Question

Take two marbles drawn one after another without replacement (order matters). There are 6 × 5 = 30 equally likely ordered outcomes.

Define
- A = "the first marble is red."

B = "the second marble is blue."

Then A ∩ B = "first is red AND second is blue" which is exactly one ordered outcome (red, blue), so
P(A ∩ B) = 1/30.

Compute marginal probabilities: P(A) = number of outcomes with first = red = 5/30 = 1/6. P(B) = number of outcomes with second = blue = 5/30 = 1/6. Check for independence: P(A)P(B)= (1/6)(1/6) = 1/36 ≠ 1/30 = P(A ∩ B), so A and B are dependent (the outcome of the first draw changes the probability for the second draw).

Answer 1

Your work is correct.

Quick check:
- Total ordered outcomes = 6×5 = 30.
- If there is exactly one red and one blue among the 6 marbles, then
- P(A) = P(first is red) = 5/30 = 1/6.
- P(B) = P(second is blue) = 5/30 = 1/6.
- P(A ∩ B) = P(first red and second blue) = 1/30.
- P(A)P(B) = (1/6)(1/6) = 1/36 ≠ 1/30 = P(A ∩ B), so A and B are dependent.

Also note P(B | A) = P(A ∩ B)/P(A) = (1/30)/(1/6) = 1/5, which differs from P(B)=1/6 — a direct sign of dependence (removing the first marble changes the composition for the second draw).