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If \(A\) and \(B\) are equally likely events and we require that the probability of their intersection be at least.98, what is \(P(A) ?\)

Short Answer

Expert verified
\(P(A) \geq 0.99\)

Step by step solution

01

Understanding the Problem

We need to determine the probability of event \(A\), knowing that events \(A\) and \(B\) are equally likely and the probability of their intersection \(P(A \cap B)\) is at least 0.98.
02

Equally Likely Events

Since \(A\) and \(B\) are equally likely, we have \(P(A) = P(B)\). Denote this common probability as \(p\). Thus, \(P(A) = p\) and \(P(B) = p\).
03

Expressing the Intersection

The probability of the intersection of \(A\) and \(B\) is given by \(P(A \cap B) = P(A) \times P(B)\) because we often assume independence for equally likely intersections. Thus, \(P(A \cap B) = p^2\).
04

Setting Up the Inequality

We know that \(P(A \cap B) \geq 0.98\). Therefore, we set up the inequality: \(p^2 \geq 0.98\).
05

Solving the Inequality

Solve the inequality \(p^2 \geq 0.98\). We take the square root of both sides: \(p \geq \sqrt{0.98}\).
06

Calculating the Square Root

Calculate \(\sqrt{0.98}\). Using a calculator, \(\sqrt{0.98} \approx 0.99\). Thus, \(p \geq 0.99\).
07

Conclusion

Since \(P(A) = p\) and \(p \geq 0.99\), we conclude that \(P(A) \geq 0.99\).

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Key Concepts

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

Equally Likely Events
In probability theory, when we say that two events, like \(A\) and \(B\), are equally likely, it means that the probability of each event occurring is the same. In mathematical terms, this is expressed as \(P(A) = P(B)\).
For example, when flipping a fair coin, getting "heads" is equally likely as getting "tails," both having a probability of \(0.5\).
Equally likely events allow us to simplify complex probability problems by assuming the same likelihood for each event.
In our exercise, as \(A\) and \(B\) are equally likely, we use a common symbol, \(p\), for their probabilities: \(P(A) = p\) and \(P(B) = p\).
This simplification helps streamline the equations when determining the intersection or union of such events.
Intersection of Events
The intersection of events in probability refers to the situation where two or more events happen at the same time.
Mathematically, for two events \(A\) and \(B\), their intersection is defined as \(P(A \cap B)\). It represents the probability that both \(A\) and \(B\) occur simultaneously.
When dealing with equally likely events and assuming independence, the probability of their intersection can be found by multiplying their individual probabilities: \(P(A \cap B) = P(A) \times P(B)\).
In our example, since \(A\) and \(B\) are equally likely, we use \(p\) for both, resulting in \(P(A \cap B) = p^2\).
Understanding intersection is crucial for solving problems where you need to find overlapping probabilities, especially when constraints, like a minimum value, are placed on the intersection.
Inequality in Probability
Inequality in probability requires us to find probabilities that satisfy a specific condition or constraint. In the given exercise, we have the inequality \(p^2 \geq 0.98\), reflecting a condition on the intersection of \(A\) and \(B\).
Solving this requires mathematical manipulation, often involving algebraic or calculus-based techniques, depending on the problem's complexity.
Here, to solve \(p^2 \geq 0.98\), take the square root of both sides. This yields \(p \geq \sqrt{0.98}\), simplifying the inequality to determine the acceptable range for \(p\).
Inequalities are common in probability, helping define constraints that ensure the validity of scenarios under specific conditions and guiding towards informed conclusions.

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Most popular questions from this chapter

A sample space consists of five simple events, \(E_{1}, E_{2}, E_{3}, E_{4},\) and \(E_{5}\) a. If \(P\left(E_{1}\right)=P\left(E_{2}\right)=0.15, P\left(E_{3}\right)=0.4,\) and \(P\left(E_{4}\right)=2 P\left(E_{5}\right),\) find the probabilities of \(E_{4}\) and \(E_{5}\) b. If \(P\left(E_{1}\right)=3 P\left(E_{2}\right)=0.3,\) find the probabilities of the remaining simple events if you know that the remaining simple events are equally probable.

A communications network has a built-in safeguard system against failures. In this system if line I fails, it is bypassed and line II is used. If line II also fails, it is bypassed and line III is used. The probability of failure of any one of these three lines is \(.01,\) and the failures of these lines are independent events. What is the probability that this system of three lines does not completely fail?

The following game was played on a popular television show. The host showed a contestant three large curtains. Behind one of the curtains was a nice prize (maybe a new car) and behind the other two curtains were worthless prizes (duds). The contestant was asked to choose one curtain. If the curtains are identified by their prizes, they could be labeled \(G, D_{1},\) and \(D_{2}\) (Good Prize, Dud1, and Dud2). Thus, the sample space for the contestants choice is \(S=\left\\{G, D_{1}, D_{2}\right\\}\). a. If the contestant has no idea which curtains hide the various prizes and selects a curtain at random, assign reasonable probabilities to the simple events and calculate the probability that the contestant selects the curtain hiding the nice prize. b. Before showing the contestant what was behind the curtain initially chosen, the game show host would open one of the curtains and show the contestant one of the duds (he could always do this because he knew the curtain hiding the good prize). He then offered the contestant the option of changing from the curtain initially selected to the other remaining unopened curtain. Which strategy maximizes the contestant's probability of winning the good prize: stay with the initial choice or switch to the other curtain? In answering the following sequence of questions, you will discover that, perhaps surprisingly, this question can be answered by considering only the sample space above and using the probabilities that you assigned to answer part (a). i. If the contestant choses to stay with her initial choice, she wins the good prize if and only if she initially chose curtain \(G\). If she stays with her initial choice, what is the probability that she wins the good prize? ii. If the host shows her one of the duds and she switches to the other unopened curtain, what will be the result if she had initially selected \(G\) ? iii. Answer the question in part.(ii) if she had initially selected one of the duds. iv. If the contestant switches from her initial choice (as the result of being shown one of the duds), what is the probability that the contestant wins the good prize? v. Which strategy maximizes the contestant's probability of winning the good prize: stay with the initial choice or switch to the other curtain?

Two events \(A\) and \(B\) are such that \(P(A)=.2, P(B)=.3,\) and \(P(A \cup B)=.4 .\) Find the following: a. \(P(A \cap B)\) b. \(P(\bar{A} \cup \bar{B})\) c. \(P(\bar{A} \cap \bar{B})\) d. \(P(\bar{A} | B)\)

An experimenter wishes to investigate the effect of three variables \(-\) pressure, temperature, and the type of catalyst-on the yield in a refining process. If the experimenter intends to use three settings each for temperature and pressure and two types of catalysts, how many experimental runs will have to be conducted if he wishes to run all possible combinations of pressure, temperature, and types of catalysts?

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