Am I justified in using the binomial distribution?

[kosovo dave]

Homework Statement

12 non-distinguishable attacks from President Snow land in Panem’s 12 districts in a particular week. Assume the attacks are located randomly, with each configuration of attacks equally likely. What is the probability that some district had more than 1 attack?

Homework Equations

$$P = \binom{n}{k}p^{k}*(1-p)^{n-k}$$ or, possibly, $$P=\frac{λ^{k}}{k!}*e^{-λ}$$

The Attempt at a Solution

$$1-\sum_{k=0}^{1}\binom{12}{k}(\frac{1}{12})^{k}(\frac{11}{12})^{12-k}≈0.65$$
Should I have used a poisson distribution instead?

 

[member 587159]

Binomial distribution seems justified. The stochastic variable X = "an attack happens" is binomially distributed, since the events are independent.

 

[kosovo dave]

Thank you for the quick response. Would a poisson distribution be an appropriate approximation if the number of attacks was much larger? Say, 120000 instead of 12?

 

[member 587159]

Thank you for the quick response. Would a poisson distribution be an appropriate approximation if the number of attacks was much larger? Say, 120000 instead of 12?

No idea. I'm not yet very familiar with poison distrubution. I am learning all this stuff myself this semester so take my answer with a grain of salt :)

 

[Orodruin]

Thank you for the quick response. Would a poisson distribution be an appropriate approximation if the number of attacks was much larger? Say, 120000 instead of 12?

For a fixed number of expected hits while the number of attacks goes to infinity, the limit of the binomial distribution is a Poisson distribution with the parameter equal to the number of expected hits. This is the Poisson limit theorem.

 

[Ray Vickson]

Thank you for the quick response. Would a poisson distribution be an appropriate approximation if the number of attacks was much larger? Say, 120000 instead of 12?

No: the normal approximation to the binomial would be appropriate.

As Orodruin has indicated in #5, the Poisson is the "large $N$, small $p$" limit of the binomial when the mean $\mu = Np$ is held constant. The normal approximation is the a "large $N$, fixed $p$" limit of the binomial when the mean $\mu = Np$ goes to $\infty$, and in that case we really need to be looking at an average such as $X/N$ whose mean remains constant.

 

[tnich]

12 non-distinguishable attacks from President Snow land in Panem’s 12 districts in a particular week.

Since the attacks are indistinguishable, swapping two attacks in any configuration results in the same configuration. So a binomial distribution is not suitable. You need to figure out how many ways there are to allocate 12 items to 12 distinct subsets.

What is the probability that some district had more than 1 attack?

Another hint: What is the probability that no district had more than one attack?

 

[kosovo dave]

Since the attacks are indistinguishable, swapping two attacks in any configuration results in the same configuration. So a binomial distribution is not suitable. You need to figure out how many ways there are to allocate 12 items to 12 distinct subsets.

Another hint: What is the probability that no district had more than one attack?

I'm not sure I follow. I think the only other distribution we've learned so far is the geometric distribution but that doesn't seem appropriate here?

 

[tnich]

I'm not sure I follow. I think the only other distribution we've learned so far is the geometric distribution but that doesn't seem appropriate here?

The answer will be related to the binomial distribution, but not in the way you have used it. Have you done stars and bars?

 

[kosovo dave]

The answer will be related to the binomial distribution, but not in the way you have used it. Have you done stars and bars?

Do I want to use the positive case or the non-negative case?

 

[tnich]

Do I want to use the positive case or the non-negative case?

Try it and see what makes sense.

 

[kosovo dave]

Try it and see what makes sense.

Since my sum is running from k=0, I think I should use the case $$\binom{n+k-1}{n}$$.

If I used $$\binom{n-1}{k-1}$$ my sum would be running from k=-1.

 

[tnich]

Since my sum is running from k=0, I think I should use the case $$\binom{n+k-1}{n}$$.

If I used $$\binom{n-1}{k-1}$$ my sum would be running from k=-1.

I think you are right.

 

[Ray Vickson]

I'm not sure I follow. I think the only other distribution we've learned so far is the geometric distribution but that doesn't seem appropriate here?

The answer depends on exactly you interpret the question---that is, the meaning of "some district had more than 1 attack". The binomial distribution would be OK if you interpret that as asking about any specific, but un-named district, because each attack either lands in that district or not. I suspect that most people would view that interpretation as somewhat artificial.

The alternative interpretation would be that "at least one district suffers more than one attack", and in that case looking at it as the complement of "every district has exactly one attack" would be the easiest way, as suggested in Post #7.

This would have an easy solution if you use the so-called "Multinomial" distribution, but you say you have not seen that yet. However, you can also do it using the ordinary Binomial distribution and conditional probabilities: if we let $X_i =$ number of hits in district $i$ for $i = 1,2, \ldots, 12$, then we want to know $P(E)$, where $E$ is the event
$$E = \{ (X_1,X_2, \ldots, X_{12}) = (1,1, \ldots, 1) \}$$
Note that $P(E) = P(X_1 = 1) P(E|X_1 = 1)$.

We get $P(X_1 = 1)$ from the Binomial distribution. The conditional probability $P(E| X_1 = 1)$ is just the probability that all 11 remaining districts have one hit each in 11 attacks, so is a smaller version of the original problem, and can be similarly tackled by computing $P(X_2 = 1 | X_1 = 1)$ and $P(E | X_1 = 1 \: \& \: X_2 = 1)$, etc.

 

[Eclair_de_XII]

What is the probability that some district had more than 1 attack?

Sorry if I'm needlessly bumping an old topic, but would it not be easier to calculate the probability of one district having no attacks?

Also, is that problem based on that Hunger Games novel?

 

[kosovo dave]

Sorry if I'm needlessly bumping an old topic, but would it not be easier to calculate the probability of one district having no attacks?

I believe so. My teacher is a nerd and it shows in her problem sets.

The solution ended up being 1-1/(23 choose 11)