Q: A bag has 4 red, 6 blue balls. Two drawn without replacement. Find P(both red). P(R1) = 4/10. After one red removed: P(R2|R1) = 3/9. P(both red) = (4/10)(3/9) = 12/90 = 2/15. (With replacement it would be (4/10)(4/10) = 4/25 - independent!)