A box holds 12 chocolates, 4 of which have a nut filling. Riya eats one chocolate and then a second one. What is the probability that the first chocolate she eats has a nut filling and the second does not?

The first chocolate has a nut filling with probability 4/12 = 1/3. It is gone, so 11 chocolates remain and 8 of them are plain, making the second plain with probability 8/11. Multiplying gives (4/12) × (8/11) = 32/132 = 8/33. Holding the denominator at 12 for the second bite, as though the chocolate were put back, gives (1/3)(2/3) = 2/9. The value 16/33 allows either order, nut-then-plain or plain-then-nut, but the question fixes the order; and 1/3 stops after the first bite.