# Can't get a grasp of this probability

1. Aug 22, 2005

### kioria

This is a simple example in a text book, I haven't done probability since High school and probability being all too confusing... I cannot seem to overcome this problem. Here's the problem:

a) Assume k car accidents occured in n days. Assume that accidents are equally likely on any day. Let A = event that one accident occured each day. What is P(A)?

The solution is given as below:
Solution: $$P(A) = \frac{n(n-1)...(n-k+1)}_{n^k}$$

Can someone explain this solution or the process of obtaining this solution in plain english? Thanks.

2. Aug 22, 2005

### EnumaElish

$$P(A) = \frac {\text{number of ways event A could have happened}}{\text{total number of ways k accidents could have happened in n days}}$$

$$={\text{number of days the first accident could have happened}\times...$$
$$...\times\text{ number of days the k'th accident could have happened, which otherwise would have been accident-free}}$$
$$\left/{\text{(I need to think a little more about the denominator here)}}\right.$$

Last edited: Aug 22, 2005
3. Aug 22, 2005

### kioria

Perhaps,

P(A) = the probability of accident happening on first day AND the probability of accident happening on second day AND ... AND the probability of accident happening on the last day.

This turns out to be:
$$P(A) = \frac{n(n-1)...(n-k+1)}_{n^k}$$, since only 1 accident per days needs to happen for all n days, AND is a likely clause... and I think its right. Unless someone corrects me!

4. Aug 22, 2005

### EnumaElish

Yes, AND is right. I've been thinking about the denominator, though... Is it the number of subsets with k elements each, out of a total of n elements?

5. Aug 22, 2005

### EnumaElish

It's more like, the number of days that the 1st accident can happen all by itself: since it's the 1st accident, it could happen any day, so the 1st accident has n "choices." Then, the 2nd accident has n-1 "choices" because one of the days has been "reserved" by the 1st accident, and so on.

P.S. In this post, the order 1st, 2nd, ... does not necessarily refer to temporal priority. "1st accident" does not necessarily mean "earliest accident." It just means "the first accident being looked at." And it could have happened in any of the n days, so in principle it could have happened on Wednesday whereas the 2nd accident being looked at could have happened the day before (Tuesday).

Last edited: Aug 22, 2005
6. Aug 22, 2005

### kioria

I belive the question says:

$$N = \{k_1, k_2, ... k_n\}$$ where n is an Integer for n days. Let k denote $$k_1 + k_2 + ... + k_n$$.

I am puzzled as to why $$\frac{(n - k)}_{n}$$ is omitted. I can't seem to picture the relationship between the final result and k.

7. Aug 22, 2005

### kioria

I see what you mean there... :surprised

[EDIT] But shouldn't the choice be chosen from total number of accidents that have happened over the period of n days? :uhh:

Last edited: Aug 22, 2005
8. Aug 22, 2005

### EnumaElish

You are right!

$$P(A) = \frac{n}{n}\times...\times\frac{n-k+1}n$$

Of course!

The reason why n-1 was left out is, if you have 7 days (Mon-Sun) and 3 accidents, then the 1st accident might happen on Wed., and second on Tue. Number of days left for the 3rd accident = 5 = 7 - 2 = 7 - (3 - 1) = 7 - 3 + 1. That's because the last accident will have k - 1 days previously "reserved" by k - 1 accidents before it has a chance to "decide" which day it's going to happen.

9. Aug 22, 2005

### EnumaElish

Exactly. So if you had 3 accidents (Tue, Wed, Thu) in 7 days (Mon-Sun) then you might say, let me see on how many days the accident that happened on Tue could have happened? The answer is 7 days. Next, having reserved the day on which the 1st accident COULD HAVE happened, what is the number of days that the 2nd accident could have happened? Since I know that the "first" accident happened on ONE DAY, I have 6 days left for the 2nd one.

10. Aug 22, 2005

### kioria

Ahhh I see it now. So k acts to restrict the cardinality of the set S in relation to the final answer. Since you start off with $$\frac{n}_{n}$$ which is probability of an accident happening on any of the n days, as days go by you can only have accidents if k is sufficiently big enough for n. Otherwise if ($$k >= n$$) then the solution would be simple $$\frac{n!}_{n^k}$$.

Thanks for that.