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Suppose that the proportions of blood phenotypes in a particular population are as follows: \(\begin{array}{cccc}\text { A } & \text { B } & \text { AB } & \text { O } \\\ .42 & .10 & .04 & .44\end{array}\) Assuming that the phenotypes of two randomly selected individuals are independent of one another, 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
Probability both O: 0.1936; Probability of matching: 0.3816.

Step by step solution

01

Understanding the Blood Phenotypes and Probabilities

In the population, blood phenotypes have the following probabilities: Phenotype A \(0.42\), B \(0.10\), AB \(0.04\), and O \(0.44\). We need to find probabilities related to independent selections of these phenotypes.
02

Calculate Probability that Both Phenotypes Are O

Assuming independence, the probability that both individuals have the \(O\) phenotype is the product of the individual probabilities: \[ P(O_1 \, \text{and} \, O_2) = P(O_1) \times P(O_2) = 0.44 \times 0.44 = 0.1936. \] Therefore, the probability that both individuals have blood type O is \(0.1936\).
03

Calculate Probability of Matching Phenotypes

To find the probability that both individuals have the same phenotype, we add the probabilities for matching phenotypes: \[ P( ext{match}) = P(A_1 \text{ and } A_2) + P(B_1 \text{ and } B_2) + P(AB_1 \text{ and } AB_2) + P(O_1 \text{ and } O_2) \] Calculating each: - \( P(A_1 \text{ and } A_2) = 0.42 ^ 2 = 0.1764 \)- \( P(B_1 \text{ and } B_2) = 0.10 ^ 2 = 0.01 \)- \( P(AB_1 \text{ and } AB_2) = 0.04 ^ 2 = 0.0016 \)- \( P(O_1 \text{ and } O_2) = 0.44 ^ 2 = 0.1936 \)Thus, the total probability is: \[ P( ext{match}) = 0.1764 + 0.01 + 0.0016 + 0.1936 = 0.3816. \]
04

Conclude with the Calculated Probabilities

The probability that both selected individuals have blood phenotype \(O\) is \(0.1936\). The probability that they have matching phenotypes, regardless of type, is \(0.3816\).

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Key Concepts

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

Blood Phenotypes
Blood phenotypes refer to the different blood types found within a human population. These types, known as A, B, AB, and O, result from specific antigens present on red blood cells. Each phenotype has its own frequency in any given population, which can change based on genetic diversity and other factors. In our example:
  • Phenotype A occurs with a probability of 0.42
  • Phenotype B with 0.10
  • Phenotype AB with 0.04
  • Phenotype O with 0.44
The sum of all these probabilities must equal 1, as these are all the potential outcomes for a blood type when randomly selecting an individual from the population. Understanding these proportions is key to solving probability problems involving blood phenotypes.
Independent Events
Working with probabilities often involves understanding the concept of independent events. Two events are independent if the outcome of one does not affect the outcome of the other. In the context of our exercise, this means that knowing the blood phenotype of one individual gives us no information about the blood phenotype of another.When dealing with probabilities involving independent events, we use the multiplication rule. If the events are independent, the probability that both occur is the product of their individual probabilities. For example, if we want to know the chance of both individuals having the O blood phenotype, we calculate:\[ P(O_1 \, \text{and} \, O_2) = P(O_1) \times P(O_2) = 0.44 \times 0.44 = 0.1936 \]This multiplication rule is fundamental when dealing with scenarios that require evaluating the likelihood of multiple independent events happening together.
Matching Probabilities
Matching probabilities involve calculating the likelihood that two separate events result in the same outcome. In the context of blood phenotypes, we are interested in the probability that two randomly selected individuals share the same blood type. To find this probability, we sum the probabilities of each possible matching scenario:
  • Both having phenotype A: \(0.42^2 = 0.1764\)
  • Both having phenotype B: \(0.10^2 = 0.01\)
  • Both having phenotype AB: \(0.04^2 = 0.0016\)
  • Both having phenotype O: \(0.44^2 = 0.1936\)
Adding these probabilities gives the total likelihood of a match: \[ P(\text{match}) = 0.1764 + 0.01 + 0.0016 + 0.1936 = 0.3816 \]This approach shows how combining separate probabilities provides insights into the overall chance of certain events co-occurring within a defined context, like blood types in this instance.

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Most popular questions from this chapter

A certain system can experience three different types of defects. Let \(A_{i}(i=1,2,3)\) denote the event that the system has a defect of type \(i\). Suppose that $$ \begin{aligned} &P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05 \\ &P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14 \\\ &P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 \end{aligned} $$ a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2,000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\) [the first board is green] and \(B=\) [the second board is green ). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think that an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+\) \(P\left(A^{\prime} \mid B\right)=1\).

Use Venn diagrams to verify the following two relationships for any events \(A\) and \(B\) (these are called De Morgan's laws): a. \((A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}\) b. \((A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\)

A boiler has five identical relief valves. The probability that any particular valve will open on demand is .95. Assuming independent operation of the valves, calculate \(P\) (at least one valve opens) and \(P\) (at least one valve fails to open).

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