11

The probability to see a falling star in the sky over the course of one hour is 0.64.

What is the probability to see it over the course of half an hour?

janos
  • 113
Maxime Vernier
  • 213

2 Answers2

25

An hour is two half hours. If the probability in each half hour is $p$, the probability not to see a falling star in the entire hour is $(1-p)^2=1-0.64=0.36$, so $1-p=0.6$ and $p=0.4$. Note that $0.64$ was chosen such that the square root could be easily drawn with or without taking complements, so you wouldn't have noticed if you hadn't taken complements.

joriki
  • 238,052
  • 4
    This assumes that the probabilities of seeing a shooting star in two consecutive half-hours are independent. However, both cloud cover and meteor showers can last many hours, so if one of the half-hours is in a period of increased meteor activity/visibility, chances are that the second one is too. (Of course, this reasoning has the disadvantage that it means the question does not have enough data to answer it). – hmakholm left over Monica Feb 14 '13 at 17:18
8

If you think you will either see one and only one falling star in the hour or not, the chance would be $0.32$ as the $0.64$ is equally spread. The other answer assumes there is a constant density of falling stars in time and you might see more than one. This shows how the assumptions change the answer.

If you think you might see one falling star in the first minute, then are guaranteed not to see one at any other time in the hour, it would depend which half hour the question is asking about.

Ross Millikan
  • 374,822
  • 2
    Equally spread? Wouldn't that imply that in 1/0.64 hours the probability will become 1, and you are guaranteed to see one? – janos Feb 14 '13 at 12:28
  • @janos: Or maybe the distribution is zero after the end of the hour. We are not told. – Ross Millikan Feb 14 '13 at 12:31
  • 2
    So if you wait for 1.5 hours then the probability to see a shooting star is 0.96? And if you wait 2 hours its 1.28? Sorry, but this is rubbish. – fgysin Feb 14 '13 at 14:52
  • 2
    -1 simply wrong – KaptajnKold Feb 14 '13 at 15:16
  • 1
    @fgysin: you are right you cannot have this distribution continue longer than 1/0.64 hour. I just want to show that there are alternate distributions that lead to alternate answers over 30 min. joriki assumed a constant rate of meteors, a reasonable assumption, and answered on that basis. – Ross Millikan Feb 14 '13 at 15:19
  • 1
    @KaptajnKold: I don't understand why there are two downvotes on this answer. It's a very good answer that pointed out an assumption I made. The assumption was reasonable but wasn't made in the question, so Ross pointed out some other assumptions one might make and the different results they lead to. That was a helpful addition to my answer -- what do you think is "simply wrong" about it? – joriki Feb 14 '13 at 15:57
  • 1
    Imagine the following 2 situations. 1) We know 100% sure a falling star will be in the sky in the next hour, but the sky is 0.36 clouded, so the odds we see it are 0.64. 2) Based on previous seen falling stars, we know there is a change of 0.64 to see one again in the next hour. Situation 1 would result in Ross answer, since the falling star can be there in the first or the second half of the hour. In situation 2 the falling stars follow poison distribution, and the other answer would be correct. – Dorus Feb 14 '13 at 17:58
  • @RossMillikan You are right. Dorus's comment made the penny drop for me. Now I wish I could undo my -1, but I can't unless the answer is edited. I will change my vote to +1 if it is. – KaptajnKold Feb 15 '13 at 09:16
  • @KaptajnKold: No worries. – Ross Millikan Feb 15 '13 at 15:37