Can't get a grasp of this probability

In summary, the probability of one accident occurring each day out of n days, where there are a total of k accidents, is given by P(A) = \frac{n(n-1)...(n-k+1)}_{n^k}. This can be understood by considering the number of days each accident can happen, starting with n choices for the first accident and decreasing by one for each subsequent accident. The solution also takes into account the restrictions imposed by k accidents occurring in n days.
  • #1
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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  • #2
[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]
 
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  • #3
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!
 
  • #4
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
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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  • #6
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.
 
  • #7
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:
 
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  • #8
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.
 
  • #9
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.
 

FAQ: Can't get a grasp of this probability

What is probability and why is it important in science?

Probability is the measure of how likely an event is to occur. It is important in science because it allows us to make predictions and understand the likelihood of different outcomes in experiments and real-world situations.

How is probability calculated?

Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This can be represented as a fraction, decimal, or percentage.

What is the difference between theoretical and experimental probability?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability is based on actual data from experiments or real-world events.

How can I improve my understanding of probability?

Practice is key to improving your understanding of probability. This can include solving problems, conducting experiments, and analyzing data. It can also be helpful to break down complex problems into smaller, more manageable parts.

What are some real-world applications of probability in science?

Probability is used in a wide range of scientific fields, including genetics, physics, and psychology. It can help predict the likelihood of genetic traits being passed down, the probability of certain particles being present in a given space, and the likelihood of certain behaviors occurring in individuals.

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