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[9 minutes] An election was held to choose the leader of a political party. - Candidate \( A \) received \( 50 \% \) of all the votes, and \( 60 \% \) of \( A^{\prime} \) s votes were cast by males. - Candidate \( B \) received \( 35 \% \) of all the votes, and \( 40 \% \) of \( B^{\prime} \) s votes were cast by males. - Candidate \( C \) received \( 15 \% \) of all the votes, and \( 20 \% \) of \( C \) s votes were cast by males. A person \( V \), who voted in the election, is selected at random. Find the probability that \( V \) (i) voted for \( A \) and is male, (ii) is female, (iii) voted for \( C \), given that \( V \) is male.

Ask by Perkins Boone. in Singapore
Mar 13,2025

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**(i)** The probability that \( V \) voted for \( A \) and is male is \( 30\% \). **(ii)** The probability that \( V \) is female is \( 53\% \). **(iii)** The probability that \( V \) voted for \( C \) given that \( V \) is male is approximately \( 6.38\% \).

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**(i)** The probability that \( V \) voted for \( A \) is \( 50\% \) or \( 0.5 \). Since \( 60\% \) of \( A \)'s votes were cast by males, the probability that \( V \) voted for \( A \) and is male is \[ P(A \text{ and male}) = 0.5 \times 0.6 = 0.3. \] **(ii)** First, compute the overall probability that a randomly selected voter is male by considering all candidates. - For candidate \( A \): \[ P(A \text{ and male}) = 0.5 \times 0.6 = 0.3. \] - For candidate \( B \): \[ P(B \text{ and male}) = 0.35 \times 0.4 = 0.14. \] - For candidate \( C \): \[ P(C \text{ and male}) = 0.15 \times 0.2 = 0.03. \] The total probability that \( V \) is male is \[ P(\text{male}) = 0.3 + 0.14 + 0.03 = 0.47. \] Thus, the probability that \( V \) is female is \[ P(\text{female}) = 1 - P(\text{male}) = 1 - 0.47 = 0.53. \] **(iii)** We are to find the probability that \( V \) voted for \( C \), given that \( V \) is male. Using the conditional probability formula: \[ P(C \mid \text{male}) = \frac{P(C \text{ and male})}{P(\text{male})}. \] We already computed \[ P(C \text{ and male}) = 0.03 \quad \text{and} \quad P(\text{male}) = 0.47. \] Thus, \[ P(C \mid \text{male}) = \frac{0.03}{0.47} \approx 0.0638. \]

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Let the total number of votes be \( x \). Candidate \( A \): - Votes for \( A = 50\% \text{ of } x = 0.5x \) - Votes for \( A \) by males = \( 60\% \text{ of } (0.5x) = 0.6 \times 0.5x = 0.3x \) Candidate \( B \): - Votes for \( B = 35\% \text{ of } x = 0.35x \) - Votes for \( B \) by males = \( 40\% \text{ of } (0.35x) = 0.4 \times 0.35x = 0.14x \) Candidate \( C \): - Votes for \( C = 15\% \text{ of } x = 0.15x \) - Votes for \( C \) by males = \( 20\% \text{ of } (0.15x) = 0.2 \times 0.15x = 0.03x \) First, we will calculate the total number of male votes: \[ \text{Total male votes} = \text{Votes for } A \text{ by males} + \text{Votes for } B \text{ by males} + \text{Votes for } C \text{ by males} \] \[ = 0.3x + 0.14x + 0.03x = 0.47x \] Now, the total number of female votes can be found by subtracting the number of male votes from the total votes: \[ \text{Total female votes} = x - 0.47x = 0.53x \] Now, let's answer the parts of the question: (i) The probability that \( V \) voted for \( A \) and is male: The number of males who voted for \( A \) is \( 0.3x \). The probability, \( P(A \text{ and male}) \): \[ P(A \text{ and male}) = \frac{\text{Votes for } A \text{ by males}}{\text{Total votes}} = \frac{0.3x}{x} = 0.3 \] (ii) The probability that \( V \) is female: The total number of females who voted is \( 0.53x \). The probability, \( P(female) \): \[ P(female) = \frac{\text{Total female votes}}{\text{Total votes}} = \frac{0.53x}{x} = 0.53 \] (iii) The probability that \( V \) voted for \( C \), given that \( V \) is male: The number of males who voted for \( C \) is \( 0.03x \). Thus, using conditional probability: \[ P(C | \text{ male}) = \frac{\text{Votes for } C \text{ by males}}{\text{Total male votes}} = \frac{0.03x}{0.47x} = \frac{0.03}{0.47} \] Calculating that gives us: \[ P(C | \text{ male}) \approx 0.0638 \text{ or } 6.38\% \] To summarize our findings: (i) \( P(A \text{ and male}) = 0.3 \) (ii) \( P(female) = 0.53 \) (iii) \( P(C | \text{ male}) \approx 0.0638 \)

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