Shooting Star
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Henry and Gretchen plan on sitting outside to look for shooting stars. They know from experience that if they watch for an hour, they will have a 90% chance of seeing a shooting star. It is a chilly night, though, so Gretchen says, "Let's only stay out for 10 minutes."
Henry says, "I was really hoping to see a shooting star tonight. If we are only out for 10 minutes, we will only have a 15% chance."
Gretchen replies, "Not true. We have a better chance than that."
Is Gretchen right? If so, what is the probability that they see a shooting star?
HintWhat is the probability that they don't see a shooting star over the course of an hour? Ten minutes?
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Answer
Gretchen is right. The probability that they will see a shooting star is about 32%.
We know that the probability that they don't see a shooting star over the course of an hour is 10%. This is the product of not seeing a shooting star for 6 consecutive 10minute periods. So if q is the probability of not seeing a shooting star over a 10minute period, we can say:
0.1 = q^6
q = 0.6813
We know that the probability that they do see a shooting star is just 1 minus the probability that they don't, or 1  0.6813, which equals about 32%.
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Comments
aardvark
Mar 17, 2005
 a good refresher on my probability course, thanks! 
CHUCKLZ
Mar 17, 2005
 ok...Had the right answer but I did not figure it out that way...good job 
lilbutt
Mar 18, 2005
 Me no like math. 
(user deleted)
Mar 25, 2005
 me no like math live w/ it 
changegurl
May 05, 2005
 Bye! 
nuccha
Oct 27, 2005
 Hmmm.... really nice work... although i think the distribution should be made known. Just a thought. 
nuccha
Oct 27, 2005
 Hmmm.... really nice work... although i think the distribution should be made known. Just a thought. 
tsimkin
Oct 27, 2005
 nuccha  when you say describe the distribution, do you mean say something like, "The arrival of shooting stars can be modeled as a Poisson distribution with a lambda of 2.302585 for one hour"? This could certainly be said (and indeed backed into, which is how I just came up with the number), but I thought the statement that they had a 90% chance of seeing at least one in an hour sounded cleaner than that. 
hidentreasure
Nov 10, 2005
 He is right you know 
Jimbo
Feb 27, 2006
 Nice puzzle. We call it a binomial distribution which is well defined anyway. 
tsimkin
Feb 27, 2006
 The number of shooting stars you see in an hour have a poisson distribution, not a binomial. The two are pretty close when the expected number (lambda in a poisson distribution) is high, but look pretty different (much more highly skewed for a poisson) when lambda is low. 
jsdodgers
Sep 13, 2006
 your brain teasers are confusing 
javaguru
Dec 11, 2008
 Nice problem.
You certainly don't need to describe the distribution other than the assumption that the events are independent and random. The reality with meteor strikes does not quite fit that description, but the problem didn't give enough information to make any other assumption.
To tsimkin: Poisson distribution! That would be useful to work out if you wanted to answer a question such as "What is the probability that you'll see x number of meteor strikes within 10 minutes?" (For example, the probability of seeing three meteors in 10 minutes would be ~0.0039.) It seems like overkill for this problem though. 
opqpop
Sep 29, 2010
 I've seen this before Thanks for the refresh. 
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