What is the probability that candidate A wins the election?

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Homework Help Overview

The discussion revolves around calculating the probability of candidate A winning an election with three candidates, where the likelihood of winning is defined in relation to each other: candidate A is twice as likely to win as candidate B, and candidate B is twice as likely to win as candidate C. Participants are trying to understand the reasoning behind the stated answer of 4/7.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to express the relationships between the probabilities of the candidates and are questioning the reasoning behind the provided answer. Some suggest writing out the given relationships and using the total probability rule, while others are exploring different ways to denote the probabilities.

Discussion Status

The discussion is ongoing, with various participants offering different perspectives on how to approach the problem. Some have suggested using algebraic expressions to relate the probabilities, while others emphasize the importance of guiding questions rather than providing direct solutions. There is no explicit consensus yet on the best approach to take.

Contextual Notes

Participants are grappling with the definitions and relationships of probabilities, and there are indications of confusion regarding the interpretation of candidate C's probability. The constraints of the problem, including the requirement that all probabilities must sum to 1, are being discussed.

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Homework Statement


In an election there are three candidates. Candidate A is twice as likely to win than candidate B and candidate B is twice as likely to win than candidate C.

What is the probability that candidate A wins the election?

Note: The answer is 4/7

The Attempt at a Solution



Could someone explain to me why the answer is 4/7 I don't really understand why it is the answer?
 
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Inertialforce said:

Homework Statement


In an election there are three candidates. Candidate A is twice as likely to win than candidate B and candidate B is twice as likely to win than candidate C.

What is the probability that candidate A wins the election?

Note: The answer is 4/7


The Attempt at a Solution



Could someone explain to me why the answer is 4/7 I don't really understand why it is the answer?

First write out all the givens, then realize that P(A) + P(B) + P(C) = 1
 


Inertialforce said:

Homework Statement


In an election there are three candidates. Candidate A is twice as likely to win than candidate B and candidate B is twice as likely to win than candidate C.

What is the probability that candidate A wins the election?

Note: The answer is 4/7


The Attempt at a Solution



Could someone explain to me why the answer is 4/7 I don't really understand why it is the answer?

A = 2B
B = 2C
C = 1 - (A + B)

Can you take it from here?

k
 


say that C =5
Then B=
 
Last edited:


Although giving a solution helps some, the socratic method helps more I believe. Ask guiding questions, then see their response instead of just solving the problem. Just my .02
 


alennix21 said:
say that C =5
Then B=

You can't say that "C= 5", C is a person, not a number.

(If you were to use C to mean the probability that candidate C wins, it still can't be 5: a probability must be between 0 and 1.)
 


alennix21 said:
say that C =5
Then B=
That's not quite the correct approach. Instead, denote a particular candidate probability of winning as P, and note that one of them will definitely win the election. So that means that the individual probabilities of candidates A,B,C must sum to 1. The statement of the problem let's you relate the individual probabilities of the other two candidates to the designated one whom has probability P of winning.
 


A=2B
B=2C

The probability equals: \frac{one event}{the whole number of events}

So it should be \frac{A}{A+B+C}

B=A/2
C=B/2 or C=A/4

Here is the result:

\frac{A}{A+A/2+A/4}

Can you continue from now on?

Regards.
 


I was thinking; to say C=5 then B=
is a concrete way of seeing the problem as a bag of marbles probelm.
 

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