35% in each package is red.
that means nothing else than for every M&M you pick, the probability is 35% (or 0.35) that is is red.
it also means the probability to get anything but red is 1 - 0.35 = 0.65 (65%).
if I understand you correctly, the question is for the probability that there are no more than 18 red M&Ms in a package of 52.
in other words, the probabilty that there are exactly 18, or exactly 17, or exactly 16 ... or exactly 1, or no red M&Ms.
the probability to get exactly 18 red out of 52 is
0.35¹⁸ × 0.65³⁴
for one of the combinations.
and we have C(52, 18) possible combinations :
52! / (18! × (52-18)!) = 52! / (18! × 34!)
18 red = 0.35¹⁸ × 0.65³⁴ × C(52, 18) = 0.115457378...
17 red = 0.35¹⁷ × 0.65³⁵ × C(52, 17) = 0.110273578...
16 red = 0.35¹⁶ × 0.65³⁶ × C(52, 16) = 0.096708177...
15 red = 0.35¹⁵ × 0.65³⁷ × C(52, 15) = 0.077665254...
14 red = 0.35¹⁴ × 0.65³⁸ × C(52, 14) = 0.056935055...
13 red = 0.35¹³ × 0.65³⁹ × C(52, 13) = 0.037956703...
12 red = 0.35¹² × 0.65⁴⁰ × C(52, 12) = 0.022909582...
11 red = 0.35¹¹ × 0.65⁴¹ × C(52, 11) = 0.012452595...
10 red = 0.35¹⁰ × 0.65⁴² × C(52, 10) = 0.006056874...
9 red = 0.35⁹ × 0.65⁴³ × C(52, 9) = 0.002615926...
8 red = 0.35⁸ × 0.65⁴⁴ × C(52, 8) = 0.000993712...
7 red = 0.35⁷ × 0.65⁴⁵ × C(52, 7) = 0.000328083..
6 red = 0.35⁶ × 0.65⁴⁶ × C(52, 6) = 9.271902896e-005
5 red = 0.35⁵ × 0.65⁴⁷ × C(52, 5) = 2.198201902e-005
4 red = 0.35⁴ × 0.65⁴⁸ × C(52, 4) = 4.252473918e-006
3 red = 0.35³ × 0.65⁴⁹ × C(52, 3) = 6.446899235e-007
2 red = 0.35² × 0.65⁵⁰ × C(52, 2) = 7.183687719e-008
1 red = 0.35 × 0.65⁵¹ × C (52, 1) = 5.231817386e-009
0 red = 0.35⁰ × 0.65⁵² × C(52, 0) =
= 0.65⁵² = 1.868506209e-010
the probability to have no more than 18 reds is the sum of all that :
0.540472593...