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.

Key Concepts

Probability Theory

Probability theory is a crucial aspect of understanding random events and their outcomes. At its core, it deals with the likelihood of different results occurring. In probability theory, events are considered as outcomes of random experiments. Each outcome is assigned a probability, a number between 0 and 1, representing how likely it is to occur.

The Poisson distribution is a notable concept within probability theory, particularly for modeling random events that happen independently over a fixed period. It's typically used when we want to measure the number of times an event occurs in an interval of time. Here, we're examining the number of colds a person catches in a year.

In this context, the Poisson distribution is defined by the parameter \( \lambda \), which is the average number of events (colds) expected in a time frame. The probability of observing \( k \) events is given by the formula:

• \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \)

This means the probability of getting exactly two colds (\( k = 2 \)) can be computed with \( \lambda \) values of 3 and 5, respectively, corresponding to whether the drug is effective or not.

Bayes' Theorem

Bayes' Theorem is a fundamental principle in probability theory, providing a way to update our beliefs based on new evidence. It links prior knowledge of an event with new evidence to give a revised probability.

In our exercise, we're trying to determine the probability that the drug is effective given that an individual had 2 colds. To do this, Bayes' Theorem is employed. It helps us relate the probability of an event based on two potential causes—in this case, whether the drug was effective (\( \lambda=3 \)) or not (\( \lambda=5 \))—with what is observed.

Bayes’ Theorem is expressed as:

• \( P(A|B) = \frac{P(B|A)P(A)}{P(B)} \)

Here, \( P(A|B) \) is the probability that the drug is effective (causing \( \lambda=3 \)) given 2 colds—\( B \) is the recorded evidence, and \( A \) is the condition (the drug being effective). Calculating \( P(\lambda=3 | \text{2 colds}) \) involves multiplying the probability of having 2 colds when the drug is effective by the probability of the drug being effective, and dividing by the total probability of having 2 colds.

Law of Total Probability

The Law of Total Probability extends the basic principles of probability theory. It aids in determining the total probability of an outcome that can occur from multiple causes or states.

In our scenario, the total probability we are interested in is \( P(\text{2 colds}) \), which represents the likelihood of having 2 colds regardless of whether the drug worked or not. The law tells us that this can be calculated by considering both "branches"—\( \lambda=3 \) and \( \lambda=5 \).

The formula is:

• \( P(2 \text{ colds}) = P(2 \text{ colds} | \lambda=3) \cdot P(\lambda=3) + P(2 \text{ colds} | \lambda=5) \cdot P(\lambda=5) \)

By using this law, we account for all possible scenarios of having 2 colds. This makes our calculations complete and allows us to work out the total probability needed for the Bayesian equation in the previous section. Thus, fundamental to solving this problem is combining these principles to effectively predict the situation's outcome.