1. **Problem 02: Two cards drawn with replacement** We want conditional probabilities about the second card given the first card. (a) Probability second card is a face card given first card was red: - Total cards in deck: 52. - Red cards: 26 - Face cards: Jack, Queen, King in each suit = 12 total (3 per suit * 4 suits). - Red face cards: Jack, Queen, King of hearts and diamonds = 6. Since the draw is with replacement, the second card distribution is the same as a fresh draw. $$P(\text{Face card}) = \frac{12}{52} = \frac{3}{13}$$ This is independent of first card being red. $$\Rightarrow P(\text{Second card face} | \text{First card red}) = P(\text{Face card}) = \frac{3}{13}$$ (b) Probability second card is an ace given first was face card: - Number of face cards: 12 - Number of aces: 4 - Since with replacement, second card is independent. So: $$P(\text{Second card ace} | \text{First card face}) = P(\text{Ace}) = \frac{4}{52} = \frac{1}{13}$$ (c) Probability second card is a black jack given first card was a red ace: - Red aces: 2 (hearts and diamonds). - Black jack cards: Jack of spades and Jack of clubs = 2. Again, independent draws: $$P(\text{Black Jack}) = \frac{2}{52} = \frac{1}{26}$$ 2. **Problem 03: Prison escape attempts** Given counts for each season and category, probabilities are counted by dividing category frequency by 45 (total years). (a) Probability escapes between 16 and 20 during winter: Count from table for winter in 16–20 row: 3 $$P = \frac{3}{45} = \frac{1}{15}$$ (b) Probability more than 10 escapes in summer: More than 10 means categories 11–15, 16–20, 21–25, More than 25 Sum counts for summer: 11–15: 7, 16–20: 6, 21–25: 5, More than 25: 4 Total = 7 + 6 + 5 + 4 = 22 $$P = \frac{22}{45}$$ (c) Probability escapes between 11 and 20 inclusive in any season: Combine categories 11–15 and 16–20 for each season, sum counts and divide by total possible counts (45*4=180) - Winter: 5+3 = 8 - Spring: 8+4 = 12 - Summer: 7+6 = 13 - Fall: 7+5 = 12 Sum = 8 + 12 + 13 + 12 = 45 $$P = \frac{45}{180} = \frac{1}{4}$$ 3. **Problem 04: Education survey** Total respondents: 1000 (a) Undergraduate (U): $$P(U) = \frac{300}{1000} = 0.3$$ (b) Graduate (G): $$P(G) = \frac{700}{1000} = 0.7$$ (c) Female (F): $$P(F) = \frac{400}{1000} = 0.4$$ (d) Male-Graduate (MG): Number male graduate = 450 $$P(MG) = \frac{450}{1000} = 0.45$$ (e) Undergraduate-Female (UF): Number undergraduate female = 150 $$P(UF) = \frac{150}{1000} = 0.15$$ (f) Probability randomly selected female is graduate: Using conditional probability: $$P(G|F) = \frac{P(G \cap F)}{P(F)} = \frac{250/1000}{400/1000} = \frac{250}{400} = 0.625$$ (g) Probability randomly selected undergraduate is male: $$P(M|U) = \frac{P(M \cap U)}{P(U)} = \frac{150/1000}{300/1000} = \frac{150}{300} = 0.5$$ 4. **Problem 05: Boxes of bananas fruit damage** Total boxes: 10,000 Damaged fruit: Ecuadoran: 200 Honduran: 365 Total damaged = 200 + 365 = 565 Overripe fruit: Ecuadoran: 840 Honduran: 295 Total overripe = 840 + 295 = 1,135 (a) Probability box chosen at random contains damaged fruit: $$P(D) = \frac{565}{10000} = 0.0565$$ Probability box chosen contains overripe fruit: $$P(O) = \frac{1135}{10000} = 0.1135$$ (b) Probability randomly selected box is from Ecuador or Honduras: Since these are the only two origins, $$P(\text{Ecuador}) = \frac{6000}{10000} = 0.6$$ $$P(\text{Honduras}) = 0.4$$ Sum is 1. (c) Probability box came from Honduras given overripe fruit: Using conditional probability: $$P(H|O) = \frac{P(H \cap O)}{P(O)} = \frac{295/10000}{1135/10000} = \frac{295}{1135} \approx 0.260"$$ (d) If damaged and overripe are mutually exclusive: $$P(D \cup O) = P(D) + P(O) = 0.0565 + 0.1135 = 0.17$$ If NOT mutually exclusive, assuming overlap is zero (not given data), use the formula: $$P(D \cup O) = P(D) + P(O) - P(D \cap O)$$ Since $P(D \cap O)$ unknown, cannot numerically calculate. **Final answers:** - Problem 02: (a) $\frac{3}{13}$, (b) $\frac{1}{13}$, (c) $\frac{1}{26}$ - Problem 03: (a) $\frac{1}{15}$, (b) $\frac{22}{45}$, (c) $\frac{1}{4}$ - Problem 04: (a) 0.3, (b) 0.7, (c) 0.4, (d) 0.45, (e) 0.15, (f) 0.625, (g) 0.5 - Problem 05: (a) $P(D)=0.0565$, $P(O)=0.1135$, (b) 0.6 and 0.4, (c) $\approx$0.26, (d) mutually exclusive $P(D\cup O)=0.17$, otherwise unknown.