1. **Problem statement:** A bag contains 5 yellow, 3 green, 6 blue, and 2 black discs. Four discs are chosen at random. Find the probability that: 1) All four discs are the same color. 2) All four discs are different colors. 3) Exactly two of the discs are black. 2. **Total number of discs:** $$5 + 3 + 6 + 2 = 16$$ 3. **Total ways to choose 4 discs from 16:** $$\binom{16}{4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820$$ --- ### Part 1: Probability all 4 discs are the same color - Possible colors with at least 4 discs: yellow (5), blue (6) - Number of ways to choose 4 yellow discs: $$\binom{5}{4} = 5$$ - Number of ways to choose 4 blue discs: $$\binom{6}{4} = \frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1} = 15$$ - Total favorable ways: $$5 + 15 = 20$$ - Probability: $$P = \frac{20}{1820} = \frac{\cancel{20}}{\cancel{1820}} = \frac{1}{91}$$ --- ### Part 2: Probability all 4 discs are different colors - There are 4 colors: yellow, green, blue, black - Number of ways to choose 1 disc of each color: $$\binom{5}{1} \times \binom{3}{1} \times \binom{6}{1} \times \binom{2}{1} = 5 \times 3 \times 6 \times 2 = 180$$ - Probability: $$P = \frac{180}{1820} = \frac{\cancel{180}}{\cancel{1820}} = \frac{9}{91}$$ --- ### Part 3: Probability exactly two discs are black - Number of ways to choose 2 black discs: $$\binom{2}{2} = 1$$ - Number of ways to choose remaining 2 discs from non-black discs (5+3+6=14): $$\binom{14}{2} = \frac{14 \times 13}{2} = 91$$ - Total favorable ways: $$1 \times 91 = 91$$ - Probability: $$P = \frac{91}{1820} = \frac{\cancel{91}}{\cancel{1820}} = \frac{1}{20}$$ --- **Final answers:** 1) $\frac{1}{91}$ 2) $\frac{9}{91}$ 3) $\frac{1}{20}$