Mixing
Compute the conditional probability that it is in fact a dog picture
$begingroup$
The computer program CAD takes as input a randomly selected cat or dog picture from the internet and outputs the decision 'cat' or 'dog'. It is known that CAD correctly classified a cat picture with probability $0.6$ while a dog picture is correctly classified with probability $0.7$. The probability that a random picture is a cat picture equals $0.75$. Given that CAD classifies a picture as a cat, compute the conditional probability that it is in fact a dog picture.
I'm quite sure that this exercise is solvable with Bayer's Rule but I'm not able to compute it, because I think the way that I call the probabilities is not fully correct:
$P(T|C)=0.6$ probability that classified in a true way given that is a cat
$P(T|D)=0.7$ probability that classified in a true way given that is a dog
$P(C)=0.75$ probability that is a cat
$P(T^c|C)=?$ probability that classified in a wrong way given that is a cat
$$P(T^c|C)=frac{P(C|T^c)*P(T^c)}{P(C|T^c)*P(T^c)+P(C|T)*P(T)}$$
Where is my mistake? How can I solve it?
Edit:
$$P(Clas. Wrong|Clas. Cat)=frac{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)}{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)+P(Clas. Cat|Clas. Correct)*P(Clas. Correct)}$$
probability statistics
asked Jan 26 at 14:20
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
The computer program CAD takes as input a randomly selected cat or dog picture from the internet and outputs the decision 'cat' or 'dog'. It is known that CAD correctly classified a cat picture with probability $0.6$ while a dog picture is correctly classified with probability $0.7$. The probability that a random picture is a cat picture equals $0.75$. Given that CAD classifies a picture as a cat, compute the conditional probability that it is in fact a dog picture.
I'm quite sure that this exercise is solvable with Bayer's Rule but I'm not able to compute it, because I think the way that I call the probabilities is not fully correct:
$P(T|C)=0.6$ probability that classified in a true way given that is a cat
$P(T|D)=0.7$ probability that classified in a true way given that is a dog
$P(C)=0.75$ probability that is a cat
$P(T^c|C)=?$ probability that classified in a wrong way given that is a cat
$$P(T^c|C)=frac{P(C|T^c)*P(T^c)}{P(C|T^c)*P(T^c)+P(C|T)*P(T)}$$
Where is my mistake? How can I solve it?
Edit:
$$P(Clas. Wrong|Clas. Cat)=frac{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)}{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)+P(Clas. Cat|Clas. Correct)*P(Clas. Correct)}$$
probability statistics
asked Jan 26 at 14:20
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
The computer program CAD takes as input a randomly selected cat or dog picture from the internet and outputs the decision 'cat' or 'dog'. It is known that CAD correctly classified a cat picture with probability $0.6$ while a dog picture is correctly classified with probability $0.7$. The probability that a random picture is a cat picture equals $0.75$. Given that CAD classifies a picture as a cat, compute the conditional probability that it is in fact a dog picture.
I'm quite sure that this exercise is solvable with Bayer's Rule but I'm not able to compute it, because I think the way that I call the probabilities is not fully correct:
$P(T|C)=0.6$ probability that classified in a true way given that is a cat
$P(T|D)=0.7$ probability that classified in a true way given that is a dog
$P(C)=0.75$ probability that is a cat
$P(T^c|C)=?$ probability that classified in a wrong way given that is a cat
$$P(T^c|C)=frac{P(C|T^c)*P(T^c)}{P(C|T^c)*P(T^c)+P(C|T)*P(T)}$$
Where is my mistake? How can I solve it?
Edit:
$$P(Clas. Wrong|Clas. Cat)=frac{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)}{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)+P(Clas. Cat|Clas. Correct)*P(Clas. Correct)}$$
probability statistics
asked Jan 26 at 14:20
Mark JaconMark Jacon
1127
$endgroup$
The computer program CAD takes as input a randomly selected cat or dog picture from the internet and outputs the decision 'cat' or 'dog'. It is known that CAD correctly classified a cat picture with probability $0.6$ while a dog picture is correctly classified with probability $0.7$. The probability that a random picture is a cat picture equals $0.75$. Given that CAD classifies a picture as a cat, compute the conditional probability that it is in fact a dog picture.
I'm quite sure that this exercise is solvable with Bayer's Rule but I'm not able to compute it, because I think the way that I call the probabilities is not fully correct:
$P(T|C)=0.6$ probability that classified in a true way given that is a cat
$P(T|D)=0.7$ probability that classified in a true way given that is a dog
$P(C)=0.75$ probability that is a cat
$P(T^c|C)=?$ probability that classified in a wrong way given that is a cat
$$P(T^c|C)=frac{P(C|T^c)*P(T^c)}{P(C|T^c)*P(T^c)+P(C|T)*P(T)}$$
Where is my mistake? How can I solve it?
Edit:
$$P(Clas. Wrong|Clas. Cat)=frac{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)}{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)+P(Clas. Cat|Clas. Correct)*P(Clas. Correct)}$$
probability statistics
probability statistics
asked Jan 26 at 14:20
Mark JaconMark Jacon
1127
asked Jan 26 at 14:20
Mark JaconMark Jacon
1127
edited Jan 26 at 18:24
Mark Jacon
asked Jan 26 at 14:20
Mark JaconMark Jacon
1127
asked Jan 26 at 14:20
Mark JaconMark Jacon
1127
asked Jan 26 at 14:20
Mark JaconMark Jacon
1127
1127
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your mistake seems to be in thinking that the problem asks for (in your notation), $P(T^c|C)$, i.e., the probability that the classification is wrong (event $T^c$) given that the picture is actually a cat (event $C$). The question actually asks for the probability that the classification is wrong given that the picture is classified as a cat. In other words, you were computing the probability of a cat picture being classified as a dog, whereas the problem is about the reverse eerror.
Once you correct that issue, Bayes's theorem should do the job for you. When I did the calculation, I got a probability of $1/7$ (which you're apparently supposed to round off to $0.14$).
answered Jan 26 at 18:11
Andreas BlassAndreas Blass
50.3k452109
$endgroup$
$begingroup$
Thank you for the answer, I edited the question, but now I have problems to visualize the probabilities with this notation, can you help me?
$endgroup$
– Mark Jacon
Jan 26 at 18:24
1
$begingroup$
The probability that a picture is really a dog but is classified as a cat is $0.25times0.3=0.075$. The probability that a picture is really a cat and is classified as a cat is $0.75times0.6=0.45$. So the total probability of being classified as a cat is $0.075+0.45=0.525$. So the conditional probability of being really a dog, given that it's classified as a cat, is $0.075/0.525$, which reduces to $1/7$.
$endgroup$
– Andreas Blass
Jan 26 at 18:41
$begingroup$
Now I understand where I was wrong in the notation, thank you a lot!
$endgroup$
– Mark Jacon
Jan 26 at 18:59
add a comment |
$begingroup$
Applying Bayes' Theorem
P( Picuture is CAT/Classified as CAT) $= 0.6times 0.75 = 0.45tag1$
P( Picture is DOG/Classfied as CAT)$=0.3times 0.25 = 0.075tag2$
What is asked is P(Classified as CAT/Picture is Dog) $= frac{(2)}{(1)+(2)}$
answered Jan 26 at 15:07
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
Thank you but I can't find the correct answer with this formula
$endgroup$
– Mark Jacon
Jan 26 at 17:04
1
$begingroup$
Could you tell me what the answer is?
$endgroup$
– Satish Ramanathan
Jan 26 at 17:33
$begingroup$
is one of these $0.14$, $0.38$, $0.08$, $0.25$, $0.4$, $0.3$ and with your method I have $0.105$
$endgroup$
– Mark Jacon
Jan 26 at 17:35
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
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active
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$begingroup$
Your mistake seems to be in thinking that the problem asks for (in your notation), $P(T^c|C)$, i.e., the probability that the classification is wrong (event $T^c$) given that the picture is actually a cat (event $C$). The question actually asks for the probability that the classification is wrong given that the picture is classified as a cat. In other words, you were computing the probability of a cat picture being classified as a dog, whereas the problem is about the reverse eerror.
Once you correct that issue, Bayes's theorem should do the job for you. When I did the calculation, I got a probability of $1/7$ (which you're apparently supposed to round off to $0.14$).
answered Jan 26 at 18:11
Andreas BlassAndreas Blass
50.3k452109
$endgroup$
$begingroup$
Thank you for the answer, I edited the question, but now I have problems to visualize the probabilities with this notation, can you help me?
$endgroup$
– Mark Jacon
Jan 26 at 18:24
1
$begingroup$
The probability that a picture is really a dog but is classified as a cat is $0.25times0.3=0.075$. The probability that a picture is really a cat and is classified as a cat is $0.75times0.6=0.45$. So the total probability of being classified as a cat is $0.075+0.45=0.525$. So the conditional probability of being really a dog, given that it's classified as a cat, is $0.075/0.525$, which reduces to $1/7$.
$endgroup$
– Andreas Blass
Jan 26 at 18:41
$begingroup$
Now I understand where I was wrong in the notation, thank you a lot!
$endgroup$
– Mark Jacon
Jan 26 at 18:59
add a comment |
$begingroup$
Your mistake seems to be in thinking that the problem asks for (in your notation), $P(T^c|C)$, i.e., the probability that the classification is wrong (event $T^c$) given that the picture is actually a cat (event $C$). The question actually asks for the probability that the classification is wrong given that the picture is classified as a cat. In other words, you were computing the probability of a cat picture being classified as a dog, whereas the problem is about the reverse eerror.
Once you correct that issue, Bayes's theorem should do the job for you. When I did the calculation, I got a probability of $1/7$ (which you're apparently supposed to round off to $0.14$).
answered Jan 26 at 18:11
Andreas BlassAndreas Blass
50.3k452109
$endgroup$
$begingroup$
Thank you for the answer, I edited the question, but now I have problems to visualize the probabilities with this notation, can you help me?
$endgroup$
– Mark Jacon
Jan 26 at 18:24
1
$begingroup$
The probability that a picture is really a dog but is classified as a cat is $0.25times0.3=0.075$. The probability that a picture is really a cat and is classified as a cat is $0.75times0.6=0.45$. So the total probability of being classified as a cat is $0.075+0.45=0.525$. So the conditional probability of being really a dog, given that it's classified as a cat, is $0.075/0.525$, which reduces to $1/7$.
$endgroup$
– Andreas Blass
Jan 26 at 18:41
$begingroup$
Now I understand where I was wrong in the notation, thank you a lot!
$endgroup$
– Mark Jacon
Jan 26 at 18:59
add a comment |
$begingroup$
Your mistake seems to be in thinking that the problem asks for (in your notation), $P(T^c|C)$, i.e., the probability that the classification is wrong (event $T^c$) given that the picture is actually a cat (event $C$). The question actually asks for the probability that the classification is wrong given that the picture is classified as a cat. In other words, you were computing the probability of a cat picture being classified as a dog, whereas the problem is about the reverse eerror.
Once you correct that issue, Bayes's theorem should do the job for you. When I did the calculation, I got a probability of $1/7$ (which you're apparently supposed to round off to $0.14$).
answered Jan 26 at 18:11
Andreas BlassAndreas Blass
50.3k452109
$endgroup$
Your mistake seems to be in thinking that the problem asks for (in your notation), $P(T^c|C)$, i.e., the probability that the classification is wrong (event $T^c$) given that the picture is actually a cat (event $C$). The question actually asks for the probability that the classification is wrong given that the picture is classified as a cat. In other words, you were computing the probability of a cat picture being classified as a dog, whereas the problem is about the reverse eerror.
Once you correct that issue, Bayes's theorem should do the job for you. When I did the calculation, I got a probability of $1/7$ (which you're apparently supposed to round off to $0.14$).
answered Jan 26 at 18:11
Andreas BlassAndreas Blass
50.3k452109
answered Jan 26 at 18:11
Andreas BlassAndreas Blass
50.3k452109
answered Jan 26 at 18:11
Andreas BlassAndreas Blass
50.3k452109
answered Jan 26 at 18:11
Andreas BlassAndreas Blass
50.3k452109
50.3k452109
$begingroup$
Thank you for the answer, I edited the question, but now I have problems to visualize the probabilities with this notation, can you help me?
$endgroup$
– Mark Jacon
Jan 26 at 18:24
1
$begingroup$
The probability that a picture is really a dog but is classified as a cat is $0.25times0.3=0.075$. The probability that a picture is really a cat and is classified as a cat is $0.75times0.6=0.45$. So the total probability of being classified as a cat is $0.075+0.45=0.525$. So the conditional probability of being really a dog, given that it's classified as a cat, is $0.075/0.525$, which reduces to $1/7$.
$endgroup$
– Andreas Blass
Jan 26 at 18:41
$begingroup$
Now I understand where I was wrong in the notation, thank you a lot!
$endgroup$
– Mark Jacon
Jan 26 at 18:59
add a comment |
$begingroup$
Thank you for the answer, I edited the question, but now I have problems to visualize the probabilities with this notation, can you help me?
$endgroup$
– Mark Jacon
Jan 26 at 18:24
1
$begingroup$
The probability that a picture is really a dog but is classified as a cat is $0.25times0.3=0.075$. The probability that a picture is really a cat and is classified as a cat is $0.75times0.6=0.45$. So the total probability of being classified as a cat is $0.075+0.45=0.525$. So the conditional probability of being really a dog, given that it's classified as a cat, is $0.075/0.525$, which reduces to $1/7$.
$endgroup$
– Andreas Blass
Jan 26 at 18:41
$begingroup$
Now I understand where I was wrong in the notation, thank you a lot!
$endgroup$
– Mark Jacon
Jan 26 at 18:59
$begingroup$
Thank you for the answer, I edited the question, but now I have problems to visualize the probabilities with this notation, can you help me?
$endgroup$
– Mark Jacon
Jan 26 at 18:24
$begingroup$
Thank you for the answer, I edited the question, but now I have problems to visualize the probabilities with this notation, can you help me?
$endgroup$
– Mark Jacon
Jan 26 at 18:24
1
1
$begingroup$
The probability that a picture is really a dog but is classified as a cat is $0.25times0.3=0.075$. The probability that a picture is really a cat and is classified as a cat is $0.75times0.6=0.45$. So the total probability of being classified as a cat is $0.075+0.45=0.525$. So the conditional probability of being really a dog, given that it's classified as a cat, is $0.075/0.525$, which reduces to $1/7$.
$endgroup$
– Andreas Blass
Jan 26 at 18:41
$begingroup$
The probability that a picture is really a dog but is classified as a cat is $0.25times0.3=0.075$. The probability that a picture is really a cat and is classified as a cat is $0.75times0.6=0.45$. So the total probability of being classified as a cat is $0.075+0.45=0.525$. So the conditional probability of being really a dog, given that it's classified as a cat, is $0.075/0.525$, which reduces to $1/7$.
$endgroup$
– Andreas Blass
Jan 26 at 18:41
$begingroup$
Now I understand where I was wrong in the notation, thank you a lot!
$endgroup$
– Mark Jacon
Jan 26 at 18:59
$begingroup$
Now I understand where I was wrong in the notation, thank you a lot!
$endgroup$
– Mark Jacon
Jan 26 at 18:59
add a comment |
$begingroup$
Applying Bayes' Theorem
P( Picuture is CAT/Classified as CAT) $= 0.6times 0.75 = 0.45tag1$
P( Picture is DOG/Classfied as CAT)$=0.3times 0.25 = 0.075tag2$
What is asked is P(Classified as CAT/Picture is Dog) $= frac{(2)}{(1)+(2)}$
answered Jan 26 at 15:07
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
Thank you but I can't find the correct answer with this formula
$endgroup$
– Mark Jacon
Jan 26 at 17:04
1
$begingroup$
Could you tell me what the answer is?
$endgroup$
– Satish Ramanathan
Jan 26 at 17:33
$begingroup$
is one of these $0.14$, $0.38$, $0.08$, $0.25$, $0.4$, $0.3$ and with your method I have $0.105$
$endgroup$
– Mark Jacon
Jan 26 at 17:35
add a comment |
$begingroup$
Applying Bayes' Theorem
P( Picuture is CAT/Classified as CAT) $= 0.6times 0.75 = 0.45tag1$
P( Picture is DOG/Classfied as CAT)$=0.3times 0.25 = 0.075tag2$
What is asked is P(Classified as CAT/Picture is Dog) $= frac{(2)}{(1)+(2)}$
answered Jan 26 at 15:07
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
Thank you but I can't find the correct answer with this formula
$endgroup$
– Mark Jacon
Jan 26 at 17:04
1
$begingroup$
Could you tell me what the answer is?
$endgroup$
– Satish Ramanathan
Jan 26 at 17:33
$begingroup$
is one of these $0.14$, $0.38$, $0.08$, $0.25$, $0.4$, $0.3$ and with your method I have $0.105$
$endgroup$
– Mark Jacon
Jan 26 at 17:35
add a comment |
$begingroup$
Applying Bayes' Theorem
P( Picuture is CAT/Classified as CAT) $= 0.6times 0.75 = 0.45tag1$
P( Picture is DOG/Classfied as CAT)$=0.3times 0.25 = 0.075tag2$
What is asked is P(Classified as CAT/Picture is Dog) $= frac{(2)}{(1)+(2)}$
answered Jan 26 at 15:07
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
Applying Bayes' Theorem
P( Picuture is CAT/Classified as CAT) $= 0.6times 0.75 = 0.45tag1$
P( Picture is DOG/Classfied as CAT)$=0.3times 0.25 = 0.075tag2$
What is asked is P(Classified as CAT/Picture is Dog) $= frac{(2)}{(1)+(2)}$
answered Jan 26 at 15:07
Satish RamanathanSatish Ramanathan
10k31323
edited Jan 27 at 2:20
answered Jan 26 at 15:07
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 26 at 15:07
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 26 at 15:07
Satish RamanathanSatish Ramanathan
10k31323
10k31323
$begingroup$
Thank you but I can't find the correct answer with this formula
$endgroup$
– Mark Jacon
Jan 26 at 17:04
1
$begingroup$
Could you tell me what the answer is?
$endgroup$
– Satish Ramanathan
Jan 26 at 17:33
$begingroup$
is one of these $0.14$, $0.38$, $0.08$, $0.25$, $0.4$, $0.3$ and with your method I have $0.105$
$endgroup$
– Mark Jacon
Jan 26 at 17:35
add a comment |
$begingroup$
Thank you but I can't find the correct answer with this formula
$endgroup$
– Mark Jacon
Jan 26 at 17:04
1
$begingroup$
Could you tell me what the answer is?
$endgroup$
– Satish Ramanathan
Jan 26 at 17:33
$begingroup$
is one of these $0.14$, $0.38$, $0.08$, $0.25$, $0.4$, $0.3$ and with your method I have $0.105$
$endgroup$
– Mark Jacon
Jan 26 at 17:35
$begingroup$
Thank you but I can't find the correct answer with this formula
$endgroup$
– Mark Jacon
Jan 26 at 17:04
$begingroup$
Thank you but I can't find the correct answer with this formula
$endgroup$
– Mark Jacon
Jan 26 at 17:04
1
1
$begingroup$
Could you tell me what the answer is?
$endgroup$
– Satish Ramanathan
Jan 26 at 17:33
$begingroup$
Could you tell me what the answer is?
$endgroup$
– Satish Ramanathan
Jan 26 at 17:33
$begingroup$
is one of these $0.14$, $0.38$, $0.08$, $0.25$, $0.4$, $0.3$ and with your method I have $0.105$
$endgroup$
– Mark Jacon
Jan 26 at 17:35
$begingroup$
is one of these $0.14$, $0.38$, $0.08$, $0.25$, $0.4$, $0.3$ and with your method I have $0.105$
$endgroup$
– Mark Jacon
Jan 26 at 17:35
add a comment |
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