/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 Four people \(-\mathrm{a}, \math... [FREE SOLUTION] | 91Ó°ÊÓ

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Four people \(-\mathrm{a}, \mathrm{b}, \mathrm{c}\), and \(\mathrm{d}-\) are waiting to give blood. Of these four, a and b have type AB blood, whereas c and d do not. An emergency call has just come in for some type AB blood. If blood samples are taken one by one from the four people in random order for blood typing and \(x\) is the number of samples taken to obtain an \(\mathrm{AB}\) individual (so possible \(x\) values are 1,2 , and 3 ), what is the probability distribution of \(x\) ?

Short Answer

Expert verified
The probability distribution of \(x\) is: \(P(x = 1) = 0.5\), \(P(x = 2) = \frac{1}{3}\), \(P(x = 3) = \frac{1}{6}\)

Step by step solution

01

Determine the total number of individuals

From the problem, we know that there are 4 individuals: a, b, c and d. Out of these 4, two (a and b) have type AB blood. So the total number of individuals is 4.
02

Calculate the probability for \(x = 1\)

If \(x = 1\), this means that the first sample taken is from an AB individual. There are two AB individuals out of the total four. So probability of selecting an AB individual in the first attempt is \(\frac{2}{4} = 0.5\)
03

Calculate the probability for \(x = 2\)

If \(x = 2\), this means that the first sample was not from an AB individual, but the second one is. The first person was either c or d (not-AB), and the second one was either a or b (AB). So the probability is the product of the probability of not selecting an AB individual in the first attempt and selecting an AB individual in the second attempt. So \(P(x = 2) = \frac{2}{4} \times \frac{2}{3} = \frac{1}{3}\)
04

Calculate the probability for \(x = 3\)

If \(x = 3\), this means that the first and second samples were not from AB individuals while the third one was. The probability for this scenario is calculated by multiplying the probabilities of: not selecting an AB individual in the first attempt, not selecting an remaining AB individual in the second attempt, and selecting an remaining AB individual in the third attempt. So \(P(x = 3) = \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} = \frac{1}{6}\)

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Key Concepts

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

Blood Types
Blood types refer to the classification of blood based on the presence or absence of certain antigens on the surface of red blood cells. There are four main types: A, B, AB, and O. Each type can be positive or negative, depending on the presence of the Rh factor. In the context of the exercise, type AB blood is significant because an emergency call specifically requested this type. Individuals with AB blood have both A and B antigens, making them universal plasma donors.

The problem involved differentiating between individuals with and without type AB blood. Understanding blood types is crucial for various reasons:
  • They determine compatibility in blood transfusions and organ transplants.
  • Knowledge of blood types helps in emergency situations where specific types might be urgently needed.
  • Blood types also have implications in population studies and genetic research.
Recognizing the properties and significance of different blood types helps in crafting solutions to statistical problems like the one presented.
Random Sampling
Random sampling is a method used to select individuals from a larger pool in such a way that every possible sample has an equal chance of being chosen. In the exercise, the concept of random sampling is crucial as the blood samples are taken one by one in random order.

This randomness ensures that:
  • There is no bias in the selection process, meaning each individual has an equal chance of being selected first, second, or third.
  • It creates the foundation for calculating probabilities, as seen by how the scenarios for picking AB blood types are structured.
  • Random sampling plays a vital role in obtaining a representative subset from a population, aiding in accuracy and validity of statistical conclusions.
Employing random sampling in this exercise helps determine the probabilistic outcomes and their respective distributions accurately by considering every possible order of sampling.
Discrete Random Variables
Discrete random variables take on distinct and separate values, often whole numbers derived from counting. In the context of the exercise, the discrete random variable is represented by the number of samples taken (\( x \)) to obtain an AB blood type. Here, \( x \) can be 1, 2, or 3, each representing scenarios with corresponding probabilities.

Discrete random variables are important because:
  • They allow us to evaluate and calculate probabilities for specific outcomes, as demonstrated in the given exercise.
  • Outcomes are defined in terms of countable, distinct events, thus lending clarity to statistical interpretations.
  • Understanding these variables offers insight into scenarios with finite or countable datasets, common in real-world problems.
Grasping how discrete random variables work gives us the ability to model and solve problems involving probability distributions, as demonstrated through the calculation of the probability for each potential value of \( x \) in the exercise.

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Most popular questions from this chapter

Seventy percent of the bicycles sold by a certain store are mountain bikes. Among 100 randomly selected bike purchases, what is the approximate probability that a. At most 75 are mountain bikes? b. Between 60 and 75 (inclusive) are mountain bikes? c. More than 80 are mountain bikes? d. At most 30 are not mountain bikes?

Suppose that \(5 \%\) of cereal boxes contain a prize and the other \(95 \%\) contain the message, "Sorry, try again." Consider the random variable \(x\), where \(x=\) number of boxes purchased until a prize is found. a. What is the probability that at most two boxes must be purchased? b. What is the probability that exactly four boxes must be purchased? c. What is the probability that more than four boxes must be purchased?

A grocery store has an express line for customers purchasing at most five items. Let \(x\) be the number of items purchased by a randomly selected customer using this line. Give examples of two different assignments of probabilities such that the resulting distributions have the same mean but quite different standard deviations.

A coin is spun 25 times. Let \(x\) be the number of spins that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. What is the probability of judging the coin fair when \(P(\mathrm{H})=.9\), so that there is a substantial bias? Repeat for \(P(\mathrm{H})=.1 .\) c. What is the probability of judging the coin fair when \(P(\mathrm{H})=.6\) ? when \(P(\mathrm{H})=.4 ?\) Why are the probabilities so large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and (b) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

Consider a large ferry that can accommodate cars and buses. The toll for cars is $$\$ 3$$, and the toll for buses is $$\$ 10 .$$ Let \(x\) and \(y\) denote the number of cars and buses, respectively, carried on a single trip. Cars and buses are accommodated on different levels of the ferry, so the number of buses accommodated on any trip is independent of the number of cars on the trip. Suppose that \(x\) and \(y\) have the following probability distributions: $$ \begin{array}{lrrrrrr} x & 0 & 1 & 2 & 3 & 4 & 5 \\ p(x) & .05 & .10 & .25 & .30 & .20 & .10 \\ y & 0 & 1 & 2 & & & \\ p(y) & .50 & .30 & .20 & & & \end{array} $$ a. Compute the mean and standard deviation of \(x\). b. Compute the mean and standard deviation of \(y\). c. Compute the mean and variance of the total amount of money collected in tolls from cars. d. Compute the mean and variance of the total amount of money collected in tolls from buses. e. Compute the mean and variance of \(z=\) total number of vehicles (cars and buses) on the ferry. f. Compute the mean and variance of \(w=\) total amount of money collected in tolls.

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