Consider an example as follows.

I am running a mobile app that allows users to apply for a loan on the app.

Say a guy signed in to my app to use his phone to apply for a loan.

Call events:

A = a person has a smartphone

B = default

First of all, I assign

`P(B)=0.8`

for this guy without any information (just be conservative).Assume that in my country

`P(A) = 0.5`

, i.e. only 50% of the population do have a smartphone.Assume

`P(A|B) = 1`

, i.e. when I look to my database, all the guys who did not pay back so far do have a smartphone, that is obvious because users need a smartphone to install my app.So apply Bayes:

`P(B|A) = P(A|B) * P (B) / P(A) = 1 * 0.8 / 0.5 = 1.6`

Two problems here indeed:

1)

`P(B|A) > 1`

. I know that more than one thread on StackExchange discussed this problem in theory to prove that`P(B|A) <= 1`

in all cases but could not find why my inference is wrong.2) Adding one more bit of information, such as “this guy has a smartphone”, according to my Bayesian inference, in fact will increase the probability of default of his case, while in my intuitive inference, it does not bring any information because I know all my customers do have smartphone already. How to explain that?

**Answer**

On the face of it your assumptions are inconsistent in that you think more people will default than have smartphones but you also think all defaulters have smartphones.

Part of the problem is that some of your assumptions are for users of your app and part for the whole population and you treat these as being for the same group

If instead you just consider users of your app, you might have $P(B)=0.8$ and $P(A)=1$ and $P(A \mid B)=1$. This will now give you $P(B \mid A)= P(A\mid B) \space P (B) \space / \space P(A) = 1 \times 0.8 / 1 = 0.8$ and there are no problems there apart from the lack of value in considering $A$ since all users of your app have smartphones

**Attribution***Source : Link , Question Author : mommomonthewind , Answer Author : Henry*