# Compute the conditional probability that it is in fact a dog picture

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: $$P(Clas. Wrong|Clas. Cat)=\frac{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)}{P(Clas. Cat|Clas. Wrong)*P(Clas. Wrong)+P(Clas. Cat|Clas. Correct)*P(Clas. Correct)}$$

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 $$0.14$$).

• 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? – Mark Jacon Jan 26 at 18:24
• The probability that a picture is really a dog but is classified as a cat is $0.25\times0.3=0.075$. The probability that a picture is really a cat and is classified as a cat is $0.75\times0.6=0.45$. 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 $1/7$. – Andreas Blass Jan 26 at 18:41
• Now I understand where I was wrong in the notation, thank you a lot! – Mark Jacon Jan 26 at 18:59

Applying Bayes' Theorem

P( Picuture is CAT/Classified as CAT) $$= 0.6\times 0.75 = 0.45\tag1$$

P( Picture is DOG/Classfied as CAT)$$=0.3\times 0.25 = 0.075\tag2$$

What is asked is P(Classified as CAT/Picture is Dog) $$= \frac{(2)}{(1)+(2)}$$

• Thank you but I can't find the correct answer with this formula – Mark Jacon Jan 26 at 17:04
• Could you tell me what the answer is? – Satish Ramanathan Jan 26 at 17:33
• is one of these $0.14$, $0.38$, $0.08$, $0.25$, $0.4$, $0.3$ and with your method I have $0.105$ – Mark Jacon Jan 26 at 17:35