I'm new to Bayes' theorem and I have to make a project explaining it, so I came up with this word problem;
A student has done 19 tests over two years. For 7 tests, she procrastinated but still passed. For 4 tests, she procrastinated but didn't pass. For 8 tests, she didn't procrastinate and she passed. There were no tests in which she didn't procrastinate and didn't pass. What is the probability that she will pass given that she procrastinates?
You could use Bayes' theorem to solve this, however it's clear that using the conditional probability rule is simpler. $P(Pass|P)= \frac{P(P∩Pass)}{P(P)} = \frac{7}{11} ≈ 64\%$
Is there a way of altering this problem to make it more complex so that using Bayes' theorem is necessary/simpler and more striaght forward?