How Does Poisson Distribution Affect the Efficacy of a Cold-Reducing Drug?

[stgermaine]

1. The number of times that a person contracts a cold in a year is a Poisson random variable with parameter lambda=5. Suppose a wonder drug reduces the Poisson parameter to lambda=3 for 75% of the population but does not affect the rest of the population. If an individual tries the drug for a year and contracts two colds in that time, how likely is it that the drug is beneficial for him or her



2. Probability mass fn is given by ((lambda^k) * e^(-lambda))/k_factorial

The Attempt at a Solution

Here's where I'm confused. The probabiliy mass fn gives the probability that a person contracts k number of colds in a given year, correct? In this problem, I solved the probability mass function with the constraints (lambda=3, k=2) and (lambda=5, k=2). When lambda=3 it means the medicine worked and the person should have a smaller chance of contracting two colds (k=2) in a given year than when lambda=5, when the drug has not worked.

However, when I calculate those two values, the probability is 0.224 when lambda=3 and 0.08422 when lambda=5.

Sometimes, the textbook has solutions to problems involving Poisson distribution where the answers are in the form 1 - e^(-n) where n is a positive integer.

Am I supposed to take the complement of this? Why is the probability of contracting two colds in a year higher when the medicine works?

I'm just having so much trouble with random variables, I'd really appreciate it if someone can provide any links to video lectures or helpful guides out there.

Thank you!

 

[IniquiTrance]

I think you can solve this using Bayes' formula.

 

[stgermaine]

I did set up a way to solve it using Bayes, but it's the calculation of the actual probabilities of catching a cold twice that gets me.

P(D) = 0.75 the chance that drug works, P(D') = 0.25 the drug doesn't work
P(C) would be the chance that one catches a cold twice in a year.
P(C|D) = calculated using Poisson dist. and lambda=3
P(C|D') = calculated using Poisson dist and lambda=5


I'm supposed to solve for P(D|C) = P(D\bigcapC / P(C)

I can't calculate P(C|D) or P(C|D') without using Poisson and I'm confused by the probability mass fn of Poisson random variable.

 

[Ray Vickson]

I did set up a way to solve it using Bayes, but it's the calculation of the actual probabilities of catching a cold twice that gets me.

P(D) = 0.75 the chance that drug works, P(D') = 0.25 the drug doesn't work
P(C) would be the chance that one catches a cold twice in a year.
P(C|D) = calculated using Poisson dist. and lambda=3
P(C|D') = calculated using Poisson dist and lambda=5


I'm supposed to solve for P(D|C) = P(D\bigcapC / P(C)

I can't calculate P(C|D) or P(C|D') without using Poisson and I'm confused by the probability mass fn of Poisson random variable.

You already computed the relevant Poisson probabilities 0.224 and 0.08422 (which I did not check). These, of course, are probabilities of an event, conditional on certain Poisson parameters, and you are told the prior probabilities of those parameters (assuming you administer the drug). It is all just straightforward Bayes.

RGV

 

[stgermaine]

I'm a bit confused as the numbers don't make sense. When the drug works, there's a higher probability of contracting the cold twice compared to when the drug doesn't work.

 

[Ray Vickson]

I'm a bit confused as the numbers don't make sense. When the drug works, there's a higher probability of contracting the cold twice compared to when the drug doesn't work.

The numbers make perfect sense: without treatment, you are less likely to have a small number of colds (that is, you are more likely to have a large number of colds). The number 2 is below both means 3 and 5, but it is farther from the mean 5 than from the mean 3, so its probability is smaller. Try plotting the probability mass functions p(k) for k = 0, 1, 2, ... for both λ = 3 and λ = 5 and you will see that for a given level of probability, the λ = 5 results are shifted to the right of the λ = 3 results; that is, for a given level of probability (on the y-axis) you get more colds (on the x-axis) without treatment than with treatment.

RGV