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Events \(A\) and \(B\) are given such that \(P(A)=\frac{3}{4}, P(A \cup B)=\frac{4}{5}\) and \(P(A \cap B)=\frac{3}{10}\) Find \(P(B)\)

Short Answer

Expert verified
\(P(B) = \frac{7}{20}\).

Step by step solution

01

Understand the Basics

We have probabilities for events A and their union and intersection with event B. We need to find \(P(B)\). We will use the formula for the union of two events: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
02

Substitute Given Values

Plug the known values into the union formula: \(\frac{4}{5} = \frac{3}{4} + P(B) - \frac{3}{10}\). This will form an equation where \(P(B)\) is the only unknown value.
03

Simplify the Equation

Start with the equation \(\frac{4}{5} = \frac{3}{4} + P(B) - \frac{3}{10}\). Combine the fractions after substituting their values for simplification.
04

Align for Common Denominator

Find a common denominator for the fractions to make addition and subtraction easy. Convert \(\frac{3}{4}\) to \(\frac{15}{20}\) and \(\frac{3}{10}\) to \(\frac{6}{20}\) and \(\frac{4}{5}\) to \(\frac{16}{20}\).
05

Solve for \(P(B)\)

Now the equation looks like \(\frac{16}{20} = \frac{15}{20} + P(B) - \frac{6}{20}\). Simplify it as \(\frac{16}{20} = \frac{9}{20} + P(B)\). Subtract \(\frac{9}{20}\) from both sides to solve for \(P(B)\): \(P(B) = \frac{16}{20} - \frac{9}{20}\).
06

Calculate \(P(B)\)

Simplify \(\frac{16}{20} - \frac{9}{20}\) which equals \(\frac{7}{20}\). Thus, \(P(B) = \frac{7}{20}\).

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Key Concepts

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

Union of Events
In probability theory, when we talk about the union of events, we refer to the possibility that at least one of the events occurs. It's about combining both happenings.
For events \(A\) and \(B\), the union is represented as \(A \cup B\). It means the occurrence of event \(A\), event \(B\), or both.
The probability of the union of two events is calculated using the formula:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula accounts for the probabilities of both events and subtracts their intersection to avoid double-counting.
When solving problems involving unions, always consider the overlap between events, which is represented by their intersection.
Intersection of Events
The intersection of events in probability refers to all outcomes shared by two or more events. When events \(A\) and \(B\) occur together, it is called their intersection, denoted as \(A \cap B\).
Think of it as the common ground where both events meet.
Intersection probabilities are crucial when finding joint probabilities, by ensuring we only count those outcomes where both conditions are met.
In the context of our original exercise, the given \( P(A \cap B) = \frac{3}{10} \) represents the probability that both events \( A \) and \( B \) happen simultaneously.
Understanding intersections helps in solving more complex probability problems where multiple events are involved.
Probability Formula
The probability formula is the backbone of calculating how likely an event is to occur. Key formulas include those for the union and intersection of events, which we used in our exercise.
For any events \(A\) and \(B\), with a known union and intersection, the primary formula applied is:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula ensures that we don't overcount the probability of events happening.
When something like \(P(B)\) is unknown, rearranging the original equation allows us to solve for it using the other probabilities given in the problem.
By substituting the known values and solving algebraically, we can find the missing probability. This method ensures we leverage every bit of available information to solve problems effectively.

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Most popular questions from this chapter

An engineering company employs three architects that are responsible for cost estimates of new projects. Antonio makes \(30 \%\) of the estimates, Richard makes \(20 \%\) and Marco \(50 \% .\) Like all estimates, there are usually some errors in these estimates. The record of percentages of'serious'errors that cost the company thousands of euros shows Antonio at \(3 \%,\) Richard at \(2 \%,\) and Marco at \(1 \% .\) Which of the three engineers is probably responsible for most of the serious errors?

Driving tests in a certain city are relatively easy to pass the first time you take them. After going through training, the percentage of new drivers passing the test the first time is \(80 \% .\) If a driver fails the first test, there is chance of passing it on a second harder test, two weeks later. \(50 \%\) of the second-chance drivers pass the test. If the second test is unsuccessful, a third attempt, a week later, is given and \(30 \%\) of the participants pass it. Otherwise, the driver has to retrain and take the test after 1 year. a) Find the probability that a randomly chosen new driver will pass the test without having to wait one year. b) Find the probability that a randomly chosen new driver that passed the test did so on the second attempt.

PIN numbers for cellular phones usually consist of four digits that are not necessarily different. a) How many possible PINs are there? b) You don't want to consider the pins that start with \(0 .\) What is the probability that a PIN chosen at random does not start with a zero? c) What is the probability that a PIN contains at least one zero? d) Given a PIN with at least one zero, what is the probability that it starts with a zero?

Fill in the missing entries in the following table.. $$\begin{array}{|l|l|l|l|l|l|} \hline \mathrm{P}(A) & \mathrm{P}(B) & \text { Conditions for events } A \text { and } B & \mathrm{P}(A \cap B) & \mathrm{P}(A \cup B) & \mathrm{P}(A | B) \\\ \hline 0.3 & 0.4 & \text { Mutually exclusive } & & & \\ \hline 0.3 & 0.4 & \text { Independent } & & & \\ \hline 0.1 & 0.5 & & & 0.6 & \\ \hline 0.2 & 0.5 & & 0.1 & & \\ \hline \end{array}$$

The probability an event A happens is 0.37 a) What is the probability that it does not happen? b) What is the probability that it may or may not happen?

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