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?
Hide
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 10-minute periods. So if q is the probability of not seeing a shooting star over a 10-minute 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%.
Hide
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. |
Back to Top
| |
|
