probability question shooting star There is a 91% chance of seeing a shooting star in the next hour, what is the probability of seeing a shooting star in the next half hour?
chance of seeing in an hour = .91
Chance of not seeing in an hour = .09 = (chance of not seeing in a half hour)^2
Chance of not seeing in a half hour = sqrt(.09) =0.3
Chance of seeing in a half hour = 0.7
Just out of curiosity, what exactly is the problem with saying that .455 is the prob of seeing a shooting star in a half hour? I realize that it's wrong because one can simply ask what is the prob of seeing a shooting star in 1.5 hours. You can't just add. 
Is the problem that we're double counting? What does .455+.455 even represent? Are we double counting the times we see 1 shooting star in the first half hour and 1 shooting star in the second?
Once we solve for chance of seeing a star in a half hour, can we check our answer?
P(shooting star in half hour) + P(shooting star in half hour) - P(2 shooting stars in 2 half hours) = .91?
.7 + .7 - .49 = .91
Yes?
 A: The number of "rare events" in a specific time interval is often modeled by the Poisson distribution. In this case, we interpret "seeing a shooting star" as meaning seeing at least one shooting star. So the probability that the number of shooting stars in the next hour is $0$ is $0.09$.  This corresponds to the Poisson with $\lambda$ given by $e^{-\lambda}=0.09$.
The number of shooting stars in a half-hour then has Poisson distribution with parameter $\frac{\lambda}{2}$. So the probability of not seeing a shooting star in a half hour is $e^{-\lambda}=\sqrt{0.09}=0.3$. The probability of seeing at least one is therefore $0.7$. 
Remark: The probability of seeing at least one shooting star in half an hour cannot be $0.455$. For on the usual assumption of independence, the probability of seeing at least one shooting star in an hour would be $0.455+0.455 -(0.455)(0.455)$, which is well short of $0.91$. 
In more technical terms, usually $\Pr(A\cup B)\ne \Pr(A)+\Pr(B)$. The formula does hold if the events $A$ and $B$ are disjoint. However, the it is not reasonable to assume that the events "seeing at least one in the first half hour" and "seeing at least one in the second half hour" are disjoint. The correct expression is $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$.
A: Yes. Under reasonable assumptions about independence etc., the probability of seeing a shooting star in the next half hour satisfies the quadratic equation $x+x-x^2=.91$, as you figured out for yourself. The roots are 0.7 and 1.3, but 1.3 is not a probability, so $0.7$ is your answer.
