Assume that the number of times a person can contract a particular viral disease in one year can be modelled by Poisson random variable with parameter a = 5. A pharmaceutical company has produced a new preventative nasal spray which can reduce this parameter to a = 2 for 60% of the population, but is not effective for the other 40% of people. If someone tries the spray for one year and contracts the disease 2 times, what is (the approximate) probability that drug is effective for that person?Answer

To solve this problem, we need to use Bayes' theorem, which is a way of finding a probability when we know certain other probabilities. The formula is:

P(A|B) = P(B|A) * P(A) / P(B)

Where:

• P(A|B

Let's denote:

E: The event that the drug is effective. N: The event that the person contracts the disease 2 times.

We want to find P(E|N), the probability that the drug is effective given that the person contracted the disease 2 times.

By Bayes' theorem, we have:

P(E|N) = P(N|E) * P(E) / P(N)

We know that:

P(E) = 0.6 (the probability that the drug is effective for a person) P(N|E) = e^(-2) * 2^2 / 2! (the probability of contracting the disease 2 times if the drug is effective, which follows a Poisson distribution with parameter a = 2)

We also need to calculate P(N), the total probability of contracting the disease 2 times. This is given by:

P(N) = P(N|E) * P(E) + P(N|~E) * P(~E)

where ~E is the event that the drug is not effective, P(~E) = 0.4, and P(N|~E) = e^(-5) * 5^2 / 2! (the probability of contracting the disease 2 times if the drug is not effective, which follows a Poisson distribution with parameter a = 5).

Substituting these values in, we get:

P(E|N) = e^(-2) * 2^2 / 2! * 0.6 / (e^(-2) * 2^2 / 2! * 0.6 + e^(-5) * 5^2 / 2! * 0.4)

After calculating the above expression, we get the approximate probability that the drug is effective for the person.

To solve this problem, we need to use Bayes' theorem, which is a way of finding a probability when we know certain other probabilities. The formula is:

P(A|B) = P(B|A) * P(A) / P(B)

Where:

• P(A|B) is the probability of event A given event B is true.
• P(B|A) is the probability of event B given event A is true.
• P(A) and P(B) are the probabilities of events A and B respectively.

In this case, we want to find the probability that the drug is effective (event A) given that the person contracted the disease 2 times (event B).

We know that:

• P(A) = 0.6 (the drug is effective for 60% of the population)
• P(A') = 0.4 (the drug is not effective for 40% of the population)
• P(B|A) is the probability of contracting the disease 2 times given the drug is effective, which follows a Poisson distribution with parameter a = 2. So, P(B|A) = e^-2 * 2^2 / 2! = 0.271
• P(B|A') is the probability of contracting the disease 2 times given the drug is not effective, which follows a Poisson distribution with parameter a = 5. So, P(B|A') = e^-5 * 5^2 / 2! = 0.084

We also need to find P(B), the probability of contracting the disease 2 times. This is equal to the sum of the probabilities of contracting the disease 2 times given the drug is effective and not effective, weighted by the probabilities of the drug being effective and not effective. So, P(B) = P(B|A) * P(A) + P(B|A') * P(A') = 0.271 * 0.6 + 0.084 * 0.4 = 0.194

Finally, we can substitute these values into Bayes' theorem to find P(A|B):

P(A|B) = P(B|A) * P(A) / P(B) = 0.271 * 0.6 / 0.194 = 0.837

So, the approximate probability that the drug is effective for that person is 0.837 or 83.7%.