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.

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.

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