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Of the volunteers donating blood in a clinic, \(80 \%\) have the Rhesus (Rh) factor present in their blood. a. If five volunteers are randomly selected, what is the probability that at least one does not have the Rh factor? b. If five volunteers are randomly selected, what is the probability that at most four have the Rh factor? c. What is the smallest number of volunteers who must be selected if we want to be at least \(90 \%\) certain that we obtain at least five donors with the Rh factor?

Short Answer

Expert verified
a) 0.67232, b) 0.67232, c) 8 volunteers.

Step by step solution

01

Understand the Event

Given the problem, we have volunteers with a probability of having the Rh factor, which is 0.80 or 80%. We need to compute different probabilities based on this data.
02

Identify Risk for Exact Opposite

To find the probability that at least one volunteer does not have the Rh factor, calculate the probability that all volunteers have the Rh factor first. In a sample of 5, this is \[(0.8)^5\] where each volunteer has an independent 0.8 chance of having the Rh factor.
03

Calculate Complement

The probability that at least one doesn't have the Rh factor is the complement of all having it. Calculate \[1 - (0.8)^5\] to find this probability.
04

Solve Part (a)

Substitute and compute:\[(0.8)^5 = 0.32768\]Thus,\[1 - 0.32768 = 0.67232\] So, at least one doesn't have the Rh factor with probability 0.67232.
05

Binomial Model for Part (b)

Reinterpret 'at most four have Rh factor' as 'at least one does not have Rh factor.' Based on Step 4, this probability for 5 volunteers remains the same as part (a): \[0.67232\].
06

Determine Min. Volunteers for Part (c)

We want to find a number \(n\) such that the probability of getting at least 5 donors with the Rh factor is at least 90%.
07

Create Equation for Part (c)

Translate this into \[1 - P(\text{fewer than 5 have the Rh factor}) \geq 0.90\]which simplifies to \[P(\text{fewer than 5 }) \leq 0.10\]. Using binomial probabilities calculates successively until this holds.
08

Analyze Required Volunteers for Part (c)

Calculate smallest \(n\), starting at 5, 6, ... where \[1 - \sum_{k=0}^{4} \binom{n}{k} (0.8)^k (0.2)^{n-k} \geq 0.90\]. For example, with 8 volunteers,\[\binom{8}{0}(0.8)^0(0.2)^8 + \binom{8}{1}(0.8)^1(0.2)^7 + \binom{8}{2}(0.8)^2(0.2)^6 + \binom{8}{3}(0.8)^3(0.2)^5 + \binom{8}{4}(0.8)^4(0.2)^4 \leq 0.10\].This approach finds the smallest \(n\) needed.
09

Determination for Part (c)

Upon computation:\[\text{For } n = 8, 1 - (\text{probability calculated for 4 or less having Rh}) \geq 0.90\]. Thus, a minimum of 8 volunteers is necessary.

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Key Concepts

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

Binomial Distribution
In probability theory, a binomial distribution represents the probability of getting a fixed number of successes in a fixed number of independent binary experiments, each with the same probability of success. This is particularly useful when dealing with scenarios where outcomes are confined to two possibilities, such as yes or no, success or failure, or in this case, having the Rhesus (Rh) factor or not having it.
Consider that we have a probability of 0.8 that a randomly chosen donor has the Rh factor. If we select five donors, the probability that exactly a certain number (say, all five) have the Rh factor can be found using the binomial probability formula:
  • \[P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}\]
Here: - \( n \) is the sample size,- \( k \) is the number of successful outcomes we are interested in,- \( p \) is the probability of success (0.8 in this instance).
This distribution can answer various questions, such as the probability of exactly four donors having the Rh factor or none at all.
Complement Rule
The complement rule is a foundational concept in probability that often simplifies the process of finding probabilities, especially when the direct calculation is complex. It states that the probability of an event occurring is one minus the probability of it not occurring.
For instance, when we want to find the probability that at least one volunteer does not have the Rh factor, it's often easier to first calculate the probability that all volunteers do have it. If five volunteers each have an 80% chance of having the Rh factor, the probability that all five have it is:
  • \[(0.8)^5 = 0.32768\]
Using the complement rule, the probability that at least one does not have the Rh factor is:
  • \[1 - 0.32768 = 0.67232\]
This approach reduces potential errors and streamlines calculations. Complement probabilities are especially powerful in solving compound probability problems.
Rhesus Factor
The Rhesus (Rh) factor is a protein found on the surface of red blood cells. Being Rh-positive means that the protein is present, while being Rh-negative means it is absent. Understanding the distribution of the Rh factor among a population is critical for various medical and genetic scenarios, including blood transfusions and pregnancy care.
In statistics, knowing whether someone is Rh-positive or Rh-negative adds a layer of binary categorization, similar to yes or no questions. This kind of binary distribution aligns perfectly with binomial probability models when assessing a population's likelihood to exhibit a certain trait, like having or lacking the Rh factor.
Sample Size Determination
Sample size determination is a critical component in statistical analysis as it affects the accuracy and reliability of the results. When dealing with random sampling or surveys, having an adequately sized sample ensures that the estimates derived from the sample can be generalized to the entire population with confidence.
In the given exercise, we wish to be at least 90% confident that five donors will have the Rh factor. This requires finding the sample size \( n \) that fulfills the condition where the probability of having fewer than five Rh-positive donors is less than or equal to 10%.
The process:
  • Consider the cumulative binomial probability for having fewer than five donors with the Rh factor in a sample and adjust \( n \) until the cumulative probability satisfies: \[1 - \sum_{k=0}^{4} \binom{n}{k} (0.8)^k (0.2)^{n-k} \geq 0.90\]
  • This summation represents the likelihood of having 0 to 4 Rh-positive individuals. Make calculations for successive values of \( n \) until reaching a total probability of more than 0.90 for five or more having the Rh factor.
In our case, a calculation reveals that a minimum sample size of 8 is necessary to reach this threshold. Mathematical tools and statistical software can ease this computation, ensuring precise sample size identification.

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