/*! 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 91 Universal blood donors People wi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Universal blood donors People with type O-negative blood are universal donors. That is, any patient can receive a transfusion of O-negative blood. Only 7.2% of the American population have O-negative blood. If 10 people appear at random to give blood, what is the probability that at least 1 of them is a universal donor? Follow the four-step process.

Short Answer

Expert verified
The probability that at least one is a universal donor is approximately 0.5205.

Step by step solution

01

Define the Random Variable

Let the random variable \( X \) represent the number of universal donors among the 10 people. Universal donors have type O-negative blood.
02

Determine the Probability Distribution

Each person independently being a universal donor follows a Bernoulli distribution with success probability \( p = 0.072 \). When considering 10 people, \( X \), the number of universal donors, follows a Binomial distribution with parameters \( n = 10 \) and \( p = 0.072 \).
03

Set Up the Complementary Event

To find the probability that at least one person is a universal donor, we first find the probability that none of them are. The complementary event is that 0 people are universal donors.
04

Calculate the Complementary Probability

Using the Binomial distribution, the probability of 0 universal donors is given by:\[P(X = 0) = \binom{10}{0} (0.072)^0 (1 - 0.072)^{10} = 1 \times 0.928^{10}\] Calculate \(0.928^{10}\) to find this probability.
05

Find the Probability of At Least One Universal Donor

Subtract the complementary probability from 1:\[P(X \geq 1) = 1 - P(X = 0) = 1 - 0.928^{10}\]Calculate this expression to obtain the probability that at least one of them is a universal donor.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Random Variable
A random variable is a concept used in statistics to denote a variable that can take different outcomes randomly. In the context of our exercise, we define a random variable, denoted as \(X\), which represents the number of people with O-negative blood among a group of 10. Here, the random variable is discrete, which means it takes on a countable set of values—in this case, from 0 up to 10.
By identifying this random variable, we can perform various calculations to determine the likelihood of certain outcomes, such as the probability that at least one person in a group of 10 is a universal donor. This representation is key for making sense of probabilities in the scenario being considered.
Bernoulli Distribution
The Bernoulli distribution is fundamental when dealing with individual trials where there are two outcomes: success or failure. In our context, 'success' refers to a person being a universal donor, while 'failure' denotes they are not.
  • Each donor check (one person's blood type) is an independent trial.
  • The probability of success (finding an O-negative person) is \( p = 0.072 \).
  • The probability can be modeled using the formula for a single Bernoulli trial outcome.
When we expand to consider multiple trials, such as checking 10 people, the appropriate model becomes a Binomial distribution. This distribution extends the Bernoulli distribution for multiple trials.
Complementary Probability
Complementary probability is a useful strategy in probability calculations. It allows us to compute the probability of one event happening by reference to its complement - that is, the event not happening.
In this situation, rather than directly computing the probability of having at least one universal donor (our event of interest), we first consider the probability of having no universal donors at all, which is easier. The complement of "at least one" is "none."
The computation involves using the complement rule:
  • Find the probability of no successes (universal donors), represented by \(P(X = 0)\).
  • Subtract this from 1 to get the desired probability: \(P(X \geq 1) = 1 - P(X = 0)\).
This simplifies the computation and often reduces the number of calculations needed.
Probability Calculation
In probability calculations, especially with a Binomial distribution, several steps are typically involved, which include using formulas related to binomial coefficients and powers of probability terms.
To find the probability of zero universal donors, we calculate:
  • Compute \( \binom{10}{0} = 1\) (this is the binomial coefficient for choosing 0 successes out of 10 trials).
  • Multiply this by \((0.072)^0 (0.928)^{10}\), clarifying that we raise the probability of failure \((1 - p)\) to the number of trials \(n\).
Thus:\[P(X = 0) = 1 \times 0.928^{10}\].
Finally, we subtract this result from 1 to find the probability that there is at least one universal donor in the ten people: \(P(X \geq 1)\). With these steps, we can determine the likelihood of certain outcomes in any binomial scenario.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bright lights? A string of Christmas lights contains 20 lights. The lights are wired in series, so that if any light fails the whole string will go dark. Each light has probability 0.02 of failing during a 3-year period. The lights fail independently of each other. Find the probability that the string of lights will remain bright for 3 years.

At the gym Suppose that 10% of adults belong to health clubs, and 40% of these health club members go to the club at least twice a week. What percent of all adults go to a health club at least twice a week? Write the information given in terms of probabilities, and use the general multiplication rule.

You read in a book about bridge that the probability that each of the four players is dealt exactly one ace is about 0.11. This means that (a) in every 100 bridge deals, each player has one ace exactly 11 times. (b) in one million bridge deals, the number of deals on which each player has one ace will be exactly 110,000. (c) in a very large number of bridge deals, the percent of deals on which each player has one ace will be very close to 11%.(d) in a very large number of bridge deals, the average number of aces in a hand will be very close to 0.11. (e) None of these

Stoplight On her drive to work every day, Ilana passes through an intersection with a traffic light. The light has probability 1/3 of being green when she gets to the intersection. Explain how you would use each chance device to simulate whether the light is red or green on a given day. (a) A six-sided die (b) Table D of random digits (c) A standard deck of playing cards

In a table of random digits such as Table D, each digit is equally likely to be any of \(0,1,2,3,4,5,6,\) \(7,8,\) or \(9 .\) What is the probability that a digit in the table is 7 or greater? $$ \begin{array}{l}{\text { (a) } 7 / 10 \quad \text { (c) } 4 / 10 \quad \text { (e) } 1 / 10} \\ {\text { (b) } 6 / 10 \quad \text { (d) } 3 / 10}\end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.