To solve this problem, consider the three different outcomes you get by picking out the combinations of coloured balls. Let $(B, G)$ represent the respective number of blue and green balls. $$ $$ If both balls are blue, then we have $$ (B, G) \rightarrow (B-1, G). $$ If both balls are green, then we have $$ (B, G) \rightarrow (B+1, G-2) $$ and if we pick out different coloured balls, then we have $$ (B, G) \rightarrow (B-1, G). $$ Since the number of green balls can only change by 2 every time we draw balls from the bag, then the last ball in the bag must always be blue.