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Let \(A\) and \(B\) be two events in a sample space \(S\) such that \(P(A)=.6\) and \(P(B \mid A)=.5 .\) Find \(P(A \cap B)\)

Short Answer

Expert verified
The probability of the intersection of events A and B is \(P(A \cap B) = 0.3\).

Step by step solution

01

Identify the given probabilities

We are given the following probabilities: - Probability of event A: \(P(A) = 0.6\) - Conditional probability of event B given event A: \(P(B \mid A) = 0.5\)
02

Use the conditional probability formula

To find the probability of the intersection of events A and B, we can use the conditional probability formula: \(P(A \cap B) = P(B \mid A) \cdot P(A)\) Now, we can plug in the given probabilities:
03

Substitute the given probabilities and calculate the answer

Plugging in the given values, we get: \(P(A \cap B) = 0.5 \cdot 0.6\) Now, calculate the result: \(P(A \cap B) = 0.3\)
04

Final answer

The probability of the intersection of events A and B is: \(P(A \cap B) = 0.3\)

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Key Concepts

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

Intersection of Events
The intersection of events is a fundamental concept in probability that refers to the occurrence of two events at the same time. It is represented symbolically as 'A intersect B', or 'A \( \cap \) B'. This notation means we're looking at the scenarios where both Event A and Event B occur together within the sample space.

When we calculate the probability of the intersection of events, we're essentially asking, 'What are the chances of both of these things happening?' For example, if Event A is 'it rains' and Event B is 'you carry an umbrella', the intersection would represent the chance that it both rains and you are carrying an umbrella.
Probability Formula
The probability formula is crucial for calculating the likelihood of different outcomes in a statistical experiment. A basic formula for probability is P(A) = Number of favorable outcomes / Total number of outcomes. However, this formula becomes more complex when dealing with dependent events.

In the context of conditional probability, we look at a modified formula: \(P(A \cap B) = P(B \mid A) \cdot P(A)\). This formula helps us find the probability of the intersection of events when we know the probability of one event occurring given that another event has already occurred. This is important when the occurrence of one event affects the probability of another, which is very common in real-world scenarios.
Sample Space
The concept of a sample space is central to understanding probability. The sample space, symbolized as 'S', represents all possible outcomes of a given experiment or event.

For instance, if we flip a coin, the sample space is {Heads, Tails} - these are the only possible outcomes. In a more complex scenario, like rolling two dice, the sample space consists of 36 outcomes, ranging from (1,1) to (6,6), each combination representing a roll of the first and second die.

Having a clear definition of the sample space allows for precise computation of probabilities and is the foundation upon which we build our understanding of random events.
Event Probability
When we talk about event probability, we're measuring the likelihood of a specific event occurring within the defined sample space. This can be represented by a number between 0 and 1, where 0 means the event is impossible, and 1 indicates certainty.

Understanding event probability involves calculating how many ways an event can happen and comparing it to the total number of outcomes in the sample space. For instance, the probability of drawing an ace from a standard deck of playing cards is \(P(Ace) = \frac{4}{52}\) since there are 4 aces and 52 total cards.

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Most popular questions from this chapter

QuaLrr ConTRoL It is estimated that \(0.80 \%\) of a large consignment of eggs in a certain supermarket is broken. a. What is the probability that a customer who randomly selects a dozen of these cggs receives at least one broken egg? b. What is the probability that a customer who selects these eggs at random will have to check three cartons before finding a carton without any broken eggs? (Each carton contains a dozen eggs.)

Refer to the following experiment: Urn A contains four white and six black balls. Urn B contains three white and five black balls. A ball is drawn from urn A and then transferred to urn B. A ball is then drawn from urn B. What is the probability that the transferred ball was black given that the second ball drawn was black?

The chief loan officer of La Crosse Home Mortgage Company summarized the housing loans extended by the company in 2007 according to type and term of the loan. Her list shows that \(70 \%\) of the loans were fixed-rate mortgages \((F), 25 \%\) were adjustable-rate mortgages \((A)\), and \(5 \%\) belong to some other category \((O)\) (mostly second trust-deed loans and loans extended under the graduated payment plan). Of the fixed-rate mortgages, \(80 \%\) were 30 -yr loans and \(20 \%\) were 15 -yr loans; of the adjustable-rate mortgages, \(40 \%\) were 30 -yr loans and \(60 \%\) were 15 -yr loans; finally, of the other loans extended, \(30 \%\) were 20 -yr loans, \(60 \%\) were 10 -yr loans, and \(10 \%\) were for a term of 5 yr or less. a. Draw a tree diagram representing these data. b. What is the probability that a home loan extended by La Crosse has an adjustable rate and is for a term of 15 yr? c. What is the probability that a home loan cxtended by La Crosse is for a term of 15 yr?

In a survey of 2000 adults \(50 \mathrm{yr}\) and older of whom \(60 \%\) were retired and \(40 \%\) were preretired, the following question was asked: Do you expect your income needs to vary from year to year in retirement? Of those who were retired, \(33 \%\) answered no, and \(67 \%\) answered yes. Of those who were pre-retired, \(28 \%\) answered no, and \(72 \%\) answered yes. If a respondent in the survey was selected at random and had answered yes to the question, what is the probability that he or she was retired?

Determine whether the given experiment has a sample space with equally likely outcomes. A ball is selected at random from an urn containing six black balls and six red balls, and the color of the ball is recorded.

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