1. Problem 4a: A bag contains 2 white, 3 black, and 4 red balls. Find the probability of drawing: i. White or red ball ii. Red or black ball iii. Black, red, or white ball Formula: Probability of event $E$ is $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$ Total balls = $2 + 3 + 4 = 9$ i. White or red means balls that are white or red: $2 + 4 = 6$ $$P(\text{White or Red}) = \frac{6}{9} = \frac{2}{3}$$ ii. Red or black means balls that are red or black: $4 + 3 = 7$ $$P(\text{Red or Black}) = \frac{7}{9}$$ iii. Black, red, or white means any ball since all colors are included: $$P(\text{Black or Red or White}) = \frac{9}{9} = 1$$ 2. Problem 4b: Find the probability that a number chosen at random from the set $\{5,6,7,\ldots,20\}$ is either even or a multiple of 5. Set size: Numbers from 5 to 20 inclusive, total numbers = $20 - 5 + 1 = 16$ Even numbers in the set: $6,8,10,12,14,16,18,20$ (8 numbers) Multiples of 5 in the set: $5,10,15,20$ (4 numbers) Numbers that are both even and multiples of 5: $10,20$ (2 numbers) Using the formula for union of two events: $$P(\text{Even or Multiple of 5}) = \frac{|E| + |M| - |E \cap M|}{16} = \frac{8 + 4 - 2}{16} = \frac{10}{16} = \frac{5}{8}$$ 3. Problem 4c: A bag contains 3 blue and 5 white balls. Two balls are drawn one after the other without replacement. Find the probabilities: Total balls = $3 + 5 = 8$ i. Two blue balls: $$P(\text{1st blue}) = \frac{3}{8}$$ After drawing one blue, remaining blue balls = 2, total balls = 7 $$P(\text{2nd blue} | \text{1st blue}) = \frac{2}{7}$$ $$P(\text{two blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$$ ii. Two white balls: $$P(\text{1st white}) = \frac{5}{8}$$ After drawing one white, remaining white balls = 4, total balls = 7 $$P(\text{2nd white} | \text{1st white}) = \frac{4}{7}$$ $$P(\text{two white}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$ iii. One blue and one white ball (either order): Case 1: Blue then white $$P = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$$ Case 2: White then blue $$P = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$$ Total probability: $$P(\text{one blue and one white}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}$$ iv. At least one blue ball = 1 - Probability of no blue balls (both white) $$P(\text{no blue}) = P(\text{two white}) = \frac{5}{14}$$ $$P(\text{at least one blue}) = 1 - \frac{5}{14} = \frac{9}{14}$$ v. Two balls of the same color = Probability(two blue) + Probability(two white) $$= \frac{3}{28} + \frac{5}{14} = \frac{3}{28} + \frac{10}{28} = \frac{13}{28}$$