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Given that \(P(B)=.29\) and \(P(A\) and \(B)=.24\), find \(P(A \mid B)\).

Short Answer

Expert verified
The conditional probability \(P(A \mid B) is approximately 0.83

Step by step solution

01

Understand the Problem

The aim is to find the conditional probability of \(A\) given \(B\), denoted as \(P(A \mid B)\). This is the probability of event \(A\) occurring given that event \(B\) has already occurred.
02

Apply the formula for conditional probability

The formula for conditional probability is \(P(A \mid B) = \frac{P(A and B)}{P(B)}\).
03

Substitute Values into the Formula

Substituting the given values: \(P(B)=.29\) and \(P(A and B)=.24\) into the conditional probability formula, we get \(P(A \mid B) = \frac{.24}{.29}\).
04

Calculate

By doing the division, we get \(P(A \mid B) \approx 0.827586207\). We can round this value to two decimal places for ease of presentation, which results in \(P(A \mid B) \approx 0.83\).

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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 the mathematical framework that allows us to quantify uncertainty. It's used to determine how likely it is that an event will happen. In this context, an "event" is any outcome or set of outcomes of a random phenomenon. Examples of events could include rolling a die and getting a six, flipping a coin and getting heads, or more complex scenarios involving multiple conditions.

The concept revolves around the idea of assigning numbers to outcomes to represent their likelihood of occurring, with numbers ranging from 0 (indicating impossibility) to 1 (signifying certainty). Probability theory provides various rules and formulas, including the addition and multiplication rules, to calculate the probability of combinations of events. By understanding these foundations, we can deal effectively with more complex concepts such as conditional probability.
Event Occurrence
An event occurrence in probability addresses the question: "What are the chances that a given event happens?" Real-life situations often present cases where the occurrence of one event affects the likelihood of another. This is especially pertinent in compound events where multiple outcomes are possible, and one event may rely on or relate to others.

Conditional probability comes into play when we want to find the probability of an event given that another event has already occurred. For instance, if we know an event B has happened, we can ask, "What's the probability that event A will happen too?" This is calculated using conditional probability formulas that help us refine our expectations based on known information about previous events.
Formula Application
Applying the formula for conditional probability simplifies complex problems by allowing us to calculate the likelihood of an event under specific circumstances. The formula for conditional probability is expressed as \(P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}\). This formula tells us how to adjust the probability of event A happening if event B is known to have occurred.

  • \(P(A \mid B)\) represents the conditional probability of event A occurring given that B has already occurred.
  • \(P(A \text{ and } B)\) is the joint probability of both A and B happening together.
  • \(P(B)\) is the probability that event B occurs on its own.

Using the given exercise values \(P(B) = .29\) and \(P(A \text{ and } B) = .24\), we substitute these into our formula to get \(P(A \mid B) = \frac{.24}{.29}\). By doing the necessary calculation, we find that \(P(A \mid B) \approx 0.83\). This means there's approximately an 83% chance that event A occurs if event B has occurred.

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Most popular questions from this chapter

A Wired Equivalent Privacy (WEP) key is a security code that one must enter in order to access a secure WiFi network. The characters in the key are used from the numbers 0 to 9 and letters from \(\mathrm{A}\) to \(\mathrm{F}\), which gives 16 possibilities for each character of the key. Note that repeats are allowed, that is, the same letter or number can be used more than once in a key. A WEP key for a WiFi network with 64 -bit security is 10 characters long. a. How many different 64 -bit WEP keys can be made by using the given numbers and letters? b. A specific 64 -bit network has a WEP key in which the 2 nd, 5th, 8 th, and 9 th characters are numbers and the other 6 characters are letters. How many different WEP keys are possible for this network? c. A hacker has determined that the WiFi network mentioned in part b will lock him out if he makes 20,000 unsuccessful attempts to break into the network. What is the probability that the hacker will be locked out of the network?

Given that \(A, B\), and \(C\) are three independent events, find their joint probability for the following. a. \(P(A)=.81, \quad P(B)=.49\), and \(P(C)=.36\) b. \(P(A)=.02, \quad P(B)=.03, \quad\) and \(\quad P(C)=.05\)

An economist says that the probability is \(.47\) that a randomly selected adult is in favor of keeping the Social Security system as it is, \(.32\) that this adult is in favor of totally abolishing the Social Security system, and .21 that this adult does not have any opinion or is in favor of other options. Were these probabilities obtained using the classical approach, relative frequency approach, or the subjective probability approach? Explain your answer.

A car rental agency currently has 44 cars available, 28 of which have a GPS navigating system. Two cars are selected at random from these 44 cars. Find the probability that both of these cars have GPS navigation systems.

A thief has stolen Roger's automatic teller machine (ATM) card. The card has a four-digit personal identification number (PIN). The thief knows that the first two digits are 3 and 5 , but he does not know the last two digits. Thus, the PIN could be any number from 3500 to \(3599 .\) To protect the customer, the automatic teller machine will not allow more than three unsuccessful attempts to enter the PIN. After the third wrong PIN, the machine keeps the card and allows no further attempts. a. What is the probability that the thief will find the correct PIN within three tries? (Assume that the thief will not try the same wrong PIN twice.) b. If the thief knew that the first two digits were 3 and 5 and that the third digit was either 1 or 7 , what is the probability of the thief guessing the correct PIN in three attempts?

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