Probability of Winning Election if Outcomes not Equally Likely

I just started learning probability, so my level is not very high.
I am doing a homework problem, and my answer is different than the book's. I can't understand why. I see how the answer in the book makes sense, but I also see how my procedure makes sense. Could someone please point at what I am doing wrong?

Problem:

Four candidates, A, B, C, and D, are running in an election. A is twice as likely to be elected as B. B and C are equally likely. C is twice as likely as D. What is the probability that C will win?

My answer:

The sample space is $A,B,C,D$. Since all the events are mutually exclusive, $S = Acup Bcup Ccup D$. So, $P(S) = P(A)+P(B)+P(C)+P(D)$. Since $P(S) = 1$, $P(A)+P(B)+P(C)+P(D) = 1$. Since A is twice as likely as B, $P(A) =2times P(B)$. Since B and C are equally likely, $P(B)=P(C)$. Thus, $P(A) = 2times P(C)$. Since C is twice as likely as D, $P(C) = 2times P(D)$. Therefore, the probability of the sample space in terms of C is: $1 = 6P(C)$. So, $P(C) = frac16$.

The book's answer:

Let $p = P(D)$. Since C is twice as likely as D, $P(C) = 2p$. Since B and C are equally likely, $P(B) = 2x$. Since A is twice as likely as B, $P(A)= 4p$. Since $P(S) = 1$, $p + 2p + 2p + 4p = 1$. So, $p = frac19$ and $P(C) = frac29$.

Thanks!

3 Answers
3

You correctly found that $Pr(A) = 2Pr(C)$, $Pr(B) = Pr(C)$, and that $Pr(C) = 2Pr(D)$. Since these events are mutually exclusive and exhaustive,
$$Pr(A) + Pr(B) + Pr(C) + Pr(D) = 1$$
Substituting $2Pr(C)$ for $Pr(A)$, $Pr(C)$ for $Pr(B)$, and $frac12Pr(C)$ for $Pr(D)$ gives
beginalign*
2Pr(C) + Pr(C) + Pr(C) + frac12Pr(C) & = 1\
4Pr(C) + 2Pr(C) + 2Pr(C) + Pr(C) & = 2\
9Pr(C) & = 2\
Pr(C) & = frac29
endalign*
I suspect that you made an incorrect substitution for $Pr(D)$. If you wrote $2Pr(C)$ for $Pr(D)$, you would have obtained the incorrect answer $Pr(C) = 1/6$.

You correctly have $P(B)=P(C)$ and $P(A)=2P(C)$ but you have accidentally misused $P(C)=2P(D)$, which is equivalent to $P(D)=frac12P(C)$

Done correctly you would then have $$2P(C)+P(C)+P(C)+frac12P(C)=1$$ which, since $2+1+1+frac12 = frac92$, would lead to $$frac92P(C)=1$$ and thus $$P(C)=frac29$$

The probabilities are in the ratio of $2,1,1 textand 0.5$ and must sum to $1$.
$$P(A) = frac22+1+1+0.5 = frac49$$

$$P(C) = frac14.5 = frac29$$

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