An electronics store has 10 phones in stock and 4 of them are defective. A customer buys 2 phones chosen at random. What is the probability that neither phone is defective?

1. Count the good phones: 10 - 4 = 6 phones are not defective. 2. Favorable selections pick 2 phones from those 6: C(6, 2) = 15 ways. 3. Total selections pick any 2 phones from all 10: C(10, 2) = 45 ways. 4. The probability is the ratio of counts: 15/45. 5. Simplify the fraction: 15/45 = 1/3. 6. So there is a 1 in 3 chance that both phones work. 7. Option B, 3/5, is the chance a single phone is good, applied once instead of to both picks. 8. Option A squares nothing and option D treats the two draws as if 4 of 10 were good, both failing the combination count. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 13 "Sequences, Probability and Counting Theory", section 13.7 PROBABILITY_