Four distinct balls are randomly distributed into three distinct boxes. If it is given that at least one box remains empty, then the conditional probability that exactly one box remains empty is:

Step-by-step solution

Let E be the event that at least one box remains empty, and F be the event that exactly one box remains empty. We need to find the conditional probability P(F|E) = n(F E) n(E) . The event E occurs if exactly one box is empty or exactly two boxes are empty. (All three cannot be empty since there are 4 balls). Case 1: Exactly two boxes are empty. This means all 4 balls go into exactly 1 box. We choose the box in ³C₁ = 3 ways. The number of ways to put all balls in it is 1^4 = 1 . So, there are 3 1 = 3 ways. Case 2: E