1. **Problem 1: Reading habits and conditional probabilities** Given probabilities: $P(A)=0.14$, $P(B)=0.23$, $P(C)=0.37$, $P(A\cap B)=0.08$, $P(A\cap C)=0.09$, $P(B\cap C)=0.13$, $P(A\cap B\cap C)=0.05$. From Venn diagram inside parts (excluding triple intersections), note: - $P(A \text{ only})=0.02$ - $P(B \text{ only})=0.07$ - $P(C \text{ only})=0.20$ - $P(A\cap B \text{ only})=0.03$ - $P(A\cap C \text{ only})=0.04$ - $P(B\cap C \text{ only})=0.08$ - $P(A\cap B\cap C)=0.05$ - Outside all three = 0.51 (Or use given values for intersections directly.) **(a) Find $P(A|B)$:** By definition, $$ P(A|B) = \frac{P(A\cap B)}{P(B)}. $$ We have $P(A\cap B) = 0.08$, $P(B) = 0.23$, so $$ P(A|B) = \frac{0.08}{0.23} \approx 0.3478. $$ **(b) Find $P(A|B \cup C)$:** First find $P(B \cup C)$: $$ P(B\cup C) = P(B) + P(C) - P(B\cap C) = 0.23 + 0.37 - 0.13 = 0.47. $$ Next find $P(A \cap (B \cup C)) = P((A \cap B) \cup (A \cap C))$. By inclusion-exclusion: $$ P(A \cap (B \cup C)) = P(A\cap B) + P(A\cap C) - P(A\cap B \cap C) = 0.08 + 0.09 - 0.05 = 0.12. $$ So, $$ P(A|B \cup C) = \frac{0.12}{0.47} \approx 0.2553. $$ **(c) Find $P(A|\text{reads at least one column})$:** The event "reads at least one" is $A \cup B \cup C$. Calculate $P(A \cup B \cup C)$ using triple union formula: $$ P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C). $$ Substitute: $$ = 0.14 + 0.23 + 0.37 - 0.08 - 0.09 - 0.13 + 0.05 = 0.49. $$ Calculate $P(A \cap (A \cup B \cup C)) = P(A)$ since $A$ is subset of $A \cup B \cup C$. So, $$ P(A | \text{reads at least one}) = \frac{P(A)}{P(A\cup B\cup C)} = \frac{0.14}{0.49} \approx 0.2857. $$ 2. **Problem 2: Voting probabilities for couples** Given: $P(H) = 0.21$ (husband votes), $P(W) = 0.28$ (wife votes), $P(H \cap W) = 0.15$ (both vote). **(a) Find probability at least one votes:** $$ P(H \cup W) = P(H) + P(W) - P(H \cap W) = 0.21 + 0.28 - 0.15 = 0.34. $$ **(b) Probability husband votes given wife votes:** $$ P(H|W) = \frac{P(H \cap W)}{P(W)} = \frac{0.15}{0.28} \approx 0.5357. $$ 3. **Problem 3: Disease testing and Bayes' theorem** Given: - $P(D) = 1/1000 = 0.001$ (has disease), - $P(\neg D) = 0.999$ (no disease), - $P(+|D) = 0.99$ (test positive if diseased), - $P(+|\neg D) = 0.02$ (false positive rate). Find $P(D|+)$ the probability the person has disease given positive test. By Bayes' theorem: $$ P(D|+) = \frac{P(+|D)P(D)}{P(+|D)P(D) + P(+|\neg D)P(\neg D)} = \frac{0.99 \times 0.001}{0.99 \times 0.001 + 0.02 \times 0.999}. $$ Calculate numerator: $0.00099$ Calculate denominator: $0.00099 + 0.01998 = 0.02097$ So, $$ P(D|+) = \frac{0.00099}{0.02097} \approx 0.0472. $$ 4. **Problem 4: Gas station customer behavior** Given probabilities: $$ P(A_1) = 0.40,\quad P(A_2) = 0.35,\quad P(A_3) = 0.25, $$ where $A_i$ indicate gas type (1=regular, 2=extra, 3=premium). Filling tank probabilities given gas type: $$ P(B|A_1) = 0.30,\quad P(B|A_2) = 0.60,\quad P(B|A_3) = 0.50, $$ where $B$ means "fills tank". **(a) Probability next customer uses extra gas and fills tank:** $$ P(A_2 \cap B) = P(A_2)P(B|A_2) = 0.35 \times 0.60 = 0.21. $$ **(b) Probability next customer fills tank (total):** Use law of total probability: $$ P(B) = \sum_{i=1}^3 P(A_i) P(B|A_i) = 0.40 \times 0.30 + 0.35 \times 0.60 + 0.25 \times 0.50 = 0.12 + 0.21 + 0.125 = 0.455. $$ **(c) If tank filled, find probability gas used is regular, extra, premium:** Use Bayes' theorem: $$ P(A_i|B) = \frac{P(A_i) P(B|A_i)}{P(B)}. $$ So, $$ P(A_1|B) = \frac{0.40 \times 0.30}{0.455} = \frac{0.12}{0.455} \approx 0.2637, $$ $$ P(A_2|B) = \frac{0.35 \times 0.60}{0.455} = \frac{0.21}{0.455} \approx 0.4615, $$ $$ P(A_3|B) = \frac{0.25 \times 0.50}{0.455} = \frac{0.125}{0.455} \approx 0.2747. $$ **Final answers:** 1(a) $\approx 0.348$ 1(b) $\approx 0.255$ 1(c) $\approx 0.286$ 2(a) $= 0.34$ 2(b) $\approx 0.536$ 3) $\approx 0.047$ 4(a) $= 0.21$ 4(b) $= 0.455$ 4(c) $\approx (0.264, 0.462, 0.275)$