# Passing a test with multiple attempts (conditional probability)

You have two attempts available. Must pass once to pass overall. Probability of passing on the first attempt is $$\mathbb{P}(P_1) = 0.7$$, and the probability of passing (language used in the question; I assume this is passing 'overall') is $$0.8$$. So if I denote failing by $$F$$, this would mean $$\mathbb{P}(P_1\cup P_2\cap F_1) = 0.8$$. I'm assuming that if you pass the first time, you don't take the test again.

Question. Given an individual passes, what is the probability they only passed on the second try?

My working. Using the information provided, we have $$\mathbb{P}(P_1\cup P_2\cap F_1) = \mathbb{P}(P_1) + \mathbb{P}(P_2\cap F_1) - \mathbb{P}(P_1\cap P_2\cap F_1).$$ Since you can't fail and pass the first attempt, that last term is $$0$$. Therefore, $$\mathbb{P}(P_2\cap F_1) = \mathbb{P}(P_1\cup P_2\cap F_1) - \mathbb{P}(P_1) = 0.8-0.7 = 0.1.$$ Then $$\mathbb{P}(P_2\cap F_1 | P) = \frac{\mathbb{P}(P_2\cap F_1\cap P)}{\mathbb{P}(P)} = \frac{0.1}{0.8} = 0.125.$$

I can't really see any other way to do this, but this seems way too simple; I feel like the 'trick' part here is that $$\mathbb{P}(P_2\cap F_1\cap P)\neq \mathbb{P}(P_2\cap F_1)$$, but I don't understand why not (note that I do not have an 'official answer' for this).

• Do you mean $P(P_2\cap F_1)$ in the very last expression? Commented Feb 23, 2023 at 3:51
• Note that, $P_2 \cap F_1 \cap P \subseteq P_2 \cap F_1$ also, $P_2 \cap F_1 \subseteq P$, so, $P_2\cap F_1\cap P = P_2 \cap F_1$. Commented Feb 23, 2023 at 4:00
There are no tricks at all here. $$\mathbb P(P_2\cap F_1\cap P)$$ is equal to $$\mathbb P(P_2\cap F_1)$$ because the event $$P$$ (passing at all) is a superset of $$P_2\cap F_1$$ (passing the second time); if you pass the second time you have passed, and that is that.