frac{4}{8} \times \frac{4}{8} = \frac{1}{4}. Therefore, the probability that the product is even is 1−14=341 - \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 48×48=14\frac{4}{8} \times \frac{4}{8} = \frac{1}{4}. The probability both are odd is also 14\frac{1}{4}. So the probability the sum is even is 14+14=12\frac{1}{4} + \frac{1}{4} = \frac{1}{2}. Since 34>12\frac{3}{4} > \frac{1}{2}, answer A is correct. Answer B incorrectly reverses the comparison. Answer C suggests they're equal, but 34≠12\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 12

Two cards are drawn from a standard 52-card deck without replacement. Which probability is greater: drawing two hearts or drawing one heart and one spade?

  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 1352\frac{13}{52}. After removing one heart, the probability of the second heart is 1251\frac{12}{51}. So the probability of two hearts is 1352×1251=1562652=117≈0.059\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: 1352×1351=1692652\frac{13}{52} \times \frac{13}{51} = \frac{169}{2652}. Spade then heart: 1352×1351=1692652\frac{13}{52} \times \frac{13}{51} = \frac{169}{2652}. Total probability: 1692652+1692652=3382652=217≈0.118\frac{169}{2652} + \frac{169}{2652} = \frac{338}{2652} = \frac{2}{17} \approx 0.118. Since 217>117\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 13

A fair coin is flipped 4 times. Compare the probability of getting exactly 3 heads to the probability of getting at least 3 heads.

  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: (43)=4\binom{4}{3} = 4 ways. Since each outcome has probability (12)4=116(\frac{1}{2})^4 = \frac{1}{16}, the probability is 416=14\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 14\frac{1}{4}. For exactly 4 heads: (44)=1\binom{4}{4} = 1 way, so probability is 116\frac{1}{16}. Therefore, P(at least 3 heads) = 14+116=416+116=516\frac{1}{4} + \frac{1}{16} = \frac{4}{16} + \frac{1}{16} = \frac{5}{16}. Since 516>416\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 14

A jar contains 5 red balls and 3 blue balls. Three balls are drawn without replacement. Which is more likely: drawing exactly 2 red balls or drawing exactly 2 blue balls?

  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 (52)×(31)=10×3=30\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 (83)=56\binom{8}{3} = 56. So P(exactly 2 red) = 3056=1528\frac{30}{56} = \frac{15}{28}. For exactly 2 blue balls, you need 2 blue and 1 red. There are (32)×(51)=3×5=15\binom{3}{2} \times \binom{5}{1} = 3 \times 5 = 15 ways to do this. So P(exactly 2 blue) = 1556\frac{15}{56}. Comparing these probabilities: 1528=3056\frac{15}{28} = \frac{30}{56} versus 1556\frac{15}{56}. Since 3056>1556\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.