What is the probability that candidate A wins the election?

1. Jan 12, 2009

Inertialforce

1. The problem statement, all variables and given/known data
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?

3. 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?

2. Jan 13, 2009

NoMoreExams

Re: Probability

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

3. Jan 13, 2009

kenewbie

Re: Probability

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

Can you take it from here?

k

4. Jan 13, 2009

alennix21

Re: Probability

say that C =5
Then B=

Last edited: Jan 13, 2009
5. Jan 13, 2009

NoMoreExams

Re: Probability

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

6. Jan 14, 2009

HallsofIvy

Staff Emeritus
Re: Probability

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.)

7. Jan 14, 2009

Defennder

Re: Probability

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.

8. Jan 14, 2009

Дьявол

Re: Probability

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.

9. Jan 14, 2009

alennix21

Re: Probability

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