Bumping due to first attempt at problem.

Average number of pulls to draw either r or b from 1 bag is 34 pulls.

Therefore, assuming on average if 34 pulls then 66 balls remaining with one of desired it makes more sense on average to continue pulling through same bag for last marble for average of ~33 more pulls? With this the possible sets are possible within population.

rrbbb, rbrbb, rbbrb, rbbbr, brrbb, brbrb, brbbr, bbrrb, bbrbr, bbbrr

bbrrr, brbrr, brrbr, brrrb, rbbrr, rbrbr, rbrrb, rrbbr, rrbrb, rrrbb

Assuming right order of b & r drawn, 5 times 34 equals 170 pulls total. However, half of the time you will have one the bottom row sets. In which case you would need to continue drawing from a pool of 66 marbles since early 3rd red was drawn. Worst case scenario I have to fish 1 more blue marble out of 66, or 33 more pulls on average for a total of 203 pulls on average from second row. Since both rows have an equal chance of happening I average the two pulls needed from both scenarios (170 +203)/2 = 187 pulls from 5 bags on average needed to draw 2 red and 3 blue marbles?