Topic: Problem with Probability (Read 950 times)

random.org

truly random numbers

Roll a die. Note the number and go that many streets west. Now roll for left or right turn, even or odd. Now the last roll is how many houses to pass until your "random" house.

... uhm t... here is no book about how to find a house at random. 

The hidden point in my question is:
You go to a house via a way. So is any house reached with the same probability?

I hope not. "break- ins for bitcoin"?

http://www.randomhouse.co.uk/

How do you choose a house at random? 

Just to add to this. The number of houses in the suburb (n) has to be divisible by 10 otherwise it's invalid as we'd have fractional houses.

ie 0.7(0.7n-1)/(n-1), where n/10=|n/10|

Thanks for now this sounds right, confusing stuff this...

Sounds more like BTC Grand Theft Auto.

They most certainly are not independent. In order for neither house to have a security system, two things must happen, the second being conditional on the first: the first house you pick must not have a security system (which obviously has a probability of 70%), AND, having already picked a house that doesn't have security system, you must pick another house that also doesn't have a security system (that much is obvious, but what is not obvious is that the probability of this is less than 70%).

A different scenario will make the reason clear: suppose you have two shoes, 50% of which are left shoes and 50% of which are right shoes. If you pick a shoe at random, what is the probability of getting a left shoe? 50%, right? Now, if you pick two shoes, what is the probability that both of them are left shoes? I'll give you a hint: it's less than 25%.

To go back to the original question, after you've picked the first house, the second time you pick a house, there is one less house to pick from (you can't pick the same house again because the question specifically requires you to pick two houses, not one house twice). And, because the question requires the first house you picked to not have a security system, of the houses that remain, you still have the same number that have security systems. For example, if there are 100 houses, 30 of which have a security system, then after picking a house that doesn't have a security system, there are now only 99 houses left to pick from, but still 30 which have a security system (and therefore 69 that don't). So the probability of the second house not having a security system is 69/99, or 69.697%. And so the total probability of both the first house not having a security system (70%), AND the second house also not having a security system (69.697%) is 70% x 69.697% = 48.788% The fewer houses there are, the lower the probability, with the probability reaching zero when there only two to chose from in the first place (as in the shoe example above).

Thanks for the answer, if one does not know the number of houses in the suburb?

 I only have one problem with answer b.
I think the events are probabilistically independent and cannot be settled with mathematical arguments and can only be settled with impirical data to decide wheater independence is reasonalbe.

What is this? Research for a new BTC business?

a. 70%
b. less than 49%; how much less depends on how many houses are in the suburb, eg if there are 100 houses, the probability is 48.788%

What do I win?

In a suburb 30% of the households have installed security systems,
a. If a household is chosen at random from this suburb what is the probability that this household has not installed an security system?
b. If two households are choosen at random from this suburb what is the probability that neither has installed an security system?