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athlete suspected of having used steroids is given two tests that operate independently of each other. Test A has probability \(0.9\) of being positive if steroids have been used. Test B has probability \(0.8\) of being positive if steroids have been used. What is the probability that at least one test is positive if steroids have been used? a. \(0.98\) b. \(0.72\) C. \(0.28\)

Short Answer

Expert verified
The probability is 0.98.

Step by step solution

01

Understanding the Problem

We know two tests, Test A and Test B, are used to detect steroids usage independently. Given the probabilities of each test detecting steroids, we must determine the probability that at least one test returns a positive result if steroids have been used.
02

Defining Probabilities

Let the probability of Test A being positive if steroids used be denoted as \( P(A) = 0.9 \). Similarly, the probability of Test B being positive is \( P(B) = 0.8 \). These numbers denote the chances of each test being positive independently.
03

Calculating the Probability of Both Tests Being Negative

To find the probability that both tests give a negative result, we calculate the complement of each test being positive: \( P(A') = 1 - P(A) = 0.1 \) and \( P(B') = 1 - P(B) = 0.2 \). The probability that both tests are negative is \( P(A' \cap B') = P(A') \times P(B') = 0.1 \times 0.2 = 0.02 \).
04

Determining the Probability of At Least One Positive Test

The probability of at least one test being positive is the complement of both tests being negative. So, \( P( ext{at least one positive}) = 1 - P(A' \cap B') = 1 - 0.02 = 0.98 \).

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Key Concepts

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

Independent Events
In probability, two events are considered independent when the occurrence of one event does not affect the probability of the other event taking place. This is a crucial concept to understand in statistics. In our exercise, Test A and Test B are independent events, meaning the result of Test A has no influence on the result of Test B. To put it simply, knowing the result of one test provides no information about the likelihood of the result of the other test.

Understanding independent events helps us analyze situations where multiple factors or events occur without affecting each other's outcomes. The independence of events is mathematically represented as:
  • \( P(A \cap B) = P(A) \cdot P(B) \)
This formula tells us that the probability of both events A and B happening at the same time is the product of their individual probabilities. It's important for statistical methods because it simplifies the calculation of combined probabilities in complex scenarios.
Complementary Probability
Complementary probability is a fundamental concept that deals with the likelihood of an event not occurring. Essentially, it looks at the flip side of a particular event taking place. This concept is useful when calculating probabilities that are simpler to derive by looking at what doesn't happen.

Looking at our exercise, we considered the probabilities of Test A and Test B being positive, which were given as \( P(A) = 0.9 \) and \( P(B) = 0.8 \), respectively. Their complementary probabilities, the chance that these tests are negative, were calculated as follows:
  • For Test A being negative, \( P(A') = 1 - P(A) = 0.1 \)
  • For Test B being negative, \( P(B') = 1 - P(B) = 0.2 \)
Complementary probability is typically used when you're looking at the situation where an event should not occur, providing a complete view of all possible outcomes.
Probability of Union
The probability of the union of two events is the probability that either one or both of the events occur. In simpler terms, it is the chance that at least one of several possible events happens. In our exercise, we needed to find the probability that at least one of the tests (either Test A or Test B) is positive.

Calculating the union probability is easier once we have the complementary probability of both events being negative, which is more straightforward to determine. The number we are interested in, the probability that at least one test is positive, is given by:
  • \( P(\text{at least one positive}) = 1 - P(A' \cap B') \)
From our previous calculation, \( P(A' \cap B') = 0.02 \), therefore the probability that at least one test is positive is \( 0.98 \). The probability of union helps streamline calculations, especially when dealing with complex and multiple intersecting events.
Statistics Education
Statistics education is essential for developing a deep understanding of probability and how it applies to real-world situations. Through exercises like this one, students learn how to interpret data, calculate probabilities, and develop critical thinking skills related to uncertain outcomes.

Teaching statistics involves explaining concepts like independent events, complementary probabilities, and the probability of unions, all of which build a solid foundation for more advanced statistical analysis. By practicing with varied exercises:
  • Students gain confidence in handling statistical scenarios.
  • They become proficient in evaluating data and determining the probabilities of different results.
  • This skill set is increasingly relevant in today’s data-driven world.
Statistics education not only enhances mathematical reasoning but also prepares students for careers where data interpretation and decision-making based on statistical information are crucial.

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Most popular questions from this chapter

Older College Students. Government data show that \(4 \%\) of adults are full- time college students and that \(37 \%\) of adults are aged 55 or older. Nonetheless, we can't conclude that because \((0.04)(0.37)=0.015\), about \(1.5 \%\) of adults are college students 55 or older. Why not?

Independent? The 2017 update to the Report on the UC Berkeley Faculty Salary Equity Study shows that 94 of the university's 253 assistant professors were women, along with 134 of the 314 associate professors and 244 of the 949 full professors. Note that the study only classified faculty members as a man or a woman. a. What is the probability that a randomly chosen Berkeley professor (of any rank) is a woman? b. What is the conditional probability that a randomly chosen professor is a woman, given that the person chosen is a full professor? c. Are the rank and sex of Berkeley professors independent? How do you know?

The Probability of a Flush. A poker player holds a flush when all five cards in the hand belong to the same suit (clubs, diamonds, hearts, or spades). We will find the probability of a flush when five cards are drawn in succession from the top of the deck. Remember that a deck contains 52 cards, 13 of each suit, and that when the deck is well shuffled, each card drawn is equally likely to be any of those that remain in the deck. a. Concentrate on spades. What is the probability that the first card drawn is a spade? What is the conditional probability that the second card drawn is a spade, given that the first is a spade? (Hint: How many cards remain? How many of these are spades?) b. Continue to count the remaining cards to find the conditional probabilities of a spade for the third, the fourth, and the fifth card drawn, given in each case that all previous cards are spades. c. The probability of drawing five spades in succession from the top of the deck is the product of the five probabilities you have found. Why? What is this probability? d. The probability of drawing five hearts or five diamonds or five clubs is the same as the probability of drawing five spades. What is the probability that the five cards drawn all belong to the same suit?

Winning at Tennis. A player serving in tennis has two chances to get a serve into play. If the first serve is out, the player serves again. If the second serve is also out, the player loses the point. Here are probabilities based on four years of the Wimbledon Championship:2z $$ \begin{aligned} P(\text { lst serve in) }&=0.59 \\ P(\text { win point } \mid \text { lst serve in) }&=0.73 \\ P(2 \text { st serve in } \mid \text { 1st serve out) }&=0.86 \\ P \text { (win point } \mid \text { 1st serve out and 2nd serve in) } &=0.59 \end{aligned} $$ Make a tree diagram for the results of the two serves and the outcome (win or lose) of the point. (The branches in your tree have different numbers of stages, depending on the outcome of the first serve.) What is the probability that the serving player wins the point?

Universal Blood Donors. People with type O-negative blood are referred to as universal donors, although if you give type O-negative blood to any patient, you run the risk of a transfusion reaction due to certain antibodies present in the blood. However, any patient can receive a transfusion of O-negative red blood cells. 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 one of them is a universal donor?

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