You are standing at a bus stop watching cars go by. The probability of observing at least one accident in an hour interval is $3/4$
What is the probability of observing at least one accident within thirty minutes? Assume that the probability of observing an accident at any moment within an hour interval is uniform.
I understand that we can consider $2$ subsequent intervals of $30$ minutes, and we can denote the probability of not observing an accident as $x$. The probability of not observing any accident within an hour interval is $1/4$. Therefore we can simply write:
$x^2 = 1/4$
$x = 1/2$
Thus to find the probability of witnessing at least one accident in a $30$ minute interval is $1 - x = 1 - 1/2 = 1/2$.
My question is why doesn't this work when we take $x$ as the probability that we observe at least one accident in a $30$ minute interval and write:
$x^2 = 3/4$ (then solve for $x$)
Thanks :D
Given two timestamps $a$ and $b$ (in hours) with $a<b$, let $A(a,b)$ be the event of an accident occuring in the time interval from $a$ to $b$. Let $\overline{A(a,b)}$ denote the complementary event, i.e. there is no accident in the interval from $a$ to $b$. Now, we have $$\overline{A(0,1)} = \overline{A(0,0.5)}\cap \overline{A(0.5,1)}$$ Hence, assuming independence of $\overline{A(0,0.5)}$ and $\overline{A(0.5,1)}$ we obtain $$P\left(\overline{A(0,1)}\right) = P\left(\overline{A(0,0.5)}\right)P\left(\overline{A(0.5,1)}\right)$$ This is the reasoning you are using under the hood in your first calculation. However, it is not true that $${A(0,1)} = {A(0,0.5)}\cap {A(0.5,1)}$$ since for $A(0,1)$ to happen, you only need at least one of $A(0,0.5)$ and $A(0.5,1)$ to happen, not necessarily both. In other words, what is true is $${A(0,1)} = {A(0,0.5)}\cup {A(0.5,1)}$$ Hence, the method you proposed at the end doesn't work.