Problem 51 - A Graphical Approach to College Algebra

Question

What will happen when an infectious disease is introduced into a family? Suppose a family has \(I\) infected members and \(S\) members who are not infected, but are susceptible to contracting the disease. The probability \(P\) of \(k\) people not contracting the disease during a 1-week period can be calculated by the formula. $$P=\left(\begin{array}{l}S \\\k\end{array}\right) q^{k}(1-q)^{S-k}$$ where \(q=(1-p)^{l}\) and \(p\) is the probability that a susceptible person contracts the disease from an infectious person. For example, if \(p=0.5,\) then there is a \(50 \%\) chance that a susceptible person exposed to one infectious person for 1 week will contract the disease. (Source: Hoppensteadt, F. and C. Peskin, Mathematics in Medicine and the Life Sciences, Springer-Verlag.) (a) Approximate the probability \(P\) of 3 family members not becoming infected within 1 week if there are currently 2 infected and 4 susceptible members. Assume that \(p=0.1\) (b) A highly infectious disease can have \(p=0.5 .\) Repeat part (a) with this value of \(p\) (c) Approximate the probability that everyone would become sick in a large family if initially \(I=1, S=9\) and \(p=0.5\)

Step-by-Step Solution

Answer

(a) P ≈ 0.4041, (b) P ≈ 0.046875, (c) P ≈ 0.00098.

Step 1: Understanding the Problem

We are given a family with infected and susceptible members. We are asked to calculate the probability of a specific number of family members not contracting an infectious disease over a given time period using the provided formula.

Step 2: Formulating Parameters for Part (a)

For part (a), we have 2 infected members ( I = 2) and 4 susceptible members ( S = 4). We need to calculate the probability for 3 ( k = 3) family members not contracting the disease. The given probability of becoming infected is p = 0.1, hence q = (1 - 0.1)^2.

Step 3: Calculating q for Part (a)

Calculate q using the given p value: q = (1 - p)^I = (1 - 0.1)^2 = 0.9^2 = 0.81.

Step 4: Applying the Formula for Part (a)

Substitute S, k and q into the formula to calculate P:\[P = \binom{4}{3} \cdot 0.81^3 \cdot (1 - 0.81)^{4-3} \]Calculate \( \binom{4}{3} \) which is 4, so:\[P = 4 \cdot 0.81^3 \cdot 0.19 = 4 \cdot 0.531441 \cdot 0.19 \approx 0.4041\]

Step 5: Formulating Parameters for Part (b)

For part (b), we adjust the probability p to 0.5 while keeping I = 2, S = 4, and k = 3, therefore compute q with the new p value.

Step 6: Calculating q for Part (b)

Calculate q with p = 0.5:\[q = (1 - 0.5)^2 = 0.5^2 = 0.25\]

Step 7: Applying the Formula for Part (b)

Substitute the values into the formula:\[P = \binom{4}{3} \cdot 0.25^3 \cdot (1 - 0.25)^{4-3} \]Calculate \( \binom{4}{3} \), which is 4, so:\[P = 4 \cdot 0.015625 \cdot 0.75 = 0.046875\]

Step 8: Formulating Parameters for Part (c)

For part (c), we have 1 infected member ( I = 1), 9 susceptible members ( S = 9), and p = 0.5. We need to find the probability that all 9 susceptible members become sick, so we set k = 0.

Step 9: Calculating q for Part (c)

Compute q with p = 0.5:\[q = (1 - 0.5)^1 = 0.5\]

Step 10: Applying the Formula for Part (c)

Since we want the probability of 0 people not contracting the disease (all are sick), we calculate:\[P = \binom{9}{0} \cdot 0.5^0 \cdot (1 - 0.5)^9\]\[P = 1 \cdot 1 \cdot 0.0009765625 = 0.0009765625\]

Step 11: Conclusion

We have calculated the probabilities for each case using the given conditions and formula: 0.4041 for part (a), 0.046875 for part (b), and 0.0009765625 for part (c).

Key Concepts

Infectious Disease Modeling

Infectious disease modeling is crucial for understanding how diseases spread within populations, such as a family unit in our exercise. These models help predict the likelihood of disease transmission under various scenarios by using mathematical formulas. In our example, the probability of infection is influenced by the interaction between infected and susceptible individuals within a defined time frame.

Two key factors in this model are the number of infected ( I ) and susceptible ( S ) individuals, and the transmission probability p . A higher transmission probability indicates a higher chance of the disease spreading. Meanwhile, I influences q , the probability that a susceptible person remains uninfected after interacting with an infected person. For example, with p = 0.1 , each interaction has a small chance of spreading the infection, which is part of the model's input to the formula we use. This modeling provides a systematic approach to predicting outcomes like how likely multiple family members stay healthy over a week.

Binomial Probability

Binomial probability is central to calculating the likelihood of different outcomes when disease exposure occurs. It deals with scenarios where there are two possible results - in our case, whether a family member contracts the disease or not. The formula includes the probability of a certain number k of events (non-infections in this case) happening.

In the formula provided, P= \( \binom{S}{k} q^k (1-q)^{S-k} \), each component has a specific role:

• \( \binom{S}{k} \) is a binomial coefficient representing the number of ways k non-infections can happen among S individuals.
• \( q^k \) accounts for each susceptible individual remaining uninfected.
• \((1-q)^{S-k}\) is the probability of those not part of the k remaining contract the disease.

Using this setup, our exercise calculates the likelihood of specific scenarios, like none of the susceptible family members becoming infected.

Mathematics Education

Mathematics education empowers students to tackle complex real-world problems like infectious disease modeling. Understanding probability, particularly within the context of real-life applications, can demystify abstract concepts and reveal their practical uses.

In educational settings, problems like these encourage students to:

• Link mathematical theories to public health issues;
• Use formulas and interpret outcomes;
• Think critically about how changes in variables, such as infection probability p , impact results.

This exercise is an example of applied mathematics, promoting skills beyond simple calculation to include analysis and interpretation, key skills that are useful across various scientific and statistical fields. Integrating such applications into curricula helps prepare students for challenges in both academic and professional environments.