1. Problem: Find the probability of rolling a 3 twice on a number cube. Formula: Probability of independent events both occurring is $$P(A \text{ and } B) = P(A) \times P(B)$$. Since each roll is independent and the probability of rolling a 3 on a number cube (6 sides) is $$\frac{1}{6}$$, $$P(3 \text{ and } 3) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$$. 2. Problem: Find the probability of rolling an even number and a 5. Even numbers on a cube are 2, 4, 6, so $$P(\text{even}) = \frac{3}{6} = \frac{1}{2}$$. Probability of rolling a 5 is $$\frac{1}{6}$$. $$P(\text{even and } 5) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$. 3. Problem: Find the probability of rolling an odd number and a 2 or a 4. Odd numbers: 1, 3, 5 so $$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$$. Numbers 2 or 4: 2 outcomes, so $$P(2 \text{ or } 4) = \frac{2}{6} = \frac{1}{3}$$. $$P(\text{odd and } (2 \text{ or } 4)) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$$. 4. Problem: Find the probability of rolling a number less than 6 and a 3 or a 1. Numbers less than 6: 1, 2, 3, 4, 5 so $$P(<6) = \frac{5}{6}$$. Numbers 3 or 1: 2 outcomes, so $$P(3 \text{ or } 1) = \frac{2}{6} = \frac{1}{3}$$. $$P(<6 \text{ and } (3 \text{ or } 1)) = \frac{5}{6} \times \frac{1}{3} = \frac{5}{18}$$. 5. Problem: Choose an H and then a D from letters A-J without replacement. Total letters: 10. $$P(H) = \frac{1}{10}$$. After choosing H, letters left: 9. $$P(D|H) = \frac{1}{9}$$. $$P(H \text{ then } D) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90}$$. 6. Problem: Choose a consonant and then an E or an I without replacement. Consonants in A-J: B, C, D, F, G, H, J (7 consonants). $$P(\text{consonant}) = \frac{7}{10}$$. After choosing consonant, letters left: 9. E and I are vowels, so 2 letters. $$P(E \text{ or } I | \text{consonant}) = \frac{2}{9}$$. $$P(\text{consonant then } E \text{ or } I) = \frac{7}{10} \times \frac{2}{9} = \frac{14}{90} = \frac{7}{45}$$. 7. Problem: Choose a vowel and then an F without replacement. Vowels in A-J: A, E, I (3 vowels). $$P(\text{vowel}) = \frac{3}{10}$$. After choosing vowel, letters left: 9. $$P(F|\text{vowel}) = \frac{1}{9}$$. $$P(\text{vowel then } F) = \frac{3}{10} \times \frac{1}{9} = \frac{3}{90} = \frac{1}{30}$$. 8. Problem: Choose a vowel and then a consonant without replacement. $$P(\text{vowel}) = \frac{3}{10}$$. After vowel chosen, letters left: 9. Consonants left: 7 (since vowels removed 1 letter). $$P(\text{consonant}|\text{vowel}) = \frac{7}{9}$$. $$P(\text{vowel then consonant}) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90} = \frac{7}{30}$$. 9. Problem: Probability of choosing 2 watches from 3 clasp bracelets, 4 watches, 5 stretch bracelets. Total jewelry: $$3 + 4 + 5 = 12$$. Number of ways to choose 2 watches: $$\binom{4}{2} = 6$$. Total ways to choose any 2: $$\binom{12}{2} = \frac{12 \times 11}{2} = 66$$. $$P(2 \text{ watches}) = \frac{6}{66} = \frac{1}{11}$$. 10. Problem: Flip heads, roll a 5, roll a 2. $$P(\text{heads}) = \frac{1}{2}$$. $$P(5) = \frac{1}{6}$$. $$P(2) = \frac{1}{6}$$. $$P(\text{heads, } 5, 2) = \frac{1}{2} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{72}$$. 11. Problem: Flip tails, roll odd number, roll 4. $$P(\text{tails}) = \frac{1}{2}$$. Odd numbers: 1, 3, 5 so $$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$$. $$P(4) = \frac{1}{6}$$. $$P(\text{tails, odd, } 4) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{6} = \frac{1}{24}$$. 12. Problem: Flip tails, roll 6 or 1, roll 3. $$P(\text{tails}) = \frac{1}{2}$$. $$P(6 \text{ or } 1) = \frac{2}{6} = \frac{1}{3}$$. $$P(3) = \frac{1}{6}$$. $$P(\text{tails, } 6 \text{ or } 1, 3) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{6} = \frac{1}{36}$$. 13. Problem: Flip heads, not roll a 2, roll an even number. $$P(\text{heads}) = \frac{1}{2}$$. Not rolling a 2 means rolling any of 1,3,4,5,6 so $$P(\text{not } 2) = \frac{5}{6}$$. Even numbers: 2,4,6 so $$P(\text{even}) = \frac{3}{6} = \frac{1}{2}$$. $$P(\text{heads, not } 2, \text{even}) = \frac{1}{2} \times \frac{5}{6} \times \frac{1}{2} = \frac{5}{24}$$.