1. **Problem statement:** There are 36 people at a gathering from two families: 25 named Lee and 11 named Chan. We select two people at random. 2. **(a) Probability both have the name Lee:** - Total people: 36 - Number of Lee: 25 - Number of ways to choose 2 people: $\binom{36}{2} = \frac{36 \times 35}{2} = 630$ - Number of ways to choose 2 Lee: $\binom{25}{2} = \frac{25 \times 24}{2} = 300$ - Probability both are Lee: $$P = \frac{\binom{25}{2}}{\binom{36}{2}} = \frac{300}{630} = \frac{10}{21}$$ 3. **(b) Probability they are married to each other:** - Lee married couples: 8 couples (16 people) - Chan married couples: 3 couples (6 people) - Total married couples: 8 + 3 = 11 couples - Number of ways to choose 2 people: 630 (as above) - Number of ways to choose a married couple: 11 (each couple is a unique pair) - Probability: $$P = \frac{11}{630}$$ 4. **(c) Probability they are a man and a woman with the same name:** - Lee single men: 4, single women: 5 - Lee married couples: 8 couples (each couple has 1 man and 1 woman) - Chan single men: 2, single women: 3 - Chan married couples: 3 couples - Number of man-woman pairs with same name: - Lee singles: $4 \times 5 = 20$ - Lee married couples: 8 (each couple is man-woman pair) - Chan singles: $2 \times 3 = 6$ - Chan married couples: 3 - Total man-woman pairs with same name: $$20 + 8 + 6 + 3 = 37$$ - Total ways to choose 2 people: 630 - Probability: $$P = \frac{37}{630}$$