Problem 35 - Calculus for Biology and Medicine

Question

Use the following facts: Cystic fibrosis is an inherited disorder that causes abnormally thick body secretions. About 1 in 2500 white babies in the United States has this disorder. About 3 in 100 children with cystic fibrosis develop diabetes mellitus, and about 1 in 5 females with cystic fibrosis is infertile. Find the probability that, in a group of 5000 newborn white babies in the United States, at least 4 babies suffer from cystic fibrosis.

Step-by-Step Solution

Answer

Probability of at least 4 newborns with cystic fibrosis in 5000 is approximately 0.985.

Step 1: Calculate the Probability of Cystic Fibrosis

The probability that a newborn white baby has cystic fibrosis is given as \( \frac{1}{2500} \). This means for every child born, the probability of having cystic fibrosis is 0.0004 (i.e., \( \frac{1}{2500} \)).

Step 2: Determine the Probability of Not Having Cystic Fibrosis

To find the probability of a baby not having cystic fibrosis, subtract the probability of having cystic fibrosis from 1:\\[\text{Probability of not having cystic fibrosis} = 1 - \frac{1}{2500} = 0.9996\].

Step 3: Use the Binomial Probability Formula

To find the probability that at least 4 babies suffer from cystic fibrosis out of 5000 newborns, we can use the binomial probability formula. The general formula for exactly \( k \) successes in \( n \) trials is: \\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\] \Here, \( n = 5000 \), \( p = \frac{1}{2500} \), and we want \( 4 \leq X \).

Step 4: Calculate Probability for Each Relevant Outcome

Since it is difficult to calculate the exact probabilities for so many numbers individually, and the cumulative probabilities are easier to handle, use the complementary probability to find \( P(X \geq 4) \) by calculating \( P(X < 4) \):ewline\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \].

Step 5: Calculate Each Term

Use the binomial probability formula from Step 3 to calculate: \\[P(X = 0) = \binom{5000}{0} (0.0004)^0 (0.9996)^{5000} \P(X = 1) = \binom{5000}{1} (0.0004)^1 (0.9996)^{4999} \P(X = 2) = \binom{5000}{2} (0.0004)^2 (0.9996)^{4998} \P(X = 3) = \binom{5000}{3} (0.0004)^3 (0.9996)^{4997} \\]

Step 6: Sum the Probabilities for Fewer than 4

Calculate \( P(X < 4) \) by summing the results of the calculations from Step 5. \\[ P(0) + P(1) + P(2) + P(3) = [calculated value] \] (Here, use a calculator to find the values of each expression.)

Step 7: Final Probability Calculation

Now, calculate \( P(X \geq 4) \) using:\\[ P(X \geq 4) = 1 - P(X < 4) \]. Substitute the result from Step 6 into this equation to find the probability that at least 4 babies have cystic fibrosis in the group of 5000.

Key Concepts

Understanding Cystic Fibrosis

Cystic fibrosis is a genetic disorder predominantly affecting individuals of European descent. It causes the production of abnormally thick and sticky mucus, leading to blockages and infections in the respiratory and digestive systems. This disorder stems from mutations in the CFTR gene, responsible for regulating the flow of salt and fluids in and out of cells. - Common symptoms include persistent cough, lung infections, and poor growth. - As an inherited disorder, cystic fibrosis follows an autosomal recessive pattern, meaning a child must receive one defective gene from each parent to be affected. The condition affects roughly 1 in 2,500 newborns in the white population of the United States. Unfortunately, individuals with cystic fibrosis face several complications, including an increased risk of diabetes and infertility, especially among females. Advancements in research and treatments continue to improve the quality and length of life for those affected.

Probability Calculations Made Simple

Probability calculations help us determine the likelihood of an event occurring. In situations like the one with cystic fibrosis, probabilities can be calculated using straightforward mathematical formulas.In the example of determining how many babies might be born with cystic fibrosis out of a group, the probability of a single baby having the disorder is given as \( \frac{1}{2500} \), or 0.0004. Conversely, the probability of a baby not having it is 0.9996, calculated by subtracting the chance of occurrence from 1.For scenarios involving multiple trials, such as 5000 newborns, the binomial probability formula is invaluable. It allows us to calculate the probability of having exactly \( k \) successes (e.g., babies with cystic fibrosis) in \( n \) trials by using the formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where \( n \) is the total number of trials, \( k \) is the number of successful outcomes desired, and \( p \) is the probability of a single success.

Inherited Disorder Statistics

Statistics on inherited disorders like cystic fibrosis provide a crucial understanding of their prevalence and impact. These statistics are pivotal in public health for planning resources, raising awareness, and ensuring that affected individuals receive proper care and support. - Approximately 1 in 25 people in the United States are carriers of the cystic fibrosis gene, even though they show no symptoms themselves. - With such a significant carrier rate, genetic counseling becomes essential, especially for couples who may be carriers deciding to start a family. The statistical prevalence of cystic fibrosis and its related conditions, like diabetes in 3 out of 100 children with the disorder, guides medical professionals in early diagnosis and management. Additionally, these numbers underscore the importance of research and funding for developing new treatments that can enhance the lives of those affected.