# What is the probability that candidate A wins the election?

## 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?

## 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?

## 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?

## 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

## 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?

## 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=

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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

HallsofIvy
Homework Helper

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.)

Defennder
Homework Helper

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 lets 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.