The box contains 4 + 6 = 10 balls total, with 4 red balls as favorable outcomes. Using the probability rule P(event) = favorable / total, we get P(red) = 4/10 = 2/5. Choice B (1/3) would be incorrect, and choice A (2/3) represents the probability of picking a black ball instead of red.

This is a probability question testing complementary probability. Choice C (3/4) is correct — there are 6 + 5 + 9 = 20 total marbles. "Not blue" means red or green: 6 + 9 = 15 marbles. P(not blue) = 15/20 = 3/4. Choice A (1/4) inverts the logic — it represents the probability of an outcome that is NOT "not blue," essentially giving P(blue) reframed. Choice B (9/20) counts only the green marbles (9 out of 20), ignoring the 6 red marbles that are also "not blue." Choice D (11/20) adds blue + green (5 + 9 = 14)... wait — actually reflects adding blue to something incorrectly. Pro tip: For "not" probability questions, count everything EXCEPT the excluded group, then divide by the total. It's often faster than computing P(blue) and subtracting from 1.

The correct answer is A (19/66). Without replacement, compute P(both same color) for each color separately, then add. P(both red) = (5/12) × (4/11) = 20/132. P(both blue) = (4/12) × (3/11) = 12/132. P(both yellow) = (3/12) × (2/11) = 6/132. Total = (20 + 12 + 6)/132 = 38/132 = 19/66. B (1/3) estimates loosely from 3 colors. C (19/144) treats the draws as independent (with replacement), using 12² = 144 as the denominator. D (47/132) overcounts, possibly including the same pair twice. The key is using without replacement: after the first draw, there are only 11 marbles left, and one fewer marble of that color.

This is a probability question testing the addition rule for mutually exclusive events. Choice C (0.80) is correct — for mutually exclusive events (which cannot both occur), P(A or B) = P(A) + P(B) = 0.5 + 0.3 = 0.8. Choice A (0.15) multiplies the probabilities: P(A) × P(B) = 0.5 × 0.3 = 0.15 — this is the formula for P(A and B) when events are INDEPENDENT, not the formula for P(A or B) when events are mutually exclusive. Choice B (0.20) subtracts: 0.5 − 0.3 = 0.2. Choice D (1.00) assumes mutually exclusive events together cover the entire sample space — but two mutually exclusive events can have probabilities that sum to less than 1 (there can be other outcomes). Pro tip: Mutually exclusive means the events cannot happen at the same time — like rolling a 2 and rolling a 5 on the same die. For mutually exclusive events: P(A or B) = P(A) + P(B). For non-mutually-exclusive events: P(A or B) = P(A) + P(B) − P(A and B). The simpler formula here is a gift — just add.

This is an expected value question testing the weighted average of outcomes. Choice A ($1.00) is correct — P(even) = P(odd) = 1/2. Expected value = (win amount × P(win)) + (loss amount × P(loss)) = ($4 × 0.5) + (−$2 × 0.5) = $2.00 − $1.00 = $1.00. Note: the loss is entered as a negative value. Choice B ($1.50) likely comes from computing (4 × 0.5) + (2 × 0.5) = 2 + 1 = 3, then dividing by 2: 3/2 = $1.50 — treating the loss as positive and then halving. Choice C ($2.00) computes only the winning term: 4 × 0.5 = 2, ignoring the loss entirely. Choice D ($3.00) adds the outcomes instead of weighting: 4 + (−2) = 2... or adds the expected gain and expected loss as positives: 2 + 1 = 3. Pro tip: Expected value = Σ(outcome × probability). For each possible outcome, multiply its value by its probability, then sum all products. Losses must be entered as negative values. A positive EV means the game favors the player; negative EV means it favors the house. Here EV = $1.00 means a player would average $1 profit per roll in the long run.

This is a probability question testing careful counting within a bounded set. Choice C (3/10) is correct — list the multiples of 3 between 1 and 20: 3, 6, 9, 12, 15, 18 — exactly 6 numbers. P = 6/20 = 3/10. Choice A (3/20) uses a numerator of 3, perhaps counting only the first three multiples (3, 6, 9) and stopping, or confusing the divisor with the count. Choice B (1/5) uses a numerator of 4, suggesting the student counted 4 multiples or computed ⌊20/3⌋ with an error. Choice D (1/3) applies a shortcut: "dividing by 3 means a 1/3 chance" — this would be true if the numbers went from 1 to infinity, but within the bounded set 1–20, there are exactly 6 multiples of 3 out of 20 numbers, not 20/3 ÷ 20 = 1/3. Pro tip: For "multiples of n from 1 to N" problems, the count is ⌊N/n⌋ — divide and take the integer part. Here: ⌊20/3⌋ = 6. Always list them to verify, especially near the boundary (18 qualifies, 21 does not).

The sample space involves two independent events: flipping a coin (2 outcomes) and rolling a die (6 outcomes). The favorable outcomes are getting heads (probability $1/2$) AND rolling an even number (2, 4, or 6, so probability $3/6 = 1/2$). Since the events are independent, we multiply: $P(\text{heads and even}) = P(\text{heads}) \times P(\text{even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Choice B incorrectly uses 1/12, possibly by multiplying $1/2 \times 1/6$ instead of recognizing there are 3 even numbers on a die.

For any event with probability P(event), the complement rule states that P(not event) = 1 - P(event). Given that P(event) = 0.25, we calculate P(not event) = 1 - 0.25 = 0.75. This represents the probability that the event does not occur. Choice A (0.25) incorrectly gives the original probability instead of its complement.

The sample space initially has 10 balls (4 white, 6 black). For the first draw, P(white) = 4/10. After drawing one white ball without replacement, 9 balls remain with 3 white. P(second white | first white) = 3/9 = 1/3. P(both white) = (4/10) × (3/9) = 12/90 = 2/15.

The probability of an event and its complement must sum to 1. If P(event) = 0.7, then P(not event) = 1 - P(event) = 1 - 0.7 = 0.3. This is a fundamental rule of probability where complementary events are mutually exclusive and exhaustive. Choice B (0.7) incorrectly gives the probability of the original event instead of its complement.