Can't get a grasp of this probability

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    Probability
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Discussion Overview

The discussion revolves around a probability problem involving car accidents occurring over a set number of days. Participants explore the calculation of the probability of having one accident each day given a total of k accidents over n days, discussing the reasoning behind the provided solution and the interpretation of the formula.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the probability problem and seeks clarification on the solution provided in a textbook.
  • Another participant proposes that the probability P(A) can be expressed as the number of ways event A could occur divided by the total number of ways k accidents could happen in n days.
  • Some participants suggest that P(A) can be calculated as the product of probabilities for each day, leading to the formula P(A) = \frac{n(n-1)...(n-k+1)}_{n^k}.
  • There is a discussion about the denominator and whether it represents the number of subsets with k elements out of n elements.
  • One participant clarifies that the first accident has n choices, the second has n-1 choices, and so on, emphasizing that the order of accidents does not imply temporal priority.
  • Another participant questions the omission of \frac{(n - k)}_{n} in the final result and expresses confusion about the relationship between k and the solution.
  • Further clarification is provided regarding how the number of choices decreases as accidents are assigned to specific days.
  • One participant concludes that k restricts the cardinality of the set of days available for accidents, noting that if k is greater than or equal to n, the solution simplifies to \frac{n!}{n^k}.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and interpretation of the probability problem, with some agreeing on the reasoning behind the formula while others remain uncertain about specific aspects, particularly regarding the denominator and the implications of k in the context of the problem. The discussion does not reach a consensus on all points raised.

Contextual Notes

Participants highlight limitations in their understanding of the problem, particularly regarding the assumptions about the distribution of accidents over the days and the mathematical steps involved in deriving the solution.

kioria
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This is a simple example in a textbook, 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 occurred in n days. Assume that accidents are equally likely on any day. Let A = event that one accident occurred each day. What is P(A)?

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


Can someone explain this solution or the process of obtaining this solution in plain english? Thanks.
 
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[tex]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}}[/tex]

[tex]={\text{number of days the first accident could have happened}\times...[/tex]
[tex]...\times\text{ number of days the k'th accident could have happened, which otherwise would have been accident-free}}[/tex]
[tex]\left/{\text{(I need to think a little more about the denominator here)}}\right.[/tex]
 
Last edited:
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:
[tex]P(A) = \frac{n(n-1)...(n-k+1)}_{n^k}[/tex], 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!
 
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?
 
kioria said:
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.
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).
 
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I believe the question says:

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

I am puzzled as to why [tex]\frac{(n - k)}_{n}[/tex] is omitted. I can't seem to picture the relationship between the final result and k.
 
EnumaElish said:
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).
I see what you mean there...

[EDIT] But shouldn't the choice be chosen from total number of accidents that have happened over the period of n days? :rolleyes:
 
Last edited:
You are right!

[tex]P(A) = \frac{n}{n}\times...\times\frac{n-k+1}n[/tex]

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.
 
kioria said:
I see what you mean there...

[EDIT] But shouldn't the choice be chosen from total number of accidents that have happened over the period of n days? :rolleyes:
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
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 [tex]\frac{n}_{n}[/tex] 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 ([tex]k >= n[/tex]) then the solution would be simple [tex]\frac{n!}_{n^k}[/tex].

Thanks for that.
 

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