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Chapter 4: Problem 85

The probability that an intermediate cell will mutate to become a malignant cell is \(5 \times 10^{-7}\) per year. Suppose a woman has 300 intermediate cells by age 45. What is the probability that she develops breast cancer by age 46 ? By age 50 ? (Hint: Use the Poisson approximation to the binomial distribution.)

Short Answer

Expert verified
The probability of developing cancer by age 46 is approximately \(1.5 \times 10^{-4}\), and by age 50 is approximately \(0.00075\).

Step by step solution

01

Understanding the Problem

We need to calculate the probability that at least one of the intermediate cells mutates within a specific period of time, namely the one-year spans from age 45 to 46 and from age 45 to 50, given a mutation probability per cell per year.
02

Using Poisson Approximation

The mutation events are rare with a small probability but affect many cells, a situation suitable for Poisson approximation. The Poisson distribution parameter, \( \lambda \), is calculated as \( \lambda = n \times p \) where \( n \) is the number of trials (or cells) and \( p \) is the probability of mutation per trial.
03

Calculating \( \lambda \) for Ages 45 to 46

For one year (from 45 to 46), \( n = 300 \) cells and \( p = 5 \times 10^{-7} \). Thus, \( \lambda = 300 \times 5 \times 10^{-7} = 1.5 \times 10^{-4} \).
04

Probability Calculation from Age 45 to 46

The probability of no mutation (0 mutations) is \( P(X = 0) = \frac{e^{-\lambda} \cdot \lambda^0}{0!} = e^{-1.5 \times 10^{-4}} \). Since we want at least one mutation, \( P(X \geq 1) = 1 - P(X = 0) \).
05

Calculating and Interpreting Result for Ages 45 to 46

Using \( e^{-1.5 \times 10^{-4}} \approx 1 - 1.5 \times 10^{-4} \) for small values, the probability is approximately \( 1.5 \times 10^{-4} \). This represents the chance of at least one mutation happening within a year.
06

Calculating \( \lambda \) for Ages 45 to 50

For five years (from age 45 to 50), \( n = 300 \times 5 = 1500 \), so \( \lambda = 1500 \times 5 \times 10^{-7} = 0.00075 \).
07

Probability Calculation from Age 45 to 50

The probability of no mutations in five years is \( P(X = 0) = e^{-0.00075} \). The probability of at least one mutation happening, \( P(X \geq 1) = 1 - e^{-0.00075} \).
08

Calculating and Interpreting Result for Ages 45 to 50

Using \( e^{-0.00075} \approx 1 - 0.00075 \) for small values, the probability is approximately \( 0.00075 \). This gives the chance of at least one mutation happening in five years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Binomial Distribution
The binomial distribution is a way to calculate the probability of a specific number of successful outcomes, like mutations, across a fixed number of trials鈥攈ere, the trials are the 300 intermediate cells. Each cell has the same chance of mutating, or success. The binomial distribution formula requires two key inputs:
  • The number of trials, which is 300 in this example.
  • The probability of success (mutation) per trial, given as \(5 \times 10^{-7}\) per year.

The aim is to find the probability that none to multiple mutations occur during specific time periods. While the binomial distribution is perfect for small numbers and fairly common events, it becomes complex and resource-intensive for many trials with low probability events, like mutations in many cells. That's where the Poisson approximation comes in handy.
Calculating Probabilities
When dealing with probabilities, especially for rare events, it's essential to determine the likelihood of different outcomes, like mutations happening or not.
  • For the given scenario, you look at two periods: from age 45 to 46 and from age 45 to 50.
  • The key is to find the probability of having at least one mutation within these periods.

First, calculate the probability of zero mutations occurring. This helps us understand the complement of the event of interest. For each period, employ the Poisson distribution, which simplifies calculations when dealing with a high number of trials and low probabilities.
Understanding Mutation Probability
Mutation probability refers to the chance of a genetic change occurring in a cell. Here, it's pointed out as \(5 \times 10^{-7}\) per cell per year. Given a scenario where a woman has 300 cells, each with this tiny probability of changing per year, the collective risk needs considering.
  • Such rare events, while unlikely individually, need to be examined over time and quantity to get a broader understanding of how likely they are to happen at least once.
  • Even though each mutation event is rare, with many cells at play, the overall probability can become notable.

This context helps in calculating the chance of cancers or other inherited conditions that may arise from such mutations over the stated periods.
Poisson Distribution for Approximation
The Poisson distribution often approximates the binomial distribution when you're dealing with numerous trials and very low probability, like in this example of cell mutations. The parameters are easier to compute:
  • Calculate \(\lambda\), the average rate (or expected number) of mutations, by multiplying the number of cells by the mutation probability.
  • For one year, \(\lambda\) is set as \(300 \times 5 \times 10^{-7} = 1.5 \times 10^{-4}\). For five years, adjust appropriately by using \(1500\) as the total number of cell-year trials.

Once you have \(\lambda\), the Poisson distribution formula \( P(X = 0) = e^{-\lambda} \) calculates the probability of no mutations. It's a pivotal step because knowing this lets you easily determine the likelihood of at least one mutation over chosen timeframes.

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Most popular questions from this chapter

A study was conducted among 234 people who had expressed a desire to stop smoking but who had not yet stopped. On the day they quit smoking, their carbonmonoxide level (CO) was measured and the time was noted from the time they smoked their last cigarette to the time of the CO measurement. The CO level provides an "objective" indicator of the number of cigarettes smoked per day during the time immediately before the quit attempt. However, it is known to also be influenced by the time since the last cigarette was smoked. Thus, this time is provided as well as a "corrected CO level," which is adjusted for the time since the last cigarette was smoked. Information is also provided on the age and sex of the participants as well as each participant's self-report of the number of cigarettes smoked per day. The participants were followed for 1 year for the purpose of determining the number of days they remained abstinent. Number of days abstinent ranged from 0 for those who quit for less than 1 day to 365 for those who were abstinent for the full year. Assume all people were followed for the entire year. (TABLE CAN NOT COPY) Develop a life table similar to Table 4.16, giving the number of people who remained abstinent at \(1,2, \ldots\) 12 months of life (assume for simplicity that there are 30 days in each of the first 11 months after quitting and 35 days in the 12 th month). Plot these data on the computer using either Excel or \(R\) or some other statistical package. Compute the probability that a person will remain abstinent at \(1,3,6,\) and 12 months after quitting.

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