B.Sc Mathematics Sequences, Probability and Counting Theory — practice questions — page 4

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_

1. There are 3 + 5 + 6 = 14 toys in total, and 5 are chosen with order irrelevant. 2. The number of ways to choose any 5 toys is C(14, 5) = 2,002. 3. The bears can be chosen in C(6, 2) = 15 ways. 4. The dogs can be chosen in C(5, 3) = 10 ways. 5. Both selections happen together, so multiply: 15 times 10 = 150 favourable selections. 6. The probability is 150/2,002. 7. Dividing numerator and denominator by 2 gives 75/1,001. 8. Adding rather than multiplying the two counts would give 25 on top, which is the trap when both groups are chosen at once. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 13 "Sequences, Probability and Counting Theory", section 13.7 Probability_