/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 The number of times that a perso... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 60

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?

Short Answer

Expert verified
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 by step solution

01

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 | \(\lambda=3\)) = \(\frac{e^{-3}\cdot3^2}{2!}\)
02

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#
03

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#
04

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.

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