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Chapter 3: Problem 14

An article in Knee Surgery, Sports Traumatology, Arthroscopy ["Arthroscopic Meniscal Repair with an Absorbable Screw: Results and Surgical Technique" (2005, Vol. \(13,\) pp. \(273-279\) ) \(]\) cited a success rate of more than \(90 \%\) for meniscal tears with a rim width under \(3 \mathrm{~mm}\), but only a \(67 \%\) success rate for tears of \(3-6 \mathrm{~mm}\). If you are unlucky enough to suffer a meniscal tear of under \(3 \mathrm{~mm}\) on your left knee and one of width \(3-6 \mathrm{~mm}\) on your right knee, what is the probability mass function of the number of successful surgeries? Assume that the surgeries are independent.

Short Answer

Expert verified
The probability mass function is: \( P(X = 0) = 0.033, \; P(X = 1) = 0.364, \; P(X = 2) = 0.603. \)

Step by step solution

01

Define the Random Variable

We start by defining \( X \) as the random variable representing the number of successful surgeries. Since you have one surgery on each knee, \( X \) can take on values \( 0, 1, \) or \( 2 \), where \( 0 \) indicates both surgeries are unsuccessful, \( 1 \) indicates one successful surgery, and \( 2 \) indicates both surgeries are successful.
02

Determine Success Probabilities

Based on the article, the success probability for a meniscal tear under \( 3 \text{ mm} \) is over \( 90\% \). We assume it to be precisely \( 0.90 \) for simplicity, i.e., \( P(L) = 0.90 \). The success probability for a tear of \( 3-6 \text{ mm} \) is \( 0.67 \), i.e., \( P(R) = 0.67 \).
03

Calculate Probability of 0 Successful Surgeries

The probability of no surgery being successful (\( X = 0 \)) is the product of both surgeries failing. This is calculated as \((1 - P(L)) \times (1 - P(R)) = (1 - 0.90) \times (1 - 0.67) = 0.10 \times 0.33 = 0.033.\)
04

Calculate Probability of 1 Successful Surgery

There are two scenarios where exactly one surgery is successful: either the left knee surgery is successful, and the right is not, or vice versa. Therefore, \( P(X = 1) = P(L) \times (1 - P(R)) + (1 - P(L)) \times P(R) = 0.90 \times 0.33 + 0.10 \times 0.67 = 0.297 + 0.067 = 0.364.\)
05

Calculate Probability of 2 Successful Surgeries

The probability that both surgeries are successful (\( X = 2 \)) is the product of both being successful: \( P(L) \times P(R) = 0.90 \times 0.67 = 0.603. \)
06

Verify Probability Distribution

Ensure the probabilities sum to 1: \( P(X = 0) + P(X = 1) + P(X = 2) = 0.033 + 0.364 + 0.603 = 1.000. \) The sum is indeed 1, confirming that our probability distribution is correct.

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Key Concepts

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

Random Variable
A random variable is a core concept in probability and statistics that helps us model real-world phenomena in a structured way. In simple terms, a random variable is a numerical value determined by the outcome of a random process, like flipping a coin or rolling a die.

In our exercise, the random variable is denoted by \( X \). It represents the number of successful surgeries out of two distinct procedures: one on the left knee and one on the right knee.

### Important Characteristics of a Random Variable
  • It is well-defined, which means we can clearly state what outcomes it represents. Here, \( X \) can be \( 0, 1, \) or \( 2 \), where \( 0 \) means no surgeries are successful, and \( 2 \) means both are successful.

  • Discrete random variables can take a finite number of values. In this case, \( X \) is discrete because it can only be \( 0, 1, \) or \( 2 \).
  • The possible values represent all possible outcomes of the random process being modeled.
The random variable \( X \) gives us a useful way of calculating probabilities related to these outcomes.
Probability Distribution
Once we define our random variable, the next step is to establish its probability distribution. This distribution is a function that assigns a probability to each possible value the random variable can have.In the scenario of knee surgeries, we need to assess the likelihood of \( X \) taking each of its possible values \( \{0, 1, 2\} \). The probability distribution is derived from the probabilities of individual successful surgeries on each knee.

### Components of a Probability Distribution
  • Each value of the random variable has an associated probability. For \( X = 0 \), the probability is \( 0.033 \); for \( X = 1 \), it is \( 0.364 \); and for \( X = 2 \), it is \( 0.603 \).

  • The sum of all probabilities must equal 1, ensuring that all potential outcomes are accounted for. You can verify this for the surgery scenario, where all probabilities add up to 1, confirming the validity of our distribution.

  • The distribution provides a complete picture of all possible outcomes and their likelihoods, helping us understand the full scope of the random process.
The probability distribution thus serves as a vital tool in quantifying expectations and making informed decisions about outcomes.
Success Probability
Success probability is a crucial aspect when evaluating events like surgeries, where an outcome can vary based on certain conditions.In the knee surgery context, success probability refers to the chance that a specific surgery will yield a positive result.

The article gives us different success probabilities for different tear widths:
  • For a tear under \( 3 \text{ mm} \) on the left knee, the success probability is \( 0.90 \).

  • For a tear between \( 3-6 \text{ mm} \) on the right knee, the success probability is \( 0.67 \).
### Calculating Combined Successes
  • The success probabilities are crucial in calculating the compound outcomes, such as both surgeries being successful.

  • They guide us in determining the likelihood of achieving one or more successful surgeries, impacting the overall assessment of treatment outcomes.

  • By knowing these probabilities, we can accurately calculate the expected number of successful surgeries and make predictions about patient outcomes.
Thus, the concept of success probability directly influences the calculation of overall probabilities for potential outcomes of medical interventions like surgeries.

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