Total ways of drawing the balls = 5C2 = 10
Note: We consider (x, y) and (y, x) as same, as order doesn't matter here.
Case 1: suppose 1 is present in the balls drawn, then the greater number could be 2, 3, 4, 5
Case 2: suppose 2 is present in the balls drawn, then the greater number could be 3, 4, 5
Case 3: suppose 3 is present in the balls drawn, then the greater number could be 4, 5
Case 3: suppose 4 is present in the balls drawn, then the greater number could be 5 only
Case 5: if 5 is present in one of the balls, this case is already considered in above parts.
Now, expected value of larger number in all the possible draws = (2 + 3 + 4 + 5 + 3 + 4 + 5 + 4 + 5 + 5) / 10 = 40 / 10 = 4