91Ó°ÊÓ

Conditional probability is a concept used to find the probability of an event occurring, given that another event has already occurred. In our problem, although not directly stated as a conditional probability problem, the principles are subtly in play. Knowing the probability of candidate A, adjustments are made for candidates B, C, and D based on given relationships. This situation subtly resembles conditional reasoning, as each candidate's probability is dependent on interrelated conditions: the initial stated probability for A and the relative ones for B, C, and D.

Understanding the relationships among the candidates is key here. The probability of C winning is described as being twice that of B. Similarly, D's probability is three times that of B's. Using these conditions, we find the probability for each candidate by solving equations derived from these relative relationships. Thus, conditional probability helps structure our understanding by providing a means to deduce one event's probability based on known information about others.

A probability distribution assigns a probability to each possible outcome in a specific situation and ensures that the sum of these probabilities equals one. In our exercise, the candidates' probabilities form a simple discrete distribution.

Let's break it down:
- Candidate A has a fixed probability of winning: - \( \operatorname{Pr}(A) = 1/5 \) - For candidates B, C, and D, the problem uses relationships to define probabilities: - \( C = 2B \) - \( D = 3B \) With these relationships, the sum of probabilities of all candidates must equal one, enabling us to solve for each unknown value. By framing this setup as a distribution, based on the equation \( 1/5 + B + 2B + 3B = 1 \), each variable contributes to the system, ensuring a complete probability model.

Mathematical problem solving involves a systematic approach to understanding and addressing a problem. By identifying knowns and unknowns, we form useful equations and explore relationships.

In solving for the candidates' probabilities, we began with \( \operatorname{Pr}(A) = 1/5 \) and used provided relationships between candidates B, C, and D. This step highlights forming equations, a crucial problem-solving skill. Here, we're aiming to express B, C, and D as algebraic relationships: \( C = 2B \) and \( D = 3B \).

The problem solution progresses by:

• Forming the equation: \( 1/5 + B + 2B + 3B = 1 \)
• Simplifying terms: \( 6B + 1/5 = 1 \)
• Isolating variables: Solving \( B = 2/15 \)
• Substituting values back: Finding \( C = 4/15 \), \( D = 2/5 \)

These steps exemplify the core of problem-solving, showing how understanding relationships and equations can lead to solutions.