Why does this equation for Bayesian probability work?

Question

Please refer to question 4 and its answer at this link: https://risingentropy.com/some-simple-probability-puzzles/ How does what is referred to as strength of evidence relate to the Bayes equation. In particular, how can you calculate the probability without assuming a prior probability?

Answer 1

This answer is an amplification of my original comment to this query.

You are given that :

If the patient has strep throat, the lab results are right 70% of the time.

If not, then the lab is right 90% of the time.

5 Lab tests were performed, resulting in 2-Y's and 3-N's
where "Y" denotes strep throat test is positive.

In this unrealistic problem, based on the above information, you are guaranteed that one of two things has occurred.

Event-1 Either the patient does have strep throat, and the results were 2-Y's and 3-N's.

Event-2 Or the patient does not have strep throat, and the results were 2-Y's and 3-N's.

Regardless of the (unspecified) normal probability that someone has Strep Throat, the question is: In the absence of any information other than that given in the problem, what are the relative probabilities of Event-1 vs Event-2.

The chance of Event-1 occurring, in the absence of any other information, is

$A = (.7)^2 \times (.3)^3.$

The chance of Event-2 occurring, in the absence of any other information, is

$B = (.1)^2 \times (.9)^3.$

Therefore, since you are guaranteed that either Event-1 or Event-2 occurred,

the chance that Event-1 occurred is $\frac{A}{A + B}$

and the chance that Event-2 occurred is $\frac{B}{A + B}.$

I can't resist this:
One way of making sense of this is to pretend that you are living in a bizarro world, where an unknown disease is of unknown origin, and all deductions re the likelihood of someone having this unknown disease will be based on the empirical results of semi-mentally-retarded testers.

Addendum

Another way of interpreting the problem is that

There is a multiple choice question with two possible answers, A and B.

Person-1 Chose choice A and Person-2 chose choice B.

The question was then graded 5 different times, by 5 different teachers, each acting independently.

It is given that when Person-1 answers a question correctly,
the probability of a teacher grading the question correctly is .7.

When Person-2 answers a question correctly,
the probability of a teacher grading the question correctly is .9.

Given that 2 teachers graded Person-1's answer as correct,
and 3 teachers graded Person-2's answer is correct,
what are the chances that Person-1's answer of "A" is correct?