1. Problem: Find the probability for two drawings of 2 balls each from a bag with 8 red and 6 blue balls, with replacement, where the first draw is 2 red balls and the second draw is 2 blue balls. Formula: Probability of independent events occurring together is the product of their individual probabilities. Step 1: Probability of drawing 2 red balls in first draw: $$P(2\text{ red}) = \frac{8}{14} \times \frac{7}{13} = \frac{56}{182} = \frac{28}{91}$$ Step 2: Since balls are replaced, probability of drawing 2 blue balls in second draw is: $$P(2\text{ blue}) = \frac{6}{14} \times \frac{5}{13} = \frac{30}{182} = \frac{15}{91}$$ Step 3: Total probability: $$P = \frac{28}{91} \times \frac{15}{91} = \frac{420}{8281} = \frac{60}{1183}$$ 2. Problem: Two draws of 2 balls each without replacement from a box with 6 red and 4 black balls. Find probability first two are red and next two are one red and one black. Step 1: Probability first two red: $$P(2\text{ red}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$$ Step 2: After removing 2 red balls, remaining balls: 4 red, 4 black. Step 3: Probability next two balls are one red and one black: $$P(1\text{ red},1\text{ black}) = \frac{4}{8} \times \frac{4}{7} + \frac{4}{8} \times \frac{4}{7} = 2 \times \frac{4}{8} \times \frac{4}{7} = \frac{32}{56} = \frac{4}{7}$$ Step 4: Total probability: $$P = \frac{1}{3} \times \frac{4}{7} = \frac{4}{21}$$ 3. Problem: Given events A and B not mutually exclusive, find: i) Probability neither A nor B occurs: $$P(\text{neither A nor B}) = 1 - P(A \cup B)$$ ii) Probability A occurs and B does not: $$P(A \cap B^c) = P(A) - P(A \cap B)$$ iii) Probability exactly one of A and B occurs: $$P(\text{exactly one}) = P(A \cup B) - P(A \cap B) = P(A) + P(B) - 2P(A \cap B)$$ 4. Problem: Probability problem solved by students A, B, C with chances 1/2, 1/3, 1/4 respectively. Find probability problem is solved. Step 1: Probability problem not solved by each: $$P(\text{not solved by A}) = \frac{1}{2}, P(\text{not solved by B}) = \frac{2}{3}, P(\text{not solved by C}) = \frac{3}{4}$$ Step 2: Probability problem not solved by all: $$P(\text{none solve}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4}$$ Step 3: Probability problem solved by at least one: $$P = 1 - \frac{1}{4} = \frac{3}{4}$$ 5. Problem: Bhutan vs India football matches, Bhutan wins probability 2/5, 3 matches played. Find: i) Bhutan loses all three: $$P(\text{lose all}) = \left(1 - \frac{2}{5}\right)^3 = \left(\frac{3}{5}\right)^3 = \frac{27}{125} = 0.216$$ ii) Bhutan wins at least one: $$P(\text{at least one win}) = 1 - P(\text{no wins}) = 1 - 0.216 = 0.784$$ Note: User answer 0.304 seems incorrect; correct is 0.784. iii) Bhutan wins at most one match: $$P(0\text{ wins}) + P(1\text{ win}) = 0.216 + \binom{3}{1} \times \frac{2}{5} \times \left(\frac{3}{5}\right)^2 = 0.216 + 3 \times \frac{2}{5} \times \frac{9}{25} = 0.216 + 3 \times \frac{18}{125} = 0.216 + \frac{54}{125} = 0.216 + 0.432 = 0.648$$ User answer 0.36 seems incorrect; correct is 0.648. 6. Problem: Given $P(A) = \frac{1}{3}$, $P(B) = \frac{3}{4}$, $P(A \cup B) = \frac{11}{12}$, find $P(A|B)$ and $P(B|A)$. Step 1: Find $P(A \cap B)$ using: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Step 2: Substitute values: $$\frac{11}{12} = \frac{1}{3} + \frac{3}{4} - P(A \cap B)$$ Step 3: Calculate sum: $$\frac{1}{3} + \frac{3}{4} = \frac{4}{12} + \frac{9}{12} = \frac{13}{12}$$ Step 4: Solve for $P(A \cap B)$: $$P(A \cap B) = \frac{13}{12} - \frac{11}{12} = \frac{2}{12} = \frac{1}{6}$$ Step 5: Calculate conditional probabilities: $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{6}}{\frac{3}{4}} = \frac{1}{6} \times \frac{4}{3} = \frac{2}{9}$$ $$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{6} \times 3 = \frac{1}{2}$$