/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 If two events \(A\) and \(B\) ar... [FREE SOLUTION] | 91Ó°ÊÓ

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If two events \(A\) and \(B\) are independent and you know that \(P(A)=0.3\), what is the value of \(P(A \mid B)\) ?

Short Answer

Expert verified
\(P(A \mid B) = 0.3\) because events \(A\) and \(B\) are independent.

Step by step solution

01

Understanding Independent Events

When two events, \(A\) and \(B\), are independent, the occurrence of one event does not affect the probability of the other. This means that \(P(A \mid B) = P(A)\).
02

Substituting Given Values

We are given that \(P(A) = 0.3\). Since events \(A\) and \(B\) are independent, \(P(A \mid B) = P(A)\). Therefore, substitute the given probability of event \(A\) to find \(P(A \mid B)\).
03

Final Calculation

With the information provided that \(P(A) = 0.3\) and using the property of independence, we conclude that \(P(A \mid B) = 0.3\).

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Key Concepts

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

Understanding Independent Events
In probability theory, independent events are those events where the occurrence of one does not influence the occurrence of the other. To grasp this concept better, imagine flipping a coin. The result of a single flip (heads or tails) has no effect on the outcome of the next flip. Similarly, if you were rolling a die, each roll is independent because what number you get on one roll does not affect the next. When we say two events, like events \(A\) and \(B\), are independent, it means that whether or not \(A\) happens doesn't change the likelihood of \(B\) occurring, and vice versa. This can be mathematically expressed as:- \(P(A \cap B) = P(A) \cdot P(B)\)- \(P(A \mid B) = P(A)\)The equation \(P(A \mid B) = P(A)\) tells us the probability of \(A\) occurring given that \(B\) has occurred, is the same as the probability of \(A\) occurring on its own. Therefore, if you know the probability of an event and its independent relationship with another, you already know the conditional probability.
Conditional Probability Clarified
Conditional probability is about finding the probability of an event given that another event has occurred. This is particularly useful when two events are not independent, as it allows us to update the likelihoods based on new information.It is denoted as \(P(A \mid B)\), which reads as the probability of \(A\) given \(B\). We can compute conditional probability using the formula:- \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) when \(P(B) > 0\)Let's break it down:- \(P(A \cap B)\) is the probability that both events \(A\) and \(B\) happen.- \(P(B)\) is the probability of event \(B\) occurring.If events \(A\) and \(B\) are independent, then their intersection \(P(A \cap B)\) simplifies to \(P(A) \cdot P(B)\), which gives us the nice result that \(P(A \mid B) = P(A)\). In this case, knowing \(B\) doesn't change the probability of \(A\).
Exploring the Properties of Probability
The properties of probability are fundamental rules that every probability measure follows. Understanding these properties helps when solving any probability-related problem.Key properties include:1. **Non-negativity:** The probability of any event is always greater than or equal to zero. This means that negative probabilities are not possible. - \(P(A) \geq 0\)2. **Normalization:** The sum of probabilities of all possible outcomes of a random experiment is 1. For example, if you roll a die, the probability of landing on any number from 1 to 6 adds up to 1. - \( \sum P(A_i) = 1\) for all possible events \(A_i\).3. **Additivity:** If two events, \(A\) and \(B\), cannot occur at the same time, the probability of either \(A\) or \(B\) occurring is the sum of their probabilities. - \(P(A \cup B) = P(A) + P(B)\) if \(A \cap B = \emptyset\)These principles form the foundation of probability theory, allowing us to calculate and understand various scenarios effectively, whether they involve independent events or a more complex relationship requiring conditional probability.

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Most popular questions from this chapter

At Litchfield College of Nursing, \(85 \%\) of incoming freshmen nursing students are female and \(15 \%\) are male. Recent records indicate that \(70 \%\) of the entering female students will graduate with a BSN degree, while \(90 \%\) of the male students will obtain a BSN degree. If an incoming freshman nursing student is selected at random, find (a) \(P\) (student will graduate I student is female). (b) \(P(\) student will graduate and student is female). (c) \(P\) (student will graduate \(\mid\) student is male). (d) \(P\) (student will graduate and student is male). (e) \(P\) (student will graduate). Note that those who will graduate are either males who will graduate or females who will graduate. (f) The events described by the phrases "will graduate and is female" and "will graduate, given female" seem to be describing the same students. Why are the probabilities \(P\) (will graduate and is female) and \(P\) (will graduate female) different?

Answer questions true or false and give a brief explanation for each answer. Hint: Review the summary of basic probability rules.If \(A\) and \(B\) are independent events, they must also be mutually exclusive events.

You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(3\) on 1 st card and 10 on 2 nd ). (c) Find \(P(10\) on 1 st card and 3 on 2 nd ). (d) Find the probability of drawing a 10 and a 3 in either order.

Answer questions true or false and give a brief explanation for each answer. Hint: Review the summary of basic probability rules.If \(A\) and \(B\) are mutually exclusive, they must also be independent.

s The state medical school has discovered a new test for tuberculosis. (If the test indicates a person has tuberculosis, the test is positive.) Experimentation has shown that the probability of a positive test is \(0.82\), given that a person has tuberculosis. The probability is \(0.09\) that the test registers positive, given that the person does not have tuberculosis. Assume that in the general population, the probability that a person has tuberculosis is \(0.04\). What is the probability that a person chosen at random will (a) have tuberculosis and have a positive test? (b) not have tuberculosis? (c) not have tuberculosis and have a positive test?

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