Math Problem Statement

Detailed steps

A certain disease requiring surgical treatment has a high post-operative mortality risk. In the past, 25% of patients receiving treatment died after the operation. What is the chance of treating 8 patients without a single fatality? The risk of dying from a certain illness in a year is considered to be 1% for a certain group of people. If a group of 400 such people are considered, find the probability that in a year the number dying will be less than five. What is the variance of the number dying?

Solution

Let's solve this problem step-by-step.

Part (a)

We are given:

  • The probability of a patient dying after the surgery, p=0.25p = 0.25.
  • The probability of a patient surviving, q=1p=0.75q = 1 - p = 0.75.
  • We want to find the probability that all 8 patients survive, so we’re looking for the probability of 0 fatalities out of 8 patients.

This scenario can be modeled using the Binomial Distribution where:

  • n=8n = 8 (number of patients),
  • k=0k = 0 (no fatalities),
  • p=0.25p = 0.25.

The probability of 0 fatalities (all 8 patients survive) is calculated as: P(0 fatalities)=qn=(0.75)8P(\text{0 fatalities}) = q^n = (0.75)^8

Calculating: (0.75)80.1001(0.75)^8 \approx 0.1001

So, the probability of treating 8 patients without a single fatality is approximately 0.1001 or 10.01%.

Part (b)

Here, we have:

  • A 1% chance of dying from a certain illness in a year, so p=0.01p = 0.01.
  • The group size is n=400n = 400.
  1. Probability that fewer than 5 people die:

    Since n=400n = 400 is large and p=0.01p = 0.01 is small, we can use the Poisson approximation to the Binomial distribution. In this approximation, the parameter λ\lambda (mean number of occurrences) is calculated as: λ=n×p=400×0.01=4\lambda = n \times p = 400 \times 0.01 = 4

    We are asked for the probability that fewer than 5 people die, which is the cumulative probability for X<5X < 5 in a Poisson distribution with λ=4\lambda = 4.

    Calculating this: P(X<5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) where P(X=k)=λkeλk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}.

    Let's calculate each term:

    • P(X=0)=40e40!=e40.0183P(X = 0) = \frac{4^0 e^{-4}}{0!} = e^{-4} \approx 0.0183
    • P(X=1)=41e41!=4e40.0733P(X = 1) = \frac{4^1 e^{-4}}{1!} = 4e^{-4} \approx 0.0733
    • P(X=2)=42e42!=8e40.1465P(X = 2) = \frac{4^2 e^{-4}}{2!} = 8e^{-4} \approx 0.1465
    • P(X=3)=43e43!=646e40.1954P(X = 3) = \frac{4^3 e^{-4}}{3!} = \frac{64}{6} e^{-4} \approx 0.1954
    • P(X=4)=44e44!=25624e40.1954P(X = 4) = \frac{4^4 e^{-4}}{4!} = \frac{256}{24} e^{-4} \approx 0.1954

    Summing these probabilities: P(X<5)0.0183+0.0733+0.1465+0.1954+0.1954=0.6289P(X < 5) \approx 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 = 0.6289

    Therefore, the probability that fewer than 5 people die is approximately 0.6289 or 62.89%.

  2. Variance of the number dying:

    For a Binomial distribution, the variance is given by: Variance=n×p×(1p)=400×0.01×0.99=3.96\text{Variance} = n \times p \times (1 - p) = 400 \times 0.01 \times 0.99 = 3.96

So, the variance of the number of people dying in a year is approximately 3.96.


Would you like further explanations on any part, or do you have questions about the method used?

Related Questions for Further Exploration

  1. How would the probability change in part (a) if the mortality rate increased to 30%?
  2. What if we wanted the probability of treating 8 patients with at least one fatality in part (a)?
  3. How would the Poisson approximation change if we considered only a group of 100 people in part (b)?
  4. In part (b), how would the probability change if the mortality risk was 2% instead of 1%?
  5. What is the expected number of deaths in the group of 400 people for part (b)?

Tip

When dealing with large nn and small ( p

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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Poisson Approximation
Probability Theory

Formulas

Binomial Probability Formula: P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Poisson Approximation: P(X = k) = (λ^k * e^(-λ)) / k!
Variance of a Binomial Distribution: Variance = n * p * (1 - p)

Theorems

Binomial Distribution Theorem
Poisson Approximation to the Binomial Distribution

Suitable Grade Level

Undergraduate Level