# Bayes Rule Disease diagnosis

I'm having trouble understanding the answer to the following Bayes rule question

1. I know that the the prompt provides comparisons between H1N1 and fever, flue and fever, and neither and fever. But why does it make sense to compare Pr(H1N1|Fever), Pr(Flue|Fever), Pr(Neither|Fever) to answer the question, "Is it more likely that you have H1N1, the flue, or neither ?" Fever could have been replaced with some other common condition right, like Pr(H1N1|wear blue socks), as long as that condition was given for everyone?

2. Using Bayes rule and the total law of probability I set up the problem like this:

$Pr(A|D) = \frac{Pr(A \cap D)}{Pr(D)}$

I understand that we can replace $Pr(A \cap D)$ with $Pr(D|A)Pr(A)$ using Bayes rule.

$Pr(D) = Pr(A \cap D) + Pr(\overline{A} \cap D)$ according to the law of total probability.

So why is it that I can't do...

$Pr(D) = Pr(D|A)Pr(A) + Pr(D|A)(1 - Pr(A))$

but instead have to sum up the probabilities of all the other scenarios?

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– Em.
May 15, 2017 at 5:34
• @Max will do that in the future sorry May 15, 2017 at 5:41
• You already have fever hence you have to calculate the probability of H1N1,flu or neither when you already have fever, hence It does make Sense.According to definantion of conditional probability It calculates the probability of occurance of one event(here H1N1) when an event already happened(here fever). May 15, 2017 at 5:45

1. Yes, it's possible to consider something else, like wearing blue socks. However, in the problem you have a high fever. In other words, having a fever is given.

2. Sure you can do $P(D) = P(D|A)P(A) +P(D|\bar A)P(\bar A)$, but the problem is that $P(D|\bar A)$ is not readily available. It wasn't given in the problem. If you try working it out, you end up partitioning over the scenarios anyway.

• Thanks I completely missed the first sentence of the problem. For the 2) why can't we just use $1-P(A)$? May 15, 2017 at 6:00
• You mean $1-P(A)$ instead of $P(\bar A)$? They're the same thing. $P(A)+P(\bar A) = 1$.
– Em.
May 15, 2017 at 6:01
• Should have clarified: We know P(A) = 0.01. P(complement A) = 0.99 - isn't that a readily available value? May 15, 2017 at 6:13
• Yes, I didn't say it wasn't. But what about $P(D|\bar A)$?
– Em.
May 15, 2017 at 6:14
• Ah I see what you mean, thanks May 15, 2017 at 6:14