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Chapter 2: Problem 70

Suppose that the proportions of blood phenotypes in a particular population are as follows: \(\begin{array}{cccc}A & B & A B & 0 \\ .42 & .10 & .04 & .44\end{array}\) Assuming that the phenotypes of two randomly selected individuals are independent of each other, what is the probability that both phenotypes are \(\mathrm{O}\) ? What is the probability that the phenotypes of two randomly selected individuals match?

Short Answer

Expert verified
Both O: 0.1936, Match: 0.3816.

Step by step solution

01

Understand the Problem

We are given the proportions of blood phenotypes in a population and are asked to find the probability of two specific events: 1) Both individuals having phenotype O and 2) Both individuals' phenotypes matching, regardless of which phenotype.
02

Calculate Probability of Both O Phenotype

The probability that one individual has phenotype O is 0.44. For two independent events (the selection of the two individuals) where both outcomes are phenotype O, we multiply their probabilities: \( P( ext{both O}) = 0.44 \times 0.44 = 0.1936 \).
03

Calculate Probability of Matching Phenotypes

For the phenotypes of the two randomly selected individuals to match, both need to have either A, B, AB, or O. This is calculated as follows: \( P( ext{both A}) = 0.42 \times 0.42 = 0.1764 \), \( P( ext{both B}) = 0.10 \times 0.10 = 0.01 \), \( P( ext{both AB}) = 0.04 \times 0.04 = 0.0016 \), and \( P( ext{both O}) = 0.44 \times 0.44 = 0.1936 \). Adding these together gives the total probability of matching phenotypes: \( P( ext{match}) = 0.1764 + 0.01 + 0.0016 + 0.1936 = 0.3816 \).

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Key Concepts

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

Blood Phenotypes
Blood phenotypes classify individuals based on the presence of specific antigens on the surface of their red blood cells. There are four main blood types: A, B, AB, and O. These phenotypes vary in frequency across different populations.

Understanding these phenotypes is crucial for practices like blood transfusions. A wrong match can lead to life-threatening reactions, making it essential to know the phenotype of both donor and recipient.

In the problem scenario, proportions are given for each phenotype in a population:
  • Type A: 42%
  • Type B: 10%
  • Type AB: 4%
  • Type O: 44%
Recognizing these proportions can help in calculating the probability of various events related to blood type.
Independent Events
Independent events imply that the outcome of one event does not affect the outcome of another. In probability theory, two events can be considered independent if the probability of one event occurring does not change the probability of the other occurring.

For example, when two individuals are selected randomly, each selection is independent of the other. Therefore, the blood phenotype of the first person doesn't influence the phenotype of the second person. This property is crucial when calculating probabilities in such scenarios.
  • To find the joint probability of two independent events, you multiply their individual probabilities.
  • In our problem, the chance for two independent selections both being Type O is calculated as \( P(\text{O}) \times P(\text{O}) \).
This concept is fundamental in understanding and calculating probabilities accurately in independent specimen selection.
Matching Probability
Matching probability refers to the likelihood that two selected items or individuals have the same characteristic, in this case, blood phenotype.

To find the matching probability in the given problem, you need to consider all possible phenotype pairs: A, B, AB, and O. For each phenotype, you calculate the probability that both individuals have this phenotype. This involves squaring the individual probability of each phenotype, as each selection is independent.
  • For Type A: \( P(\text{A}) \times P(\text{A}) = 0.42 \times 0.42 = 0.1764 \)
  • For Type B: \( P(\text{B}) \times P(\text{B}) = 0.10 \times 0.10 = 0.01 \)
  • For Type AB: \( P(\text{AB}) \times P(\text{AB}) = 0.04 \times 0.04 = 0.0016 \)
  • For Type O: \( P(\text{O}) \times P(\text{O}) = 0.44 \times 0.44 = 0.1936 \)
Add these probabilities to get the total probability of any matching phenotypes, which in our case is 0.3816. Recognizing how matching probabilities work helps in broad scenarios beyond biology, like pattern recognition and statistical matching.
Population Proportions
In statistical terms, population proportions represent the fraction of the population that exhibits a particular trait or characteristic. These proportions are invaluable in predicting outcomes in similar groups or forecasting results based on sampled data.

In this exercise, population proportions for blood phenotypes help us gauge the probability of certain blood types in a randomly selected sample from that population. Knowing proportions enables accurate probability estimations for different combinations and matches.
  • Blood Type A: 42%
  • Blood Type B: 10%
  • Blood Type AB: 4%
  • Blood Type O: 44%
These percentages are used in probability calculations to determine the likelihood of events like blood type matches. Mastery of population proportions aids in choosing representative samples for research and understanding overall trends across larger groups.

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Most popular questions from this chapter

Components of a certain type are shipped to a supplier in batches of ten. Suppose that \(50 \%\) of all such batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0,1 , and 2 defective components being in the batch under each of the following conditions? a. Neither tested component is defective. b. One of the two tested components is defective. [Hint: Draw a tree diagram with three firstgeneration branches for the three different types of batches.]

Allan and Beth currently have \(\$ 2\) and \(\$ 3\), respectively. A fair coin is tossed. If the result of the toss is \(\mathrm{H}\), Allan wins \(\$ 1\) from Beth, whereas if the coin toss results in \(\mathrm{T}\), then Beth wins \(\$ 1\) from Allan. This process is then repeated, with a coin toss followed by the exchange of \(\$ 1\), until one of the two players goes broke (one of the two gamblers is ruined). We wish to determine \(a_{2}=P\) (Allan is the winner \(\mid\) he starts with \(\$ 2\) ) To do so, let's also consider \(a_{i}=P\) (Allan wins | he starts with \(\$ i\) ) for \(i=0,1,3,4\), and 5 . a. What are the values of \(a_{0}\) and \(a_{5}\) ? b. Use the law of total probability to obtain an equation relating \(a_{2}\) to \(a_{1}\) and \(a_{3}\). [Hint: Condition on the result of the first coin toss, realizing that if it is a \(\mathrm{H}\), then from that point Allan starts with \$3.] c. Using the logic described in (b), develop a system of equations relating \(a_{i}(i=1,2,3,4)\) to \(a_{i-1}\) and \(a_{i+1}\). Then solve these equations. [Hint: Write each equation so that \(a_{i}-a_{i-1}\) is on the left hand side. Then use the result of the first equation to express each other \(a_{i}-a_{i-1}\) as a function of \(a_{1}\), and add together all four of these expressions \((i=2,3,4,5)\).] d. Generalize the result to the situation in which Allan's initial fortune is \(\$ a\) and Beth's is \(\$ b\). Note: The solution is a bit more complicated if \(p=P(\) Allan wins \(\$ 1) \neq .5 .\)

Suppose that vehicles taking a particular freeway exit can turn right \((R)\), turn left \((L)\), or go straight \((S)\). Consider observing the direction for each of three successive vehicles. a. List all outcomes in the event \(A\) that all three vehicles go in the same direction. b. List all outcomes in the event \(B\) that all three vehicles take different directions. c. List all outcomes in the event \(C\) that exactly two of the three vehicles turn right. d. List all outcomes in the event \(D\) that exactly two vehicles go in the same direction. e. List outcomes in \(D^{\prime}, C \cup D\), and \(C \cap D\).

A construction firm is currently working on three different buildings. Let \(A_{i}\) denote the event that the \(i\) th building is completed by the contract date. Use the operations of union, intersection, and complementation to describe each of the following events in terms of \(A_{1}, A_{2}\), and \(A_{3}\), draw a Venn diagram, and shade the region corresponding to each one. a. At least one building is completed by the contract date. b. All buildings are completed by the contract date. c. Only the first building is completed by the contract date. d. Exactly one building is completed by the contract date. e. Either the first building or both of the other two buildings are completed by the contract date.

Consider independently rolling two fair dice, one red and the other green. Let \(A\) be the event that the red die shows 3 dots, \(B\) be the event that the green die shows 4 dots, and \(C\) be the event that the total number of dots showing on the two dice is 7 . Are these events pairwise independent (i.e., are \(A\) and \(B\) independent events, are \(A\) and \(C\) independent, and are \(B\) and \(C\) independent)? Are the three events mutually independent?
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