Given: P(A) = 1/2, P(A' B) = 1/4, P(A B) = 2/9.

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Question 1:
Given: $P(A) = \frac{1}{2}$ , $P(A' \cap B) = \frac{1}{4}$ , $P(A \cap B) = \frac{2}{9}$ .

a) Calculate $P(B)$

Use the formula $P(B) = P(A \cap B) + P(A' \cap B)$ .
$P(B) = \frac{2}{9} + \frac{1}{4}$
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 $P(A|B)$
Use the formula for conditional probability.
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
Substitute the known values for $P(A \cap B)$ and $P(B)$ .
$P(A|B) = \frac{\frac{2}{9}}{\frac{17}{36}}$
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.
Calculate the probability of a female candidate from List A.
$P(F \cap L_A) = P(F|L_A) \times P(L_A) = 0.30 \times 0.60 = 0.18$
Calculate the probability of a female candidate from List B.
$P(F \cap L_B) = P(F|L_B) \times P(L_B) = 0.40 \times 0.40 = 0.16$
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.
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 ( $F_B$ ):
$P(F_B) = P(F|L_B) \times P(L_B) = 0.40 \times 0.40 = 0.16$
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.

Given: P(A) = (1)/(2), P(A' B) = (1)/(4), P(A B) = (2)/(9).

Step-by-step solution

Use the formula $P(B) = P(A \cap B) + P(A' \cap B)$ .

Find a common denominator and add the fractions.

Use the formula for conditional probability.

Substitute the known values for $P(A \cap B)$ and $P(B)$ .

Simplify the expression.

Calculate the probability of a female candidate from List A.

Calculate the probability of a female candidate from List B.

Sum these probabilities to find the total probability of a female candidate winning.

Calculate the overall probability for each type of candidate.

Identify the two highest probabilities.

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