Question

2 (a) Suppose you start with one Red and one Black ball in a bag. You draw a ball at random from the bag repeatedly and look at its colour each time. After each successive draw, you double the number of Red balls and triple the number of Black balls in the bag before the next draw (eg. on the third draw there will be 4 Red and 9 Black balls in the bag). Explain why you will draw only finitely many Red balls in total with probability 1. [3 marks]

(b) In a modified game, suppose instead that for before the nth draw, you first make an n by n square arrangement of balls where the perimeter balls are all Red and any interior balls are Black ball. Thus, on the second draw there are 4 Red and 0 Black in the bag (22 in total), on the third draw there are 8 Red and 1 Black in the bag (32 in total), on the fourth there are 12 Red and 4 Black in the bag (42 in total), and so on. How many Red and Black balls will there be on the nth draw? Will you still draw only finitely many red balls? Explain your answer. [4 marks]