easy  probability 
The probability of having accidents on a road in one hour is $3/4$. What is the probability of accidents in half an hour?
The joint probability of no events in two intervals of half an hour each is the same as the probability of no event in one hour.
1/2
Probability of no accident in $1$ hour = P(no accidents in the first half an hour ) * P(no accidents in the next half an hour)
$\implies 1  3/4 =p^2 \implies p=1/2$
Hence the probability of an accident in half an hour $=1  1/2 = 1/2$
For those who are more inclined towards the inner wirings
Assume that the disjoint time intervals are independent, the probability of no accidents in $[0, 2t)$ = the probability of no accidents in $[0,t)$ AND $[t,2t)$
Define $p(t)$ = probability of an accident in an interval of $t$ hours
$\implies 1p(2t) = (1p(t)) \cdot (1p(2t  t)) = (1  p(t))^2$
Given that $p(1h)=3/4 \implies p(0.5h)=1/2$ and $p(2h)=15/16$.
The way we have defined $p(t)$ above, it is the probability that at least $1$ accident happens in time interval $t$. Thus, as $t$ increases, $p(t)$ also increases.
Trivia

The assumption for the independence of disjoint intervals is called Memorylessness

The probability of having at least one accident = 3/4. But there can be any number of accidents. Typically, it follows a Poisson Distribution, with parameter $\lambda =$ average number of accidents in that period.

The waiting time for the next accident is a continuous random variable. Typically it follows an exponential distribution, with the parameter $\beta =$ average waiting time for the next event.

The occurrence rate ($\lambda$) and the average waiting time ($\beta$) are related as $\beta = 1/\lambda$. This means that if on average, there are 3 meteorites per hour, then on average we wait for (1/3) hours to see the next meteorite.