/*! 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 30 Only about \(70 \%\) of all dona... [FREE SOLUTION] | 91Ó°ÊÓ

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Only about \(70 \%\) of all donated human blood can be used in hospitals. The remaining \(30 \%\) cannot be used because of various infections in the blood. Suppose a blood bank has 10 newly donated pints of blood. Let \(r\) be a binomial random variable that represents the number of "good" pints that can be used. (a) Based on questionnaires completed by the donors, it is believed that at least 6 of the 10 pints are usable. What is the probability that at least 8 of the pints are usable, given this belief is true? Compute \(P(8 \leq r | 6 \leq r)\) (b) Assuming the belief that at least 6 of the pints are usable is true, what is the probability that all 10 pints can be used? Compute \(P(r=10 | 6 \leq r)\)

Short Answer

Expert verified
(a) 0.518; (b) 0.057

Step by step solution

01

Define the Problem

We are given that 70% of blood can be used, meaning each pint has a 0.7 probability of being usable. For part (a), we want to find the probability that at least 8 pints are usable given that at least 6 are. For part (b), we need the probability that all 10 pints are usable under the same condition.
02

Use Binomial Probability Formula

The number of usable pints follows a Binomial distribution with parameters \(n = 10\) and \(p = 0.7\). The probability of \(r\) usable pints is given by:\[ P(r) = \binom{10}{r} (0.7)^r (0.3)^{10-r} \]
03

Calculate Unconditional Probabilities

For part (a), calculate \(P(8 \leq r)\) and \(P(6 \leq r)\). For part (b), calculate \(P(r = 10)\). These are:- \(P(r \geq 8) = P(r=8) + P(r=9) + P(r=10)\)- \(P(r \geq 6) = 1 - P(r \leq 5) = 1 - \sum_{r=0}^5 P(r)\)- \(P(r=10)\)

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Key Concepts

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

Conditional Probability
Conditional probability is an important concept in probability theory that allows us to calculate the probability of one event happening, given that another event has already occurred. It's like asking, "What is the probability that event B happens if we already know event A has happened?" In this exercise, we are tasked with finding the probability of having a certain number of usable pints of blood under the condition that already at least 6 are usable.

This requires us to understand the formula for conditional probability: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]Here, \(P(A|B)\) is the probability of event A given event B.

- For part (a), we need to find \(P(8 \leq r | 6 \leq r)\). This means we need the probability of having at least 8 usable pints if we know we already have at least 6 usable pints. So, our event B is "at least 6 are usable," and event A is "at least 8 are usable."- For part (b), we compute \(P(r=10 | 6 \leq r)\). Here, event A is "all 10 pints are usable," while event B remains "at least 6 are usable."Using the given probabilities and the binomial distribution, we can calculate these probabilities effectively. The key is to first determine the probabilities of each event separately, then apply the formula to find the conditional probabilities.
Binomial Theorem
The binomial theorem is vital not only in algebra but also in probability, especially when dealing with a binomial distribution. In our exercise, the scenario of determining usable blood pints follows a binomial distribution because each pint of blood is either usable or it isn’t, making it a sequence of independent binary outcomes.

A binomial distribution can be characterized by two parameters, \( n \), the number of trials (or pints of blood, in this case), and \( p \), the probability of success (usable pint). The binomial probability formula is given by:\[ P(r) = \binom{n}{r} p^r (1-p)^{n-r} \]Where:
  • \( \binom{n}{r} \) is the binomial coefficient, "n choose r," indicating the number of combinations of \( n \) items taken \( r \) at a time.
  • \( p^r \) is the probability of having exactly \( r \) successes, and \((1-p)^{n-r} \) accounts for the failures.
In our situation, \( n = 10 \) and \( p = 0.7 \), meaning each pint has a 70% chance of being usable. This formula helps us calculate the probability of any number of successes out of 10 pints being usable.
Probability Calculation
Calculating probabilities directly from the binomial distribution involves using the formula to determine the likelihood of a certain number of successes. In the exercise, we explored different probabilities, depending on the conditions set by using the binomial theorem.

First, we calculate the probability for a given \( r \) value using:\[ P(r) = \binom{10}{r} (0.7)^r (0.3)^{10-r} \]We use different values of \( r \) for each part:
  • For \( P(r \geq 8) \), calculate \( P(8) + P(9) + P(10) \).
  • For \( P(6 \leq r) \), compute \( 1 - P(r \leq 5) \).
  • For \( P(r = 10) \), simply compute \( P(10) \).
Once these probabilities are calculated, they are used as inputs for the conditional probabilities \( P(8 \leq r | 6 \leq r) \) and \( P(r=10 | 6 \leq r) \) from the earlier section.

By doing so, we gain insights into the likely outcomes of the blood usability context, allowing us to understand the role of conditional probability in determining the usability of all blood pints."

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Most popular questions from this chapter

Much of Trail Ridge Road in Rocky Mountain National Park is over 12,000 feet high. Although it is a beautiful drive in summer months, in winter the road is closed because of severe weather conditions. Winter Wind Studies in Rocky Mountain National Park by Glidden (published by Rocky Mountain Nature Association) states that sustained galeforce winds (over 32 miles per hour and often over 90 miles per hour) occur on the average of once every 60 hours at a Trail Ridge Road weather station. (a) Let \(r=\) frequency with which gale-force winds occur in a given time interval. Explain why the Poisson probability distribution would be a good choice for the random variable \(r\) (b) For an interval of 108 hours, what are the probabilities that \(r=2,3,\) and 4? What is the probability that \(r<2 ?\) (c) For an interval of 180 hours, what are the probabilities that \(r=3,4,\) and 5? What is the probability that \(r<3 ?\)

The Wall Street Journal reported that approximately \(25 \%\) of the people who are told a product is improved will believe that it is, in fact, improved. The remaining \(75 \%\) believe that this is just hype (the same old thing with no real improvement). Suppose a marketing study consists of a random sample of eight people who are given a sales talk about a new, improved product. (a) Make a histogram showing the probability that \(r=0\) to 8 people believe the product is, in fact, improved. (b) Compute the mean and standard deviation of this probability distribution. (c) How many people are needed in the marketing study to be \(99 \%\) sure that at least one person believes the product to be improved? Hint: Note that \(P(r \geq 1)=0.99\) is equivalent to \(1-P(0)=0.99,\) or \(P(0)=0.01\).

Consider each distribution. Determine if it is a valid probability distribution or not, and explain your answer. (a)$$\begin{array}{c|ccc} \hline x & 0 & 1 & 2 \\ \hline P(x) & 0.25 & 0.60 & 0.15 \\ \hline \end{array}$$ (b)$$\begin{array}{c|ccc} \hline x & 0 & 1 & 2 \\ \hline P(x) & 0.25 & 0.60 & 0.20 \\ \hline \end{array}$$

What does the expected value of a binomial distribution with \(n\) trials tell you?

At State College all classes start on the hour, with the earliest start time at 7 A.M. and the latest at 8 p.M. A random sample of freshmen showed the percentages preferring the listed start times. $$\begin{array}{l|ccccc} \text { Start Time } & 7 \text { or } 8 \text { A.M. } & 9,10, \text { or } 11 \text { A.M. } & 12 \text { or } 1 \text { P.M. } & \text { 1 } \text { P.M. or later } & \text { After } 5 \text { P.M. } \\ \hline \% \text { preferring } & 10 \% & 35 \% & 28 \% & 25 \% & 15 \% \end{array}$$ Can this information be used to make a discrete probability distribution? Explain.

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