Probability of takeover announcement in the next hour I have encountered a problem: > The probability that at least one company is going to announce a takeover in the next hour is 84%. What is the probability that at least one announces in the next 30 minutes, assuming that the distribution of announcement times is uniformly distributed? Their solution to the problem is: > Correct answer: 60% > > If the desired probability is $p$ then the probability of no announcement in a half hour is $1 - p$ so the probability of no announcement in the next hour is $(1-p)^2 = 1-0.84$ I believe that the solution is wrong because the probabilities of non announcements in consecutive half-hours are not independent so it is not possible to use the formula $(1-p)^2 = 1-0.84$. I believe that the probability is 42 % because the CDF of uniform distribution is linear. Am I correct?

While the distribution of announcement times may be uniform, the distribution of the number of announcements per hour is Poisson with mean $\lambda$ given by $$p(X\geq1)=1-p(X=0)=1-e^{-\lambda}=0.84$$ $$\implies\lambda=-\ln0.16$$

Therefore the probability of at least one announcement in a half hour period is $$1-e^{-\frac 12\lambda}=1-e^{\ln0.4}=0.6$$