What are my mathematical chances of being shot?

[oinkoinkpig]

Just hand over your range rover to me and I'll take the bullets for you

[Synaptik]

Originally Posted by BNR32_Coupe

Hi MegaMx!!

That's a good question, thank you for bringing that up. Now, with every single particle and wave of energy in this universe, it's location relative to other particles or frequency has a precise location or quantity - to an extent! Each particles location has a small but general probability room. This means that it is possible for a particle to go through a solid wall, for example, as illustrated in this picture:



This would be an example of quantum tunneling and probability relative to everything that exists in the universe. Let's move on towards a more specific topic, being shot in your Range Rover infront of the IGA Marketplace in Burnaby. If you factor in several key variables, you can easily find out your odds of being shot versus how many visits you make to that location in your Range Rover.




1) First, we analyze the data provided for high fuel high cost SUV's relative to their related crime rates



2) Population growth for Vancouver. Burnaby shouldn't be far off. We'll use this data and pinpoint-average the location for the IGA in Burnaby to give us an accurate crime rate figure for that block


Logistic crinoid distribution count. Convert every 1mm = 1km. Factor this in to the pinpoint-average above and you have an accurate estimate of how bad the crime will be.


Success versus frame factor (bit error rate of course). This is important for later



Let's use our findings in the following formula's!






Rule of Addition If events A and B come from the same sample space, the probability that event A and/or event B occur is equal to the probability that event A occurs plus the probability that event B occurs minus the probability that both events A and B occur.
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Note: Invoking the fact that P( A ∩ B ) = P( A )P( B | A ), the Addition Rule can also be expressed as
P(A ∪ B) = P(A) + P(B) - P(A)P( B | A )

All our data suggests that the probability that event A (being shot in your range rover in IGA burnaby) occurs is equal to 1 in 4.8816507805309E06

Hehe, that was fun, wasn't it? I get really excited when we get to do formula's on revscene. Well, I hope you guys had as much fun as I do, till then, keep your calculators recharged, guys!

poasting in epic thred la~

[MegaMx]

Originally Posted by GordonTse

as long as your RR is stock and not on chrome 24's, and you don't wear all that ed hardy, CA shit.. you should be fine.

I took the chrome's off for winter and put winter tires on stock rims. I don't think I'm going to put the chrome's back on. I don't have any gay ed hardy shirts nor do i dress like a douche but still look young.

[Leopold Stotch]

Originally Posted by MegaMx

I took the chrome's off for winter and put winter tires on stock rims. I don't think I'm going to put the chrome's back on. I don't have any gay ed hardy shirts nor do i dress like a douche but still look young.

didn't you just buy a CLS55/63 or a CL?

[JHuJHu]

Originally Posted by RFlush

Hmm doing the mathematical calculations and given the variables, every time you step outside your house you have a 50% chances of being shot. You have a 33% chances of being shot if you remain in your house and you have a 95.6777%(repeating of course) chance of being shot when you are in your SUV.

My best suggestion would be to kill yourself before another person gets the chance.

reasonable answer

[carisear]

HOLY FUCK i too, nominate BNR32 for post of the year ... and it's only Feb! hell, i haven't read ANYTHING on rs in the past 5 years that was as full of win as that post!