A store has 8 phones and 3 of them are defective. A couple buys 2 phones at random. What is the probability that neither is defective?

1. Order does not matter when two phones are bought together, so count combinations. 2. Of the 8 phones, 8 - 3 = 5 are not defective. 3. The favourable selections number C(5, 2) = 5!/(2! times 3!) = 10. 4. All possible selections number C(8, 2) = 8!/(2! times 6!) = 28. 5. The probability is therefore 10/28. 6. Dividing numerator and denominator by 2 gives 5/14. 7. Leaving the answer as 10/28 unsimplified, or using C(3, 2) on top, both miss that the good phones are the ones being counted. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 13 "Sequences, Probability and Counting Theory", section 13.7 Probability_