a) If 20 cards are defective, 120 are not. There are C(120, 20) ≈ 2.946·10²² ways to choose 20 cards of which none are defective. There are C(140, 20) ≈ 8.272·10²³ ways to choose 20 cards. So, the probability of choosing 20 cards of which none are defective is about 2.946/82.72 ≈ 0.03562. This means the probability that at least one is defective is
1 - 0.03562 ≈ 0.96438
b) Similarly, 1 - C(135, 20)/C(140, 20) ≈ 0.54294
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C(n, k) = n!/(k!(n-k)!)