A bag contains one red pen, four black pens, and three blue pens. Two pens are randomly chosen from the bag and are not replaced. To the nearest hundredth, what is the probability that a black pen is chosen first and then another black pen is chosen?

There are 8 pens total, and 4 are black, so the chance the first pen is black is 4/8. Since the pen is not replaced, 7 pens remain, including 3 black pens, so the chance the second pen is black is 3/7. Multiply: (4/8) × (3/7) = 12/56 ≈ 0.21. The correct choice is 0 point 2 1.

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