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Let \(A\) and \(B\) be events such that \(P(A \cap B)=1 / 4, P(\tilde{A})=1 / 3,\) and \(P(B)=\) \(1 / 2 .\) What is \(P(A \cup B) ?\)

Short Answer

Expert verified
\(P(A \cup B) = \frac{11}{12}\).

Step by step solution

01

Understand the Problem

We are given events \(A\) and \(B\) with certain probabilities and need to find \(P(A \cup B)\). We know: \(P(A \cap B) = \frac{1}{4}\), \(P(\tilde{A}) = \frac{1}{3}\), and \(P(B) = \frac{1}{2}\).
02

Find \(P(A)\)

Using the complement rule, \(P(A) = 1 - P(\tilde{A})\). Given \(P(\tilde{A}) = \frac{1}{3}\), so \(P(A) = 1 - \frac{1}{3} = \frac{2}{3}\).
03

Use the Addition Rule

The probability of the union of two events is given by \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). We have all these probabilities from the previous steps.
04

Calculate \(P(A \cup B)\)

Substitute the known values into the equation: \(P(A \cup B) = \frac{2}{3} + \frac{1}{2} - \frac{1}{4}\).
05

Simplify the Expression

Convert each fraction to have a common denominator and perform the arithmetic: \(\frac{2}{3} = \frac{8}{12}\), \(\frac{1}{2} = \frac{6}{12}\), and \(\frac{1}{4} = \frac{3}{12}\).Therefore, \(P(A \cup B) = \frac{8}{12} + \frac{6}{12} - \frac{3}{12} = \frac{11}{12}\).

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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, the concept of the union of events is crucial for understanding how different events can combine to create new probabilistic outcomes. The union of events "A" and "B", denoted by \(P(A \cup B)\), represents the probability that either event "A" or event "B" or both will occur. It is like merging two sets where you want to know the chances of any element from either set appearing. This is not simply a matter of adding probabilities, as it can lead to overcounting shared outcomes. For example, in our problem, both "A" and "B" can occur together with a chance of \(P(A \cap B)\).
When calculating \(P(A \cup B)\), consider the overlap between these events. The correct formula is:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula ensures that the overlap \(P(A \cap B)\) isn't counted twice, providing a more precise probability of either event happening. Understanding this principle is key to mastering more complex probability problems.
Complement Rule
The complement rule in probability offers a straightforward way to find the probability of an event not happening. If you know the probability of an event happening, denoted as \(P(A)\), its complement, represented as \(P(\tilde{A})\), is simply the probability of "A" not happening. The sum of these two probabilities is always equal to 1, as these are exhaustive and mutually exclusive scenarios. Hence, the formula is:
  • \(P(A) = 1 - P(\tilde{A})\)
In the given exercise, we used this rule to find \(P(A)\) when \(P(\tilde{A}) = \frac{1}{3}\). Thus, \(P(A) = 1 - \frac{1}{3} = \frac{2}{3}\).
This method is particularly useful when the complement of an event is easier to calculate or given directly. It helps simplify the analysis and provides essential insights into the entire spectrum of possible outcomes.
Addition Rule
The addition rule is a fundamental concept used in calculating the probability of the union of two events. Unlike simple addition, this rule carefully considers overlapping instances to ensure accurate probability measures. The rule is expressed through the formula:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula accounts for the "double-counting" that occurs when both events occur simultaneously, denoted as \(P(A \cap B)\). By subtracting \(P(A \cap B)\) from the sum of \(P(A)\) and \(P(B)\), we accurately account for all possible outcomes without overlap.
Consider the problem where you need to determine \(P(A \cup B)\) with known values of \(P(A)\), \(P(B)\), and \(P(A \cap B)\). These values allowed for the use of the addition rule to solve for \(P(A \cup B) = \frac{11}{12}\). By understanding the application of this rule, students gain a deeper understanding of how events combine in probabilistic analysis.

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Most popular questions from this chapter

(from vos Savant \(^{26}\) ) A reader of Marilyn vos Savant's column wrote in with the following question: My dad heard this story on the radio. At Duke University, two students had received A's in chemistry all semester. But on the night before the final exam, they were partying in another state and didn't get back to Duke until it was over. Their excuse to the professor was that they had a flat tire, and they asked if they could take a make-up test. The professor agreed, wrote out a test and sent the two to separate rooms to take it. The first question (on one side of the paper) was worth 5 points, and they answered it easily. Then they flipped the paper over and found the second question, worth 95 points: 'Which tire was it?' What was the probability that both students would say the same thing? My dad and I think it's 1 in 16\. Is that right?" (a) Is the answer \(1 / 16 ?\) (b) The following question was asked of a class of students. "I was driving to school today, and one of my tires went flat. Which tire do you think it was?" The responses were as follows: right front, \(58 \%\), left front, \(11 \%\), right rear, \(18 \%\), left rear, \(13 \%\). Suppose that this distribution holds in the general population, and assume that the two test-takers are randomly chosen from the general population. What is the probability that they will give the same answer to the second question?

You are offered the following game. A fair coin will be tossed until the first time it comes up heads. If this occurs on the \(j\) th toss you are paid \(2^{j}\) dollars. You are sure to win at least 2 dollars so you should be willing to pay to play this game - but how much? Few people would pay as much as 10 dollars to play this game. See if you can decide, by simulation, a reasonable amount that you would be willing to pay, per game, if you will be allowed to make a large number of plays of the game. Does the amount that you would be willing to pay per game depend upon the number of plays that you will be allowed?

Tversky and his colleagues \(^{12}\) studied the records of 48 of the Philadelphia 76ers basketball games in the \(1980-81\) season to see if a player had times when he was hot and every shot went in, and other times when he was cold and barely able to hit the backboard. The players estimated that they were about 25 percent more likely to make a shot after a hit than after a miss. In fact, the opposite was true \(-\) the 76 ers were 6 percent more likely to score after a miss than after a hit. Tversky reports that the number of hot and cold streaks was about what one would expect by purely random effects. Assuming that a player has a fifty-fifty chance of making a shot and makes 20 shots a game, estimate by simulation the proportion of the games in which the player will have a streak of 5 or more hits.

If \(A, B,\) and \(C\) are any three events, show that $$ \begin{aligned} P(A \cup B \cup C)=& P(A)+P(B)+P(C) \\ &-P(A \cap B)-P(B \cap C)-P(C \cap A) \\ &+P(A \cap B \cap C) \end{aligned} $$

A die is rolled until the first time that a six turns up. We shall see that the probability that this occurs on the \(n\) th roll is \((5 / 6)^{n-1} \cdot(1 / 6)\). Using this fact, describe the appropriate infinite sample space and distribution function for the experiment of rolling a die until a six turns up for the first time. Verify that for your distribution function \(\sum_{\omega} m(\omega)=1\).

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