There's an 80% probability of a certain outcome, we get some new information that means that outcome is 4 times more likely to occur.
What's the new probability as a percentage and how do you work it out?
As I remember it the question was posed like so:
Suppose there's a student, Tom W, if you were asked to estimate the
probability that Tom is a student of computer science. Without any
other information you would only have the base rate to go by
(percentage of total students enrolled on computer science) suppose
this base rate is 80%.Then you are given a description of Tom W's personality, suppose from
this description you estimate that Tom W is 4 times more likely to be
enrolled on computer science.What is the new probability that Tom W is enrolled on computer
science.
The answer given in the book is 94.1% but I couldn't work out how to calculate it!
Another example in the book is with a base rate of 3%, 4 times more likely than this is stated as 11%.
Best Answer
The most reasonable way to match the answer in the book would be to define the likelihood to be the ratio of success over failure (aka odds): $$ q=\frac{p}{1-p} $$ then the probability as a function of the odds is $$ p=\frac{q}{1+q} $$ In your case the odds are $4:1$ so $4$ times as likely would be $16:1$ odds which has a probability of $$ \frac{16}{17}=94.1176470588235\% $$ This matches the $3\%$ to $11.0091743119266\%$ transformation, as well.
Bayes' Rule
Bayes' Rule for a single event says that $$ O(A\mid B)=\frac{P(B\mid A)}{P(B\mid\neg A)}\,O(A) $$ where the odds of $X$ is defined as earlier $$ O(X)=\frac{P(X)}{P(\neg X)}=\frac{P(X)}{1-P(X)} $$ This is exactly what is being talked about in the later addition to the question, where it is given that $$ \frac{P(B\mid A)}{P(B\mid\neg A)}=4 $$