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Use the following information to complete the contingency table: $$ \begin{array}{l} \text { - } \mathrm{P}(\mathrm{A} \text { and } \mathrm{B})=10 / 75 \\ \cdot \mathrm{P}(\mathrm{A})=40 / 75 \\ \cdot \mathrm{P}(\operatorname{not} \mathrm{B})=45 / 75 \end{array} $$

Short Answer

Expert verified
A contingency table shows: \( A \text{ and } B=10 \), \( A \text{ and not } B=30 \), \( \text{not } A \text{ and } B=20 \), \( \text{not } A \text{ and not } B=15 \).

Step by step solution

01

Understand the Data

We are given probabilities and need to convert these into a full contingency table. The probabilities are: \( P(A \text{ and } B) = \frac{10}{75} \), \( P(A) = \frac{40}{75} \), and \( P(\text{not } B) = \frac{45}{75} \). These represent certain events in a total sample space of 75 observations.
02

Introduce the Table Structure

A typical contingency table will have four entries: \( A \), \( \text{not } A \), \( B \), and \( \text{not } B \). It helps to understand how many fall into each category by using combinations: \( A \text{ and } B \), \( A \text{ and not } B \), \( \text{not } A \text{ and } B \), and \( \text{not } A \text{ and not } B \). The table also includes row and column totals.
03

Calculate Entries Involving A and B

We know \( P(A \text{ and } B) = \frac{10}{75} \), which means 10 observations fall into \( A \text{ and } B \). We also know \( P(A) = \frac{40}{75} \). So, \( A \text{ and not } B = 40 - 10 = 30 \), since 30 observations are \( A \) but not \( B \).
04

Calculate Entries Involving Not B

We are given \( P(\text{not } B) = \frac{45}{75} \). The total number of \( \text{not } B \) observations, therefore, is 45. We already calculated \( A \text{ and not } B = 30 \). Hence, \( \text{not } A \text{ and not } B = 45 - 30 = 15 \).
05

Calculate Remaining Entry Not A and B

Now, we can determine the remaining entry in the table. Calculate \( \text{not } A \text{ and } B \) by noting that the total \( B \) should be \( P(B) = 1 - P(\text{not } B) = \frac{30}{75} \). The total \( B \) is 30. Therefore, \( \text{not } A \text{ and } B = 30 - 10 = 20 \).
06

Verify Row and Column Totals

Now confirm row and column totals match the total sample space. Totals for each category: \( A \)=40, \( \text{not } A \)=35, \( B \)=30, \( \text{not } B \)=45. The total sample space is 75. Ensure that adding rows and columns still results in consistent totals.

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Key Concepts

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

Understanding Probability
Probability is a measure of how likely an event is to occur. It's an essential concept in statistics and helps us understand the likelihood of different outcomes. A probability is expressed as a number between 0 and 1.
This number represents the chance of an occurrence, where 0 means an event is impossible, and 1 indicates certainty. In the context of the problem given, probability helps in understanding how likely certain events are in the contingency table.
  • The probability of event A (\(P(A)\)), indicates the chance of A occurring out of all possible outcomes. Here, it's given as \(\frac{40}{75}\) or approximately 0.53, meaning there's a 53% chance of event A happening.
  • The probability of event B not occurring (\(P(\text{not } B)\)) shows how likely B is to not happen, given as \(\frac{45}{75}\) or 60% chance.
  • Lastly, \(P(A \text{ and } B)\) shows the joint probability where both A and B occur simultaneously, calculated as \(\frac{10}{75}\) or about a 13% chance.
Calculating these probabilities is the first step in building a contingency table.
The Art of Event Counting
Event counting is a crucial skill in probability and statistics. It involves determining how many ways an event can occur, given certain restrictions. When using a contingency table, event counting helps organize and simplify data.
For example, in the context of our example, once you have the probability of events, converting these probabilities to counts is crucial:
  • For the scenario \(P(A \text{ and not } B)\), there are 30 observations, calculated from \(P(A)\) minus \(P(A \text{ and } B)\). That means there are 30 instances where A happens, but not B.
  • Similarly, counting for "not A and B" involves subtracting the observations of "A and B" from total B, amounting to 20 observations.
Event counting turns abstract probabilities into tangible numbers that fit into the contingency table, making analysis more intuitive.
Intersection and Complement of Events
Within probability, the intersection and complement are vital concepts that describe the relationships between events. The intersection of events examines where two or more events occur simultaneously. In our problem, this is seen directly with \(P(A \text{ and } B)\), indicating the probability that both events A and B occur.
The complement of an event describes the scenario in which the event does not happen. It is often expressed as "not" the event. For instance, \(P(\text{not } B)\) is the complement of the probability of B, representing the likelihood that B does not take place.
  • For intersections, think of events overlapping. In our scenario, \(P(A \text{ and } B) = \frac{10}{75}\) describes this overlap.
  • Complements cover everything outside the event's occurrence. Here, \(P(\text{not } B)\) accounts for all scenarios where B does not happen, a crucial part of filling out a contingency table.
Understanding these concepts helps build a complete picture of all possible outcomes and their probabilities in studies involving multiple events and categories.

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Most popular questions from this chapter

A fitness center coach kept track over the last year of whether members stretched before they exercised, and whether or not they sustained an injury. Among the 400 members, 322 stretched before they exercised, 327 did not sustain an injury, and 270 both stretched and did not sustain an injury. a. Create a contingency table for the information. b. What is the probability that a member sustained an injury? c. What is the probability that a member sustained an injury and did not stretch? d. What is the probability that a member stretched or did not sustain an injury? e. What is the probability that a member sustained an injury given they stretched? f. What is the probability that a member sustained an injury given they did not stretch? g. Does it appear that sustaining an injury depends on whether the member stretches before exercising? Or are they independent? Use probability to support your elaim.

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A ball is drawn randomly from a jar containing 18 black marbles, 4 purple marbles, and 9 blue marbles. Find the probability of: a. Drawing a black marble. b. Not drawing a purple marble. c. Drawing a blue or purple marble. d. Drawing a yellow marble. e. Drawing two black marbles if you draw with replacement. f. Drawing first a blue marble then a black marble if marbles are drawn without replacement.

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