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Chapter 6: Problem 35

In kidney transplantations, compatibility between donor and receiver depends on such factors as blood type and antigens. Suppose that for a randomly selected donor from a large national kidney registry, there is a \(10 \%\) chance that he or she is compatible with a specific receiver. Three donors are randomly selected from this registry. Find the probability that \(0,1,2,\) or all 3 selected donors are compatible. a. Do this by constructing the sample space and finding the probability for each possible outcome of choosing three donors. Use these probabilities to construct the probability distribution. b. Do this using the formula for the binomial distribution.

Short Answer

Expert verified
The probabilities are 0.729 for 0 compatible, 0.243 for 1, 0.027 for 2, and 0.001 for all 3 compatible donors.

Step by step solution

01

Define the Scenario

We have three independent trials (choosing three donors), each with a probability of success (compatibility) of 0.10. We need to calculate the probability of having 0, 1, 2, and 3 compatible donors.
02

Construct the Sample Space

The sample space for three donors involves all combinations of each donor being compatible (C) or not compatible (N). The total outcomes are: N-N-N, N-N-C, N-C-N, C-N-N, N-C-C, C-N-C, C-C-N, C-C-C. This gives us 8 possible outcomes.
03

Calculate Probabilities for Each Outcome

Each donor's outcome is independent, so we calculate each outcome probability using the formula: \[ P(X) = p^k (1-p)^{n-k} \]where \( p = 0.10 \), \( 1-p = 0.90 \), \( n = 3 \), and \( k \) varies based on how many are compatible.Examples: - P(N-N-N) = \((0.90)^3 = 0.729 \)- P(N-N-C) = \((0.90)^2(0.10) = 0.081 \)
04

Construct Probability Distribution Based on Sample Space

Now, sort the outcomes based on how many compatible donors (0, 1, 2, or 3) there are: - 0 compatible: N-N-N (0.729) - 1 compatible: N-N-C, N-C-N, C-N-N (3 outcomes, each 0.081) - 2 compatible: N-C-C, C-N-C, C-C-N (3 outcomes, each 0.009) - 3 compatible: C-C-C (0.001) Add probabilities for each scenario: - P(0 compatible) = 0.729 - P(1 compatible) = 3 脳 0.081 = 0.243 - P(2 compatible) = 3 脳 0.009 = 0.027 - P(3 compatible) = 0.001
05

Use the Binomial Distribution Formula

The binomial probability formula is:\[ P(X = k) = C(n, k) p^k (1-p)^{n-k} \]where \( C(n, k) \) is the binomial coefficient: \( \frac{n!}{k! (n-k)!} \).Compute probabilities:- P(0) = \( C(3, 0) (0.10)^0 (0.90)^3 = 0.729 \)- P(1) = \( C(3, 1) (0.10)^1 (0.90)^2 = 3 \times 0.081 = 0.243 \)- P(2) = \( C(3, 2) (0.10)^2 (0.90)^1 = 3 \times 0.009 = 0.027 \)- P(3) = \( C(3, 3) (0.10)^3 (0.90)^0 = 0.001 \)

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Key Concepts

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

Probability Distribution
In probability theory, a probability distribution provides a mathematical framework to describe the likelihood of various outcomes. For example, in the kidney transplantation exercise, we have several outcomes depending on the number of compatible donors selected.
This setup gives us a way to list all possible outcomes and calculate the probability associated with each scenario.
The probability distribution allows us to understand the expected frequency of different compatibility results, helping us predict and plan better.
  • For 0 compatible donors: Probability is 0.729.
  • For 1 compatible donor: Probability sums to 0.243.
  • For 2 compatible donors: Probability sums to 0.027.
  • For all 3 compatible donors: Probability is just 0.001.
Understanding probability distribution aids in visualizing various outcomes and their probabilities, offering insight into the real-world scenario of donor-receiver compatibility.
Sample Space
The sample space in a probabilistic situation is the set of all possible outcomes. In our exercise, the sample space involves clearly defining all the ways donors can either be compatible (C) or not compatible (N) with a receiver.
For three donors, each donor can be either C or N. This results in a total of eight combinations, as each possibility for one donor can combine with each possibility for another, showing us every possible outcome in this specific scenario.
  • N-N-N
  • N-N-C
  • N-C-N
  • C-N-N
  • N-C-C
  • C-N-C
  • C-C-N
  • C-C-C
Each combination represents a distinct outcome in the sample space. Visualizing and constructing this list helps ensure that we've considered every potential outcome, key to accurate probability calculations.
Independent Trials
Independent trials in probability refer to experiments where the outcome of one does not influence the other. In the donor selection problem, each donor's compatibility does not affect the others.
Thus, selecting each donor can be regarded as an independent trial, with a separate chance of compatibility (0.10).
  • This independence ensures that the probability of each outcome can be calculated simply by multiplying the probabilities of the individual events (e.g., multiplying 0.90 for N and 0.10 for C).
  • It simplifies evaluating the overall scenario where several trials are involved.
Understanding independence is crucial when setting up problems with multiple events, clarifying how the chances unfold.
Binomial Coefficient
The binomial coefficient plays a significant role in determining the number of ways we can pick compatible donors when dealing with a fixed number of trials (say 3 donors). It's a part of the binomial distribution formula, representing the combinations aspect.
It helps calculate scenarios like selecting 1, 2, or all 3 compatible donors from the group, denoted as \(C(n, k) = \frac{n!}{k! (n-k)!}\).
This nCr function or notation identifies how many combinations or ways the compatible event can occur given the independent nature of each trial.
  • To calculate the probability of exactly 1 compatible donor out of 3, use the binomial coefficient \(C(3, 1)\).
  • Similarly, for 2 compatible donors, \(C(3, 2)\) is used.
  • Computing these coefficients helps predict probabilities using the binomial probability formula.
The binomial coefficient simplifies understanding the probability of different outcomes in trials with two possible results, success or failure.

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