Question about finding a distribution without taking into account previous events > We have 8 prisoners, each has a probability of escaping (independently) each day of $0.4$, what is the distribution of the amount of escaping prisoners on the third day? This is the answer: the probability of one escaping on the third day is $0.6^2\cdot 0.4=0.144$, so let $x$ be the wanted amount and we have $x\sim B(8,0.144)$. My question is, why we don't take into account the probability of some of the prisoners escaping before the third day? Doesn't this assume all 8 prisoners are still there by the third day?

It's the factor $0.6^2$ that takes into account the possibility of prisoners escaping before the third day. On each day, the probability of a prisoner escaping is $0.4$, so the probability of the prisoner staying is $0.6$, and this is squared because there are two days before the third day on which the prisoner might escape.