Mixing
Probability of Winning Election if Outcomes not Equally Likely
Clash Royale CLAN TAG#URR8PPP
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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!
probability
asked 3 hours ago
Sebastian Linde
183
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Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
3
down vote
favorite
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!
probability
asked 3 hours ago
Sebastian Linde
183
New contributor
Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
How did you arrive at $1=6mathsf P(C)$?
– joriki
3 hours ago
Did you use $Pr(D) = frac12P(C)$?
– N. F. Taussig
3 hours ago
@joriki By substituting $P(A), P(B), P(D)$ in terms of $P(C)$.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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!
probability
asked 3 hours ago
Sebastian Linde
183
New contributor
Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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!
probability
probability
asked 3 hours ago
Sebastian Linde
183
New contributor
Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Sebastian Linde
183
New contributor
Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Sebastian Linde
183
New contributor
Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Sebastian Linde
183
asked 3 hours ago
Sebastian Linde
183
183
New contributor
Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sebastian Linde is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
How did you arrive at $1=6mathsf P(C)$?
– joriki
3 hours ago
Did you use $Pr(D) = frac12P(C)$?
– N. F. Taussig
3 hours ago
@joriki By substituting $P(A), P(B), P(D)$ in terms of $P(C)$.
– Sebastian Linde
3 hours ago
add a comment |Â
1
How did you arrive at $1=6mathsf P(C)$?
– joriki
3 hours ago
Did you use $Pr(D) = frac12P(C)$?
– N. F. Taussig
3 hours ago
@joriki By substituting $P(A), P(B), P(D)$ in terms of $P(C)$.
– Sebastian Linde
3 hours ago
1
1
How did you arrive at $1=6mathsf P(C)$?
– joriki
3 hours ago
How did you arrive at $1=6mathsf P(C)$?
– joriki
3 hours ago
Did you use $Pr(D) = frac12P(C)$?
– N. F. Taussig
3 hours ago
Did you use $Pr(D) = frac12P(C)$?
– N. F. Taussig
3 hours ago
@joriki By substituting $P(A), P(B), P(D)$ in terms of $P(C)$.
– Sebastian Linde
3 hours ago
@joriki By substituting $P(A), P(B), P(D)$ in terms of $P(C)$.
– Sebastian Linde
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
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 $Pr(C)$, $Pr(C)$ 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 $Pr(C) = 1/6$.
answered 3 hours ago
N. F. Taussig
39.3k93153
1
Thanks so much.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
2
down vote
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 $$P(C)=frac29$$
answered 3 hours ago
Henry
93.9k471150
Thank you. I am sorry I can't give the check to both answers.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
1
down vote
The probabilities are in the ratio of $2,1,1 textand 0.5$ and must sum to $1$.
$$P(A) = frac22+1+1+0.5 = frac49$$
$$P(C) = frac14.5 = frac29$$
answered 2 hours ago
Phil H
2,0922311
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 $Pr(C)$, $Pr(C)$ 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 $Pr(C) = 1/6$.
answered 3 hours ago
N. F. Taussig
39.3k93153
1
Thanks so much.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
2
down vote
accepted
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 $Pr(C)$, $Pr(C)$ 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 $Pr(C) = 1/6$.
answered 3 hours ago
N. F. Taussig
39.3k93153
1
Thanks so much.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 $Pr(C)$, $Pr(C)$ 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 $Pr(C) = 1/6$.
answered 3 hours ago
N. F. Taussig
39.3k93153
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 $Pr(C)$, $Pr(C)$ 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 $Pr(C) = 1/6$.
answered 3 hours ago
N. F. Taussig
39.3k93153
answered 3 hours ago
N. F. Taussig
39.3k93153
answered 3 hours ago
N. F. Taussig
39.3k93153
answered 3 hours ago
N. F. Taussig
39.3k93153
39.3k93153
1
Thanks so much.
– Sebastian Linde
3 hours ago
add a comment |Â
1
Thanks so much.
– Sebastian Linde
3 hours ago
1
1
Thanks so much.
– Sebastian Linde
3 hours ago
Thanks so much.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
2
down vote
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 $$P(C)=frac29$$
answered 3 hours ago
Henry
93.9k471150
Thank you. I am sorry I can't give the check to both answers.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
2
down vote
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 $$P(C)=frac29$$
answered 3 hours ago
Henry
93.9k471150
Thank you. I am sorry I can't give the check to both answers.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
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 $$P(C)=frac29$$
answered 3 hours ago
Henry
93.9k471150
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 $$P(C)=frac29$$
answered 3 hours ago
Henry
93.9k471150
answered 3 hours ago
Henry
93.9k471150
answered 3 hours ago
Henry
93.9k471150
answered 3 hours ago
Henry
93.9k471150
93.9k471150
Thank you. I am sorry I can't give the check to both answers.
– Sebastian Linde
3 hours ago
add a comment |Â
Thank you. I am sorry I can't give the check to both answers.
– Sebastian Linde
3 hours ago
Thank you. I am sorry I can't give the check to both answers.
– Sebastian Linde
3 hours ago
Thank you. I am sorry I can't give the check to both answers.
– Sebastian Linde
3 hours ago
add a comment |Â
up vote
1
down vote
The probabilities are in the ratio of $2,1,1 textand 0.5$ and must sum to $1$.
$$P(A) = frac22+1+1+0.5 = frac49$$
$$P(C) = frac14.5 = frac29$$
answered 2 hours ago
Phil H
2,0922311
add a comment |Â
up vote
1
down vote
The probabilities are in the ratio of $2,1,1 textand 0.5$ and must sum to $1$.
$$P(A) = frac22+1+1+0.5 = frac49$$
$$P(C) = frac14.5 = frac29$$
answered 2 hours ago
Phil H
2,0922311
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The probabilities are in the ratio of $2,1,1 textand 0.5$ and must sum to $1$.
$$P(A) = frac22+1+1+0.5 = frac49$$
$$P(C) = frac14.5 = frac29$$
answered 2 hours ago
Phil H
2,0922311
The probabilities are in the ratio of $2,1,1 textand 0.5$ and must sum to $1$.
$$P(A) = frac22+1+1+0.5 = frac49$$
$$P(C) = frac14.5 = frac29$$
answered 2 hours ago
Phil H
2,0922311
answered 2 hours ago
Phil H
2,0922311
answered 2 hours ago
Phil H
2,0922311
answered 2 hours ago
Phil H
2,0922311
2,0922311
add a comment |Â
add a comment |Â
Sebastian Linde is a new contributor. Be nice, and check out our Code of Conduct.
Sebastian Linde is a new contributor. Be nice, and check out our Code of Conduct.
Sebastian Linde is a new contributor. Be nice, and check out our Code of Conduct.
Sebastian Linde is a new contributor. Be nice, and check out our Code of Conduct.
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1
How did you arrive at $1=6mathsf P(C)$?
– joriki
3 hours ago
Did you use $Pr(D) = frac12P(C)$?
– N. F. Taussig
3 hours ago
@joriki By substituting $P(A), P(B), P(D)$ in terms of $P(C)$.
– Sebastian Linde
3 hours ago