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A study of drug offenders who have been treated for drug abuse suggests that the chance of conviction within a 2 -year period after treatment may depend on the offender's education. The proportions of the total number of cases that fall into four education/conviction categories are shown in the following table. Suppose a single offender is selected from the treatment program. Here are the events of interest: \(A:\) The offender has 10 or more years of education B: The offender is convicted within 2 years after completion of treatment Find the appropriate probabilities for these events: a. \(A\) b. \(B\) c. \(A \cap B\) d. \(A \cup B\) e. \(A^{c}\) f. \(A\) given that \(B\) has occurred g. \(B\) given that \(A\) has occurred

Short Answer

Expert verified
Answer: The formula to calculate conditional probability is P(A|B) = P(A ∩ B) / P(B).

Step by step solution

01

Understanding the given information

The table provided shows the proportions of four categories, classified by the offender's education and conviction status. We can use these proportions to calculate the probabilities of events A (the offender has 10 or more years of education) and B (the offender is convicted within 2 years after completion of treatment).
02

Calculating the probability of A

To find the probability of A (the offender has 10 or more years of education), we will sum up the proportions of cases with 10 or more years of education, regardless of whether the offender is convicted or not. If we denote the proportions as \(p_{1}, p_{2}, p_{3}, p_{4}\) according to their categories, then \(P(A) = p_{1} + p_{2}\).
03

Calculating the probability of B

To find the probability of B (the offender is convicted within 2 years after completion of treatment), we sum up the proportions of cases where the offender is convicted, regardless of their education level. So, \(P(B) = p_{2} + p_{4}\).
04

Calculating the probability of A ∩ B

The probability of A ∩ B means the probability of an offender having both 10 or more years of education and being convicted within 2 years after treatment. This corresponds to category 2 in the table, so \(P(A \cap B) = p_{2}\).
05

Calculating the probability of A ∪ B

To find the probability of A ∪ B (either A or B or both), we can use the formula for the union of probabilities: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
06

Calculating the probability of \(A^{c}\)

The probability of \(A^{c}\), the complement of A, represents the probability of an offender having less than 10 years of education. We can find this by subtracting the probability of A from 1: \(P(A^{c}) = 1 - P(A)\).
07

Calculating the probability of A given B

To find the conditional probability of A given B (\(P(A|B)\)), meaning the probability of an offender having 10 or more years of education given that they were convicted within 2 years after treatment, we use the formula for conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).
08

Calculating the probability of B given A

Lastly, to find the conditional probability of B given A (\(P(B|A)\)), meaning the probability of an offender being convicted within 2 years after treatment given that they have 10 or more years of education, we use the formula for conditional probability: \(P(B|A) = \frac{P(A \cap B)}{P(A)}\).

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Key Concepts

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

Conditional Probability
Conditional probability represents the likelihood of one event happening, given that another event has already occurred. It's written as \(P(A|B)\) to express "the probability of event A given that B has occurred."

To calculate conditional probability, you use the equation: \[P(A|B) = \frac{P(A \cap B)}{P(B)}\] Here, \(P(A \cap B)\) is the probability of both events A and B happening at the same time. And \(P(B)\) is the probability of B occurring. By using the conditional probability formula, you can refine the likelihood of an event when new, related information is available.

This concept comes in handy in situations like predicting outcomes based on previous conditions or narrowing down possibilities in decision-making.
Event Probability
Event probability reflects the chance of a single outcome happening in a particular experiment or situation. Let's think of it as the basic building block of probability.

To determine event probability, you calculate: \[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\] In situations like the one presented with offenders, if you want to know the probability that the offender has 10 or more years of education, you sum up the proportions from relevant categories.

Knowing event probabilities helps in predicting certain outcomes and preparing for potential scenarios.
Union and Intersection of Events
Union and intersection describe how two events interact. The **union** of events, expressed as \(A \cup B\), involves all outcomes that belong to either event A or event B, or to both. The formula for union is:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B)\] Here, the minus part \(P(A \cap B)\) ensures that any overlap (where both events happen) is not counted twice.

The **intersection** of events, expressed as \(A \cap B\), captures only the outcomes that events A and B share, meaning both events happen together.

Understanding these concepts allows you to evaluate how events relate and influence each other. This is crucial when analyzing situations with multiple criteria or factors.
Complementary Events
Complementary events cover all the probabilities that aren't included in the desired event. If you have event A, the complement is denoted as \(A^c\) and describes everything outside of A. The two, A and \(A^c\), collectively cover all possible outcomes.

The formula is simply: \[ P(A^c) = 1 - P(A)\] Using \( (1-P(A)) \), you effectively account for the uncertainty not captured by A, ensuring complete coverage of the scenario.

Complementary events are useful in many real-world applications, such as confidence intervals and risk assessments, because they provide insight into the remainder of potential outcomes that are not originally targeted.

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Most popular questions from this chapter

A population can be divided into two subgroups that occur with probabilities \(60 \%\) and \(40 \%,\) respectively. An event A occurs \(30 \%\) of the time in the first subgroup and \(50 \%\) of the time in the second subgroup. What is the unconditional probability of the event A, regardless of which subgroup it comes from?

A man takes either a bus or the subway to work with probabilities .3 and \(.7,\) respectively. When he takes the bus, he is late \(30 \%\) of the days. When he takes the subway, he is late \(20 \%\) of the days. If the man is late for work on a particular day, what is the probability that he took the bus?

A teacher randomly selects 1 of his 25 kindergarten students and records the student's gender, as well as whether or not that student had gone to preschool. a. Construct a tree diagram for this experiment. How many simple events are there? b. The table on the next page shows the distribution of the 25 students according to gender and preschool experience. Use the table to assign probabilities to the simple events in part a. $$ \begin{array}{lcc} \hline & \text { Male } & \text { Female } \\ \hline \text { Preschool } & 8 & 9 \\ \text { No Preschool } & 6 & 2 \end{array} $$ c. What is the probability that the randomly selected student is male? d. What is the probability that the student is a female and did not go to preschool?

For the experiments, list the simple events in the sample space, assign probabilities to the simple events, and find the required probabilities. A roulette wheel contains 38 pocketsthe numbers 1 through \(36,0,\) and \(00 .\) The wheel is spun and the "winning" pocket is recorded. If observing any one pocket is just as likely as any other, what is the probability of observing either 0 or \(00 ?\)

Define the simple events for the experiments in Exercises \(16-20 .\) The grade level of a high school student is recorded.

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