Problem 97
Question
Use the Poisson approximation. About 1 in 700 births in the United States is affected by Down syndrome, a chromosomal disorder. Find the probability that there is at most one case of Down syndrome among 1000 births by (a) computing the exact probability and (b) using a Poisson approximation.
Step-by-Step Solution
Verified Answer
The probability of at most one Down syndrome case in 1000 births is about 80.2% using Poisson approximation, similar to the exact probability.
1Step 1: Identify the problem type
We have a problem involving a rare event (Down syndrome) and a large sample size (1000 births). This scenario is suitable for a Poisson approximation.
2Step 2: Define parameters for Poisson approximation
The probability of a birth being affected by Down syndrome is approximately 1/700. For 1000 births, the expected number of cases, \( \lambda \), is calculated as:\[\lambda = \frac{1000}{700} \approx 1.4286\].
3Step 3: Poisson probability calculations
The Poisson probability function is defined as:\[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]We compute the probabilities for at most one case, \( P(X \leq 1) \):\[P(X = 0) = \frac{e^{-1.4286} (1.4286)^0}{0!} \]\[P(X = 1) = \frac{e^{-1.4286} (1.4286)^1}{1!}\]Add these probabilities for \( X \leq 1 \):\[P(X \leq 1) = P(X = 0) + P(X = 1)\].
4Step 4: Calculate the exact probability using Binomial distribution
For the binomial distribution, use \( n = 1000 \) and \( p = \frac{1}{700} \). Calculate the probability for \( X = 0 \) and \( X = 1 \) and sum them:\[P(X = 0) = \binom{1000}{0} (\frac{1}{700})^0 (\frac{699}{700})^{1000}\]\[P(X = 1) = \binom{1000}{1} (\frac{1}{700})^1 (\frac{699}{700})^{999}\]\[P(X \leq 1) = P(X = 0) + P(X = 1)\].
5Step 5: Final calculations and comparisons
Calculate the numerical values from the Poisson and Binomial distributions. Compare the approximation to the exact probability to see how the Poisson approximation fares. Use a calculator or software for accurate computation of exponentials and factorials.
Key Concepts
Binomial DistributionProbability CalculationsPoisson Distribution
Binomial Distribution
The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials of a binary experiment. A binary experiment is one that has only two possible outcomes: success or failure. To model a situation using a binomial distribution, you typically need two parameters:
When using a binomial distribution to determine the probability of at most one event occurring — such as having at most one case of Down syndrome among 1000 births — you compute the probability for each possible outcome (zero and one in this case) and sum them to find the cumulative probability:
- n: The number of trials or experiments.
- p: The probability of success on each trial.
When using a binomial distribution to determine the probability of at most one event occurring — such as having at most one case of Down syndrome among 1000 births — you compute the probability for each possible outcome (zero and one in this case) and sum them to find the cumulative probability:
- \( P(X = 0) \): Probability of zero cases
- \( P(X = 1) \): Probability of exactly one case
Probability Calculations
Probability calculations involve working with numbers that express how likely an event is to occur on a scale from 0 to 1. A probability of 0 means the event is impossible, while a probability of 1 means it’s certain. For real-world applications, probabilities often indicate the expected chance of an event happening based on past data or theoretical assumptions.
Standard probability calculations for different distributions rely on using the appropriate formulas and interpreting the results accurately:
- Poisson Probability — Used for rare events over a fixed interval, like time or space.
- Binomial Probability — Used for a fixed number of binary trials.
- Identify the type of distribution that best fits the scenario.
- Use the mathematical formula associated with the distribution.
- Perform calculations, often requiring factorial or exponential operations.
Poisson Distribution
The Poisson distribution is a discrete probability distribution used to model the number of times an event occurs within a specific interval of time or space. It is particularly useful for modeling rare events, which happen relatively infrequently compared to the size of the potential sample space.To use a Poisson distribution, you need:
In the context of the original exercise, the calculation of \( \lambda = \frac{1000}{700} \approx 1.4286 \) takes into account the expected number of Down syndrome cases among 1000 births.The Poisson probability for an event's exact number, \( k \), occurring is given by:\[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]
For the scenario of at most one event, you calculate:
- \( \lambda \): The average number of occurrences in the interval.
In the context of the original exercise, the calculation of \( \lambda = \frac{1000}{700} \approx 1.4286 \) takes into account the expected number of Down syndrome cases among 1000 births.The Poisson probability for an event's exact number, \( k \), occurring is given by:\[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]
For the scenario of at most one event, you calculate:
- \( P(X = 0) \): Probability of zero cases
- \( P(X = 1) \): Probability of exactly one case
Other exercises in this chapter
Problem 95
Use the Poisson approximation. For a certain vaccine, 1 in 1000 individuals experiences some side effects. Find the probability that, in a group of 500 people,
View solution Problem 96
Use the Poisson approximation. For a certain vaccine, 1 in 500 individuals experiences some side effects. Find the probability that, in a group of 200 people, a
View solution Problem 98
Use the Poisson approximation. About 1 in 1000 boys is affected by fragile \(X\) syndrome, a genetic disorder that causes learning difficulties. Find the probab
View solution Problem 94
Suppose \(X\) and \(Y\) are independent and Poisson with mean \(\lambda\). Given that \(X+Y=n\), find the probability that \(X=k\) for \(k=0,1,2, \ldots, n\)
View solution