Problem 60 - A First Course in Probability

Question

The number of times that a person contracts a cold in a given year is a Poisson random variable with parameter \(\lambda=5 .\) Suppose that a new wonder drug (based on large quantities of vitamin \(\mathrm{C}\) ) has just been marketed that reduces the Poisson parameter to \(\lambda=3\) for 75 percent of the population. For the other 25 percent of the population, the drug has no appreciable effect on colds. If an individual tries the drug for a year and has 2 colds in that time, how likely is it that the drug is beneficial for him or her?

Step-by-Step Solution

Answer

The probability that the drug is beneficial for an individual who has 2 colds after taking it for a year is: P(\(\lambda=3\) | 2 colds) = \(\frac{\frac{e^{-3}\cdot3^2}{2!} \cdot 0.75}{\frac{e^{-3}\cdot3^2}{2!} \cdot 0.75 + \frac{e^{-5}\cdot5^2}{2!} \cdot 0.25}\)

Step 1: Probability with \(\lambda = 3\)

Compute the probability of contracting 2 colds for those with the lower Poisson parameter (\(\lambda=3\)): P(2 colds | \(\lambda=3\)) = \(\frac{e^{-\lambda}\lambda^k}{k!}\), where \(k=2\) and \(\lambda=3\). P(2 colds | \(\lambda=3\)) = \(\frac{e^{-3}\cdot3^2}{2!}\)

Step 2: Probability with \(\lambda = 5\)

Compute the probability of contracting 2 colds for those with the higher Poisson parameter (\(\lambda=5\)): P(2 colds | \(\lambda=5\)) = \(\frac{e^{-\lambda}\lambda^k}{k!}\), where \(k=2\) and \(\lambda=5\). P(2 colds | \(\lambda=5\)) = \(\frac{e^{-5}\cdot5^2}{2!}\) #Step 2: Applying Bayes' theorem#

Step 3: Bayes' theorem

Compute the probability that the drug is beneficial, P(\(\lambda=3\) | 2 colds): P(\(\lambda=3\) | 2 colds) = \(\frac{P(2 \text{ colds} | \lambda=3) \cdot P(\lambda=3)}{P(2 \text{ colds})}\) To find P(2 colds), we can use the law of total probability to consider the combined probability of having 2 colds with both \(\lambda=3\) and \(\lambda=5\): P(2 colds) = P(2 colds | \(\lambda=3\))P(\(\lambda=3\)) + P(2 colds | \(\lambda=5\))P(\(\lambda=5\)) We know the probabilities for having \(\lambda=3\) and \(\lambda=5\), which are given as, P(\(\lambda=3\))=0.75 and P(\(\lambda=5\))=0.25. Substitute these values to obtain P(2 colds). #Step 3: Calculate the final probability#

Step 4: Final probability calculation

Substitute the values obtained in the numerator and denominator and calculate P(\(\lambda=3\) | 2 colds): P(\(\lambda=3\) | 2 colds) = \(\frac{P(2 \text{ colds} | \lambda=3) \cdot P(\lambda=3)}{P(2 \text{ colds} | \lambda=3) \cdot P(\lambda=3) + P(2 \text{ colds} | \lambda=5) \cdot P(\lambda=5)}\) The result will provide the probability that the drug is beneficial for an individual who has 2 colds after taking it for a year.