Why does this solution use the simplified version of Bayes' Formula?

Question

For the following problem

In a given health insurance group $20$% of policyholders have high blood pressure and $30$% have high cholesterol. Of those with high blood pressure $25$% have high cholesterol. A policyholder is randomly selected from the group. Calculate the probability that they have high blood pressure given that they have high cholesterol.

Now, this is a simple Bayesian probability question. The answer is $$P(HBP \vert HCh) = \frac{P(HCh \vert HBP) \cdot P(HBP)}{P(HCh)} = \frac{0.25 \cdot 0.2}{0.3}$$

My question is why is this the form of Bayes' Formula that is used and not the following more general form?

$$P(HBP \vert HCh) = \frac{P(HCh \vert HBP) \cdot P(HBP)}{(P(HCh \vert HBP)P(HBP) + (P(HCh \vert HBP^{c})P(HBP^{c})}$$

I know it cannot be solved this way as we don't know the quantity $P(HCh \vert HBP^{c})$ but I want to know how to know which denominator to use (just the single probability or the total probability) from the problem statement. Is it simply the case if the denominator in Bayes' formula contains information that we can't determine then we default to the single probability version? That would be a useful working understanding of how to approach these problems but I'm hoping someone can give me a more fundamental intuition.

Apologies if this question is quite simplistic, it's been a few months since I last studied the material and I've regressed quite a bit in my understanding. 

Answer 1

You should understand the basis of Bayes' Theorem, then application is easier, also try and simplify notation as much as possible, as it is the formula can become quite fearsome looking at times.

So I take $P(B) =$ P(have high Blood pressure),
$P(C) =$ P(have high Cholesterol)

And the basis of Bayes' Theorem is

$P(B \cap C) = P(B) * P(C\mid B) = P(C) * P(B \mid C)$
which provides the means to invert a conditional probability

So don't worry about the full form of Bayes' Theorem, if the "baby" Theorem suffices. When the "whole" denominator is given, why must you break it up into parts ?

$P(B\mid C) = \Large\frac{P(B \cap C)}{P(C)}$

Answer 2

You can always use whichever of the two formulas is the more convenient. Since \begin{align} P(A\,|\,B)P(B)+P\big(A\,\big|\,B^c\big)P\big(B^c\big)&=P(A\cap B)+P\big(A\cap B^c\big)\\ &=P\big((A\cap B)\cup\big(A\cap B^c\big)\big)\\ &=P\big(A\cap\big(B\cup B^c\big)\big)\\ &=P(A) \end{align} then the formula $$ P(B\,|\,A)=\frac{P(A\,|\,B)P(B)}{P(A)} $$ will always produce exactly the same result as the formula $$ P(B\,|\,A)=\frac{P(A\,|\,B)P(B)}{P(A\,|\,B)P(B)+P\big(A\,\big|\,B^c\big)P\big(B^c\big)}\ . $$ Note also that it's not possible for you to know all the quantities $\ P(A), P(B)\ $ and $\ P(A\,|\,B)\ $ without also knowing (or at least being able to work out) the quantity $\ P\big(A\,\big|\,B^c\big)\ $, because this latter quantity satisfies the identity $$ P\big(A\,\big|\,B^c\big)=\frac{P(A)-P\big(A\,\big|\,B\big)P(B)}{1-P(B)}\ . $$ In your problem, for instance, this gives you \begin{align} P\big(HCh\,\big|\,HBP^c\big)&=\frac{0.3-0.25\cdot0.2}{0.8}\\ &=0.3125\ . \end{align} Nevertheless, it happens to be much quicker and easier for you to deduce the values of the quantities $\ P(HCh),$$\, P(HBP)\ $ and $\ P(HCh\,|\,HBP)\ $ that you need to apply the first formula from the data you've been given than it is for you to deduce all the quantities $\ P(HBP),$$\, P\big(HBP^c\big),$$\,P(HCh\,|\,HBP)\ $ and $\ P\big(HCh\,\big|\,HBP^c\big)\ $ you'd need to apply the second. For this particular problem, therefore, the first formula is much more convenient (as I suspect it will also be in most other circumstances).