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 $=1 - 1/2 = 1/2$

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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

$\implies 1-p(2t) = (1-p(t)) \cdot (1-p(2t - t)) = (1 - p(t))^2$

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.

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Trivia

1. The assumption for the independence of disjoint intervals is called Memorylessness

2. 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.

3. 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.

4. 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.