Among 10 laptop computers, five are good and five have defects. Unaware of this, a customer buys 6 laptops. (a) What is the probability of exactly 2 defective laptops among them? (b) Given that at least 2 purchased laptops are defective, what is the probability that exactly 2 are defective.

You would have to use combinations.

Use combinations, The total outcomes for this is 10C6 = (10! / (6! (10-6) !) = 210

The number of outcomes with support question "a" are 5C2*5C4 = 10*5=50 (2 from 2 defective so there are 4 not defective)

The probability is 50/210 = 0.238 = 23.8%

Part b: as they say "at least" that means greater than 2, so you need the probability for 1 defective + 2 defective, for 1 defective you solve 5C1*5C5/10C6 = 0.024. Add the probabilities for 1 and 2 defective 0.238+0.024=0.262 = 26.2% probability at least 2 laptops defectives.