Problem 1A

This question deals with the Poisson distribution, it looks like we are being asked to calculate the CDF for this distribution, but I am not entirely sure.

For reference here is the equation for the Poisson distribution:

where lambda is just how often some discrete happends within some specified time interal, and k is the times that it was observed:

\[{e^{- \lambda}\lambda^{k} \over k!}\]

So in the context of this problem, k refers to the number of purchases made in the specified time period and lambda refers to the average.

We can use the subtraction principle to figure out these probabilities. For the first probability we just subtract the probability of no purhcases from 1, to find this we just evaluate the poisson function at k=0 as shown below

\[ 1 - {e^{- \lambda}\lambda^{0} \over 0!} \]

Because 0 factorial is 1 and lambda to the 0 is 1 this expression simplifies to:

\[ P(X>= 1) = 1 - e^{-\lambda} \]

If we subtract from this imediate result the PDF evaluated at (1), we get P(>= 1)

\[ P(X>= 2) = 1 - e^{-\lambda} - {e^{-\lambda}\lambda^{1} \over 1!} \]

This simplifies to:

\[ P(X>= 2) = 1 - e^{-\lambda} - e^{- \lambda}\lambda \]

I got these problems right, but maybe you can see if they are worded incorrectly at some point or the explanation isn’t that good.




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1B

Looking at wikipedia you can see that one needs to simply divide a standard Poisson pmf by “1 - the pmf evaluated at 0”. The reason you do this is that you are just renormalizing because it is a new conditional probability we know from above that the denominator in this new pmf fill have to be:

\[ \over {1 - e^{-\lambda}} \]

And the numerator is just the poisson:

\[ {e^{- \lambda}\lambda^{k} \over k!} \]

Now it is just algebra:

\[ {e^{- \lambda}\lambda^{k} \over k!} \over {1 - e^{-\lambda}} \]

This simplifies to:

\[ P(X = k|X >=1) = {e^{- \lambda}\lambda^{k} \over k!(1-e^{\lambda})} \]

Interesting note: in the wikipedia page they do one step further of simplication, but looks like just more algebra: https://en.wikipedia.org/wiki/Zero-truncated_Poisson_distribution




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Problem 2a:

On the question of whether it is a binomial or a hypgeometric distribution, from the reading we see that the crucial differnce between these two distributions is that in the binomial distribution the individual bernouli trials are independent. In the problem statement, it tooks about how they independently read the mansuscript, this is how we know that we are dealing with the binomial distribution.

Right of the bat, my intuition tells me that you are going to have to multiply these two random variables together, becaue they are independent, however, the third answer would only reduce the probability and doesn’t make sense intuitively, the third answer would be the distribution of the probability of both of the graders catching the same typo for n typos (but I’m not sure about this).

So the only logical choice now is the **first answer* which makes sense because for at least one person to catch an air we can use the subtraction princple, and subtract from one the probability that both graders do not catch a typo: all other cases involve at least one grader catching the typo. I got this one right.




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2b

If p1 = p2, then both graders are equally good, that is, they have an equal probability of catching each error. We are told that X sub 1 and X sub 2 equals t, why would we need to add random variables like this, is this common? What purpose does this serve? I still have not figured out the correct answer.




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3 (Birthday problem)

First let’s revisit the traditional birthday problem, calculating the probability that out of n people in the room what is the probability that at least 2 people have the same birthday. The following equation leverages the idea of calculating the complement, that nobody has the same birthday in a room of n people. Also it is weird that this equation works, it seems to get slightly different numbers than just a hand calculation. Pulled equation from here: https://medium.com/@krause60/understanding-the-birthday-paradox-42d6d34c0fad

\[ P(n) = 1 - ({364 \over 365})^{n(n-1)\over2} \]

# Create the above function in R
birthday<-function(n){
 
1-(364/365)^((n*(n-1))/2)
}
# Test function for 23 people
birthday(23)
## [1] 0.5004772
birthday(3)
## [1] 0.00819668
# You can see birthday 3 does not equal what it does above
1- ((364/365)*(363/365))
## [1] 0.008204166




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4a

For this problem, X is the number of Heads in 10 fair coin tosses. We are asked to find the conditional PMF of X, given that the first two tosses both land Heads.

We are dealing with independent bernoulli trials, so the relevant distribution is the binomial distribution. Let’s first imagine that we were just dealing without a conditional probability: the number of heads in 10 fair coin tosses. The number of k heads would be equal to the following equation:

\[ {10 \choose k}.5^{k}.5^{10-k} \]

I am a little stuck in this problem, there is a hint that says "Consider a concrete example first to bild intuition, e.g., the probability of 7 heads given that the first 2 tosses were heads.




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4b

I’m also stuck on this one.

The hint says: "First think about whether the answer should be the same as the answer to the previous part. Then do the calculation, using Bayes’ rule or the definition of conditional probability.