A bag contains 5 red marbles and 3 blue marbles. You draw one marble, it is red, and you do not replace it. What is the probability the next marble drawn is also red?
A5/8 — the same probability as the first draw, because marble color doesn't change
B4/7 — one red marble was removed, leaving 4 red among 7 remaining
C5/7 — the red marble was counted once already, so we keep the same numerator
D4/8 — one marble was removed, so we reduce only the total
Without replacement, the first draw changes the available pool. After drawing one red marble, 4 red marbles remain among 7 total, giving a probability of 4/7. Option A (5/8) is the classic independence error — treating this dependent event as if nothing changed. Option D (4/8) is wrong because both the numerator (remaining red marbles) and the denominator (remaining total) must update: one marble was removed, so both change.
Question 2 Multiple Choice
Which of the following pairs of events is dependent?
ARolling a 4 on a number cube, then flipping tails on a coin
BDrawing an ace from a standard deck without replacing it, then drawing another ace
CSpinning a spinner and then rolling a die
DRandomly picking a number from 1–10 and then flipping a coin
Events are dependent when the first outcome changes the probabilities for the second. Drawing cards without replacement is the canonical example: after drawing one ace (probability 4/52), only 3 aces remain among 51 cards, so the second draw's probability (3/51) is different from the first. The other pairs all involve genuinely separate experiments — a die roll cannot affect a coin flip, and a spinner cannot affect a die — so they are independent.
Question 3 True / False
For two independent events A and B, the probability that both occur equals P(A) multiplied by P(B).
TTrue
FFalse
Answer: True
This is the multiplication rule for independent events. Independence means the outcome of A doesn't affect the probability of B, so the joint probability is simply the product of individual probabilities. For example, a coin flip (1/2) and a die roll showing 6 (1/6) together have probability (1/2)(1/6) = 1/12. Each branch in a probability tree narrows the probability multiplicatively — the final leaf is the product of all probabilities along the path.
Question 4 True / False
To find the probability that both event A and event B occur, you add their individual probabilities: P(A and B) = P(A) + P(B).
TTrue
FFalse
Answer: False
Adding probabilities gives the probability of A OR B occurring (with adjustment for overlap), not A AND B. The 'and' (joint) probability uses multiplication for independent events, not addition. Adding P(A) + P(B) will always produce a value larger than P(A and B) and can even exceed 1. The confusion between 'and' (multiply) and 'or' (add) is one of the most common errors in compound probability.
Question 5 Short Answer
A standard deck has 52 cards, 13 of which are hearts. You draw two cards without replacement. Explain why the second draw's probability is different from the first, and calculate the probability that both cards are hearts.
Think about your answer, then reveal below.
Model answer: The first draw changes the deck: if the first card is a heart, only 12 hearts remain among 51 cards. The events are dependent because sampling without replacement alters the pool for the second draw. P(both hearts) = (13/52) × (12/51) = 156/2652 = 1/17. If the first card were not a heart, the second probability would change differently — which is why the dependency must be accounted for before computing.
This problem illustrates why classifying events as dependent or independent is the essential first step. Using the original 13/52 for both draws (the independence error) would give (13/52)² = 169/2704 ≈ 1/16, which is too large — it overestimates the probability by treating the second draw as if the first never happened. The correct answer updates the numerator and denominator to reflect the changed situation.