A bin holds 3 bunnies, 5 dogs and 6 bears. A child picks 5 toys at random. What is the probability of choosing exactly 2 bears and 3 dogs?

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_