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: $$\frac{4}{18} \times \frac{3}{17} = \frac{12}{306}$$

Total probability of same color: $$\frac{56 + 30 + 12}{306} = \frac{98}{306}$$

Since probabilities must sum to 1, the probability of different colors is: $$1 - \frac{98}{306} = \frac{208}{306}$$

Since $$\frac{208}{306} > \frac{98}{306}$$, 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.

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.

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.

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)

Total: 1 + 1 + 10 + 7 = 19 numbers, so P(contains digit 7) = $$\frac{19}{100} = 0.19$$.

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: $$\lfloor 100 ÷ 8 \rfloor = 12$$ confirms there are 12 multiples of 8 up to 100.

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).

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) = $$\frac{4}{9} + \frac{1}{9} = \frac{5}{9}$$

For different colors, you need red then black OR black then red:

• 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) = $$\frac{2}{9} + \frac{2}{9} = \frac{4}{9}$$

Since $$\frac{5}{9} > \frac{4}{9}$$, 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.

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.

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.

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.

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.