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Chapter 17: Problem 15

Only \(4 \%\) of people have Type AB blood. a) On average, how many donors must be checked to find someone with Type \(\mathrm{AB}\) blood? b) What's the probability that there is a Type \(\mathrm{AB}\) donor among the first 5 people checked? c) What's the probability that the first Type \(\mathrm{AB}\) donor will be found among the first 6 people? d) What's the probability that we won't find a Type \(\mathrm{AB}\) donor before the 10 th person?

Short Answer

Expert verified
a) 25 donors. b) 18.5%. c) 21.77%. d) 66.48%.

Step by step solution

01

Determine the Probability of a Type AB Donor

The probability of a donor having Type AB blood is given as \( p = 0.04 \). This means that \( 4\% \) of the population is expected to have Type AB blood.
02

Calculate the Average Number of Donors Checked (Part a)

The average number of donors that need to be checked to find someone with Type AB blood can be modeled using the concept of 'expected value'. The expected number of trials to get the first success in a geometric distribution is given by \( \frac{1}{p} \). Thus, we calculate \( \frac{1}{0.04} = 25 \). On average, 25 donors must be checked.
03

Calculate the Probability Among First 5 (Part b)

To find the probability that there is at least one Type AB donor in the first 5 donors, we use the complement rule. First, find the probability that none of the first 5 donors have Type AB blood: \( (1 - p)^5 = 0.96^5 \). Thus, the probability of at least one is \( 1 - 0.96^5 \). This calculates to about \( 0.185 \) or \( 18.5\% \).
04

Determine Probability of First 6 Trials (Part c)

The probability that the first Type AB donor is found within the first 6 donors is \( 1 - (1 - p)^6 = 1 - 0.96^6 \), which is about \( 0.2177 \) or \( 21.77\% \).
05

Calculate Probability No AB until 10th (Part d)

The probability that we do not find a Type AB donor before the 10th person means all first 9 donors do not have Type AB blood. This is \( (1-p)^9 = 0.96^9 \), which calculates to approximately \( 0.6648 \) or \( 66.48\% \).

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Key Concepts

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

Probability Theory
Probability theory is a mathematical framework that helps us understand randomness and uncertainty in various situations. It provides the tools to quantify the likelihood of different outcomes. Using probability, we can make informed guesses about the occurrence of events.

For instance, if you know that the probability of a donor having Type AB blood is 4%, you can predict how often this blood type will appear in a group of donors. In probability theory, these chances are usually represented by the symbol \( p \).

In our exercise, \( p = 0.04 \), indicating a 4% chance. Probability is essential for calculating the likelihood of events, such as finding a Type AB blood donor among a group of potential donors.
  • Probabilities range from 0 to 1.
  • A probability of 1 means an event is certain to happen.
  • A probability of 0 means an event is impossible.
Understanding the basics of probability can aid us in various real-world scenarios, including making decisions under uncertainty.
Expected Value
The expected value is a key concept in probability and statistics, providing a measure of the center of a probability distribution. In simpler terms, it's like an average outcome you can expect from a random event.

In the context of our blood donor example, if we want to know how many donors we need to check on average to find someone with Type AB blood, we calculate the expected value. When dealing with a geometric distribution, which models the number of trials until the first success, the formula is \( \frac{1}{p} \).

For this exercise, with \( p = 0.04 \), the expected number of donors is \( \frac{1}{0.04} = 25 \). This means that on average, 25 donors need to be checked to find someone with Type AB blood.
  • The expected value helps predict long-term average outcomes.
  • It simplifies the decision-making process in uncertain situations.
Using the expected value can give us a clear picture of what to expect in repeated or similar situations.
Complement Rule
The complement rule is a handy tool in probability, especially when it's easier to calculate the likelihood of an event not happening rather than happening directly. The rule states that the probability of an event occurring is \( 1 \) minus the probability of it not occurring.

Applying this rule is crucial in problems like ours. To find the probability of at least one Type AB donor among the first 5 people checked, we first calculate the probability that none of them has Type AB blood. This is given by \( (1-p)^5 = 0.96^5 \).

Then using the complement rule, we find \( 1 - (1-p)^5 = 1 - 0.96^5 \), which equates to approximately 18.5%. This gives us the likelihood of our desired scenario happening.
  • The complement rule simplifies complex probability calculations.
  • It is particularly useful when dealing with multiple trials or events.
  • The sum of the probability of an event and its complement equals 1.
The complement rule is an essential technique for tackling probability questions efficiently and correctly.

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Most popular questions from this chapter

Police estimate that \(80 \%\) of drivers now wear their seatbelts. They set up a safety roadblock, stopping cars to check for seatbelt use. a) How many cars do they expect to stop before finding a driver whose seatbelt is not buckled? b) What's the probability that the first unbelted driver is in the 6 th car stopped? c) What's the probability that the first 10 drivers are all wearing their seatbelts? d) If they stop 30 cars during the first hour, find the mean and standard deviation of the number of drivers expected to be wearing seatbelts. e) If they stop 120 cars during this safety check, what's the probability they find at least 20 drivers not wearing their seatbelts?

Suppose \(75 \%\) of all drivers always wear their seatbelts. Let's investigate how many of the drivers might be belted among five cars waiting at a traffic light. a) Describe how you would simulate the number of seatbelt-wearing drivers among the five cars. b) Run at least 30 trials. c) Based on your simulation, estimate the probabilities there are no belted drivers, exactly one, two, etc. d) Find the actual probability model. e) Compare the distribution of outcomes in your simulation to the probability model.

Assume that \(13 \%\) of people are left-handed. If we select 5 people at random, find the probability of each outcome described below. a) The first lefty is the fifth person chosen. b) There are some lefties among the 5 people. c) The first lefty is the second or third person. d) There are exactly 3 lefties in the group. e) There are at least 3 lefties in the group. f) There are no more than 3 lefties in the group.

Vitamin D is essential for strong, healthy bones. Our bodies produce vitamin D naturally when sunlight falls upon the skin, or it can be taken as a dietary supplement. Although the bone disease rickets was largely eliminated in England during the 1950 s, some people there are concerned that this generation of children is at increased risk because they are more likely to watch TV or play computer games than spend time outdoors. Recent research indicated that about \(20 \%\) of British children are deficient in vitamin D. Suppose doctors test a group of elementary school children. a) What's the probability that the first vitamin Ddeficient child is the 8 th one tested? b) What's the probability that the first 10 children tested are all okay? c) How many kids do they expect to test before finding one who has this vitamin deficiency? d) They will test 50 students at the third-grade level. Find the mean and standard deviation of the number who may be deficient in vitamin D. e) If they test 320 children at this school, what's the probability that no more than 50 of them have the vitamin deficiency?

Do these situations involve Bernoulli trials? Explain. a) You are rolling 5 dice and need to get at least two 6 's to win the game. b) We record the distribution of eye colors found in a group of 500 people. c) A manufacturer recalls a doll because about \(3 \%\) have buttons that are not properly attached. Customers return 37 of these dolls to the local toy store. Is the manufacturer likely to find any dangerous buttons? d) A city council of 11 Republicans and 8 Democrats picks a committee of 4 at random. What's the probability they choose all Democrats? e) A 2002 Rutgers University study found that \(74 \%\) of high school students have cheated on a test at least once. Your local high school principal conducts a survey in homerooms and gets responses that admit to cheating from 322 of the 481 students.
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