The important thing to realise in this question is that the marbles are removed without replacement. This means that the probabilities are variable as the number of marbles reduces as each one is drawn. If more than 3 yellow marbles are drawn then this implies that 4 or 5 yellow marbles must be drawn. In other words, at least 4.
Suppose:
1st draw is a yellow, probability=9/33=3/11. There are 32 marbles left, 8 of which are yellow.
2nd draw is a yellow, probability=8/32=1/4. 31 marbles left 7 of which are yellow.
3rd draw is a yellow, probability=7/31. 30 marbles left of which 6 are yellow.
4th draw is a yellow, probability=6/30=1/5. 29 marbles left of which 24 are not yellow.
a) 5th draw is not yellow, probability=24/29.
b) 5th draw is yellow, probability=5/29.
If X represents a non-yellow marble and Y represents a yellow marble then the 5 draws containing 4 yellows would be represented by:
YYYYX, YYYXY, YYXYY, YXYYY, XYYYY representing the results of 5 draws in order (1st, 2nd, 3rd, 4th, 5th).
These correspond to the combined probabilities, call them A, B, C, D, E:
A=(9/33)(8/32)(7/31)(6/30)(24/29), B=(9/33)(8/32)(7/31)(24/30)(6/29),
C=(9/33)(8/32)(24/31)(7/30)(6/29), D=(9/33)(24/32)(8/31)(7/30)(6/29),
E=(24/33)(9/32)(8/31)(7/30)(6/29). Note that the denominators are the same and the numerators are the same numbers arranged differently.
Each of these is the same value so the sum (representing A or B or C or D or E) is 5(3/11)(1/4)(7/31)(1/5)(24/29)=126/9889.
The probability of 5 yellows is (9/33)(8/32)(7/31)(6/30)(5/29)=
(3/11)(1/4)(7/31)(1/5)(5/29)=21/39556.
So the probability of more than 3 yellow marbles (that is, 4 or 5) drawn without replacement is 126/9889+21/39556=525/39556=0.01327 approx or 1.327%.
This is the probability that we draw 4 or 5 yellows (without replacement), that is, more than three.