Here are the solutions to the questions, correcting Question 1 based on the provided image. Question 1: Given: $P(A) = \frac{1}{2}$, $P(A' \cap B) = \frac{1}{4}$, $P(A \cap B) = \frac{2}{9}$. a) Calculate $P(B)$ Step 1: Use the formula $P(B) = P(A \cap B) + P(A' \cap B)$. $$ P(B) = \frac{2}{9} + \frac{1}{4} $$ Step 2: Find a common denominator and add the fractions. $$ P(B) = \frac{2 \times 4}{9 \times 4} + \frac{1 \times 9}{4 \times 9} $$ $$ P(B) = \frac{8}{36} + \frac{9}{36} $$ $$ P(B) = \frac{17}{36} $$ a) $\boxed{\frac{17}{36}}$ b) Calculate $P(A|B)$ Step 1: Use the formula for conditional probability. $$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$ Step 2: Substitute the known values for $P(A \cap B)$ and $P(B)$. $$ P(A|B) = \frac{\frac{2}{9}}{\frac{17}{36}} $$ Step 3: Simplify the expression. $$ P(A|B) = \frac{2}{9} \times \frac{36}{17} $$ $$ P(A|B) = \frac{2 \times 4}{17} $$ $$ P(A|B) = \frac{8}{17} $$ b) $\boxed{\frac{8}{17}}$ c) Calculate $P(A' \cap B)$ This value is directly given in the problem statement. $$ P(A' \cap B) = \frac{1}{4} $$ c) $\boxed{\frac{1}{4}}$ Question 2: Let $L_A$ be the event that a voter preferred List A, and $L_B$ be the event that a voter preferred List B. Let $M$ be the event that a voter preferred a male candidate, and $F$ be the event that a voter preferred a female candidate. Given probabilities: $P(L_A) = 0.60$ $P(L_B) = 0.40$ $P(M|L_A) = 0.70 \implies P(F|L_A) = 1 - 0.70 = 0.30$ $P(M|L_B) = 0.60 \implies P(F|L_B) = 1 - 0.60 = 0.40$ 1) Draw a tree diagram to illustrate this information. ` Start | |-- P(L_A) = 0.60 --|-- P(M|L_A) = 0.70 --> Male from List A (M_A) | |-- P(F|L_A) = 0.30 --> Female from List A (F_A) | |-- P(L_B) = 0.40 --|-- P(M|L_B) = 0.60 --> Male from List B (M_B) |-- P(F|L_B) = 0.40 --> Female from List B (F_B) ` 2) Calculate the probability that: a) A female candidate wins. Step 1: Calculate the probability of a female candidate from List A. $$ P(F \cap L_A) = P(F|L_A) \times P(L_A) = 0.30 \times 0.60 = 0.18 $$ Step 2: Calculate the probability of a female candidate from List B. $$ P(F \cap L_B) = P(F|L_B) \times P(L_B) = 0.40 \times 0.40 = 0.16 $$ Step 3: Sum these probabilities to find the total probability of a female candidate winning. $$ P(F) = P(F \cap L_A) + P(F \cap L_B) = 0.18 + 0.16 = 0.34 $$ a) $\boxed{0.34}$ b) A female candidate wins given that she is from list B. This is asking for $P(F|L_B)$. From the given information, $P(F|L_B) = 1 - P(M|L_B) = 1 - 0.60 = 0.40$. b) $\boxed{0.40}$ 3) Determine the two most popular candidates who win in the post parliamentary primaries. Step 1: Calculate the overall probability for each type of candidate. • Probability of a Male candidate from List A ($M_A$): $$ P(M_A) = P(M|L_A) \times P(L_A) = 0.70 \times 0.60 = 0.42 $$ • Probability of a Female candidate from List A ($F_A$): $$ P(F_A) = P(F|L_A) \times P(L_A) = 0.30 \times 0.60 = 0.18 $$ • Probability of a Male candidate from List B ($M_B$): $$ P(M_B) = P(M|L_B) \times P(L_B) = 0.60 \times 0.40 = 0.24 $$ • Probability of a Female candidate from List B ($F_B$): $$ P(F_B) = P(F|L_B) \times P(L_B) = 0.40 \times 0.40 = 0.16 $$ Step 2: Identify the two highest probabilities. The probabilities are $0.42, 0.18, 0.24, 0.16$. The highest probability is $0.42$, corresponding to the Male candidate from List A. The second highest probability is $0.24$, corresponding to the Male candidate from List B. The two most popular candidates are the Male candidate from List A and the Male candidate from List B.