# Probability that an event that can only occur once will occur?

1. Jun 15, 2012

### moonman239

According to the FBI, the probability of being murdered within a year is 1/18690. Assuming that a person is alive at age 12 and will die at 82 due to natural causes if that person isn't shot before then, what is the probability that he will be murdered?

Obviously, whether or not the person is alive at a certain time between age 12 and 82 would depend on whether or not he or she was murdered before then. Therefore, a murder can only happen once.

2. Jun 15, 2012

### cheekujodhpur

Alright, here is the solution. Take 60 points to consider. It is a summation.
Event No.1-Probability of being murdered in first year(starting from 12). is (c = 1/18690).
Event No.2-Probablity of being not murdered in 1st year AND being murdered in 2nd year = (1-c)*c.
Event No.3-Probability of no murder in 1st and 2nd year but in 3rd = (1-c)*(1-c)*c
and so on....... till 60..
You have to add them all. You fill find that it is a geometric progression. Take 'c' common to get something like P = c(1+(1-c)+(1-c)^2 +(1-c)^3......till (1-c)^59)...
Add them using the summation rule for geometric progression, if u don't know, Google is your friend.
The answer that you'll get is....3.2*10^-3.
thanks..it was a good question.
P.S. In fractions, it is 2/625. The probability increases with age, yes.

3. Jun 16, 2012

### haruspex

Simplest to approximate as a negative exponential: prob of not being murdered in next 70 years is (1-1/18690)70 ≈ e-70/18690 = 1 - 0.0037.
So prob of being murdered is about 0.37%.
cheekujodhpur, what did you mean by "the probability increases with age"? If you increase the 82, sure, but if you take 82 as fixed and ask, for your current age, what is the probability that you will die violently then it decreases with age (as you'd expect).

4. Jun 16, 2012

### cheekujodhpur

"If you increase the 82, sure, but if you take 82 as fixed and ask, for your current age, what is the probability that you will die violently then it decreases with age (as you'd expect)."

Yes, that's exactly what I meant.

I got 0.32%, so that's the same. I just took 60 instead of 70, my mistake!
But I don't know much about the approximation. Can you give a link or just post why the approximation is valid. Is it for probabilities or for any geometric progression?

5. Jun 16, 2012

### haruspex

(1-x)n = e-nx + O(x2) is a standard approximation for small x, large n. Expand each and compare the first three terms.

6. Jun 16, 2012

### cheekujodhpur

Thanks!

7. Jun 16, 2012

### D H

Staff Emeritus
Or just use the binomial expansion and bypass the exponential function, (1-x)n = 1-nx+O(x2)

8. Jun 16, 2012

### moonman239

Hey, thanks for your help guys. For a minute there, I thought that this theoretical person could not safely expect to live to be 82 years old.

9. Jun 16, 2012

### SW VandeCarr

In fact, on a purely probabilistic basis, for any finite time no matter how large, there is a non zero probability that a person would survive that long. So for a sufficiently large population, there would be a theoretic person that would live 100,000 years. This, of course, has no basis in biology.

In terms of the probability of being murdered, the model would not hold for the 100,000 year old person. In terms of the model, probably the best one can do is assume the proportion of causes of death would be constant. The calculation above needs to be corrected for overall survival in terms of death from any cause.

Last edited: Jun 16, 2012
10. Jun 16, 2012

### viraltux

Interesting... but 0.37% is not that small percentage, don't you think? That means, roughly speaking, that a community of around 300 persons can expect that one of them will be murdered.

If you consider that the number of people we know plus acquaintances can easily be around 300 persons that would mean that most 82 year old persons know of someone in their circles who has been murdered. Mmm... that might be an interesting survey.

11. Jun 16, 2012

### SW VandeCarr

As I said in my previous post, this is a misapplication of statistics. You have to consider survival in terms of all cause death. If you just consider the murder rate, then at some point nearly everyone gets murdered.

12. Jun 16, 2012

### moonman239

I know that.

This person will not die until he reaches age 82, if he is not murdered. As mentioned before, this person has a 68% chance of living to be 82.

13. Jun 16, 2012

### D H

Staff Emeritus
The probability of living to 82 per this problem is 99.63%, not 68%. You missed the decimal point on the 0.37%.

14. Jun 16, 2012

### SW VandeCarr

To make more sense of this question, say the probability that the person will die before age 82 IS 0.32. Then using the usual proportional hazard model, the probability that the person will be murdered given they die is (1/18690)/(0.32). The cumulative effect is already accounted for in the probability of death before age 82. The proportion of hazards is assumed to remain constant.

If you're asking the probability of being murdered given you are otherwise immortal, you will have a ridiculously long mean life expectancy. This is why there are so many jokes about statistics.

Last edited: Jun 16, 2012
15. Jun 16, 2012

### viraltux

You are of course right, I was just suggesting that it would be interesting to make a survey among 80 years old fellas or older and see how the distribution of number of murdered acquaintances behave... I have the feeling that even accounting for all the parameters you mention the expected value for this variable might be (sadly) higher than most of us would expect.

16. Jun 16, 2012

### SW VandeCarr

Note, the proportion of murder in a younger age group is larger as a fraction of all deaths. So for a young person the expectation of dying before a certain age might be 0.001 while the murder rate remains fixed. So for a fixed murder rate, 0.00005/.001 you have 0.05 as a proportion of murders among all deaths. For the 82 year olds we have 0.00005/0.32 over a lifetime so we have 0.00015. When the entire cohort is dead, we have 0.00005/1 were murdered which is what it should be.

The mistake is trying to make murders independent of all deaths Being murdered is conditional on being dead although the reverse is not true.

Last edited: Jun 16, 2012
17. Jun 16, 2012

### D H

Staff Emeritus
Stop. That.

This is a basic probability problem whose goal is to show students how to calculate simple probabilities. It is nothing more than that. You are reading far too much into this simple problem. Suppose the problem had been written as

According to the Froboz Company, one out of 18690 Froboz widgets fail every month. Assuming such a widget is used in a device that will be dismantled in 70 months, what is the probability that the widget will fail before the device is dismantled?
The answer is exactly the same, about 0.003738, or 0.3738%. Just as the original wording ignores finer details such as the fact that the murder rate varies with age, this re-wording also ignore finer details such as the bathtub curve nature of device failures.

Those finer details are irrelevant and they are a derailment of this thread. It's obvious that this is a simplification as there are lots of ways people can die besides being murdered and dropping dead of old age at 82. It's obvious that this is a simplification because the murder rate depends on a lot of factors: Age, where one lives, income bracket, whether or not one's idea of a good time is to go pick a fight at the neighborhood biker bar, ...

Because of this derailment over whether this question is valid, nobody has pointed out that there are two ways to solve this problem:
• Calculate the probability directly as a sum of conditional probabilities, or
• Calculate the probability indirectly by solving the problem that the person won't be murdered (or the device won't fail).

The former method is rather nasty while the latter is very simple. This should be the takeaway point from this question.

18. Jun 16, 2012

### haruspex

No, there'll be a correlation. In some communities the expected value of murders of your acquaintances (before or after your own demise) will be more than 1/300, in others rather less. So in a random community of 300, the odds of one or more having been murdered may be a lot less than 1.
A second correlation arises. If you are 82, you are more likely to be in a community with a low murder rate.

19. Jun 17, 2012

### viraltux

Well, obviously, grandpa living in "Sunny Valley" community doesn't have much chances to get kill, but when I said acquaintances and circles I meant throughout the whole life of the older than 80 years old person. e.g. I would count in this survey something like an old school mate being killed.

I don't know, I think the expected value for the whole population might not be much lower than one, and I am not even considering casualties of war but just common crimes.

20. Jun 17, 2012

### SW VandeCarr

I would ask this question for anyone here, because it is relevant to the OP's question. Given a cohort of N persons (people born in the same year), and a fixed murder rate of 1/18690, what is the fraction of people murdered when the entire cohort has died?

Last edited: Jun 17, 2012
21. Jun 17, 2012

### D H

Staff Emeritus
I suspect your answer to your question is N. Per this conditions of this simple problem, that's wrong.

This is a textbook problem in which some intentional simplifications have been made so as to formulate a problem that students of an introductory probability and statistics class can reasonably be expected to answer. You are reading way too much into this problem, and you apparently are missing that there is another cause of death in this simple problem. That one other cause of death: People drop dead of old age the instant they reach their 82nd birthday.

Is this realistic? Of course not. It uses a murder rate of 5.35 people per hundred thousand people per year as the basis for this 1/18690 probability of being murdered (and this isn't quite valid). It ignores that there are lots of other causes of death besides murder and old age. It ignores that people don't die of old age before they reach 82, and then everyone does. It ignores lots of things. So what? It's a textbook problem.

22. Jun 17, 2012

### viraltux

Well, assuming a constant decease rate d and murder rate m the population will decrease each year by $d \cdot m$, so if we calculate the minimum number of years t required for everyone to die we solve $max \ t \ | \ t<-\frac{log(N)}{log((1-d) \cdot (1-m))}$ and thus the proportion of murdered people would be $\frac{d^t}{m^t+d^t}$

Last edited: Jun 17, 2012
23. Jun 17, 2012

### D H

Staff Emeritus
There's something basic that is wrong with that analysis, viraltux.

Beyond that "something basic", another problem is the use of an exponential process, which is (I think) SW's complaint. Consider the US, for example, with it's overall death rate of 793.8 per 100,000 per year. If death was an exponential process, the mean age of death would be 126 and lots of people would be living to over 1,000. Death is not an exponential process. It's discrete, and it is not uniform.

The problem as posed by the OP does not suffer this problem. There are no 2,000 year old men here because old age is a cliff. Live to 82 and you die of old age with a probability of one. Murder per the OP follows a binomial distribution with probability 1/18690 per year. The probability that a newborn will die of old age is (1-1/18690)82, or about 99.562%. The probability that a newborn will die by murder is one less this, or about 0.438%. I'll denote this quantity, 1-(1-1/18690)82, as p.

Of a group of N newborns, the probability that all of them will live to 82 is (1-p)N. That exactly one will be murdered, 82*p*(1-p)N-1, that exactly two will be murdered, 82*81/2*p2*(1-p)N-2, and so on. This is a binomial distribution. The mean of this binomial distribution is N*p, so the fraction of who are murdered is just p≈0.438%, or about 1/228. That's per the assumptions laid in the opening post, which of course are not realistic.

24. Jun 17, 2012

### viraltux

I should not post on Sundays... back to the UEFA EURO 2012 :tongue:

25. Jun 17, 2012

### D H

Staff Emeritus
So far I'm doing a really lousy job flipping between the two games; I don't have picture-in-picture.