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A single fair die is tossed. Assign probabilities to the simple events and calculate the probabilities. \(C:\) Observe a number greater than 2

Short Answer

Expert verified
Answer: The probability of observing a number greater than 2 when tossing a single fair die is \(\frac{2}{3}\).

Step by step solution

01

Identify the number of possible outcomes

There are 6 sides on a fair die, numbered 1 to 6. So, there are 6 possible simple events which we denote by {1, 2, 3, 4, 5, 6}.
02

Assign probabilities to the simple events

Since the die is fair, every side has an equal chance of occurring. Therefore, the probability of each simple event is \(\frac{1}{6}\). So, we have the following probabilities: P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = \(\frac{1}{6}\).
03

Define the event \(C\)

The event \(C\) is defined as observing a number greater than 2, i.e., {3, 4, 5, 6}.
04

Calculate the probability of the event \(C\)

To find the probability of event \(C\), add the probabilities of the simple events that make up \(C\): {3, 4, 5, 6}. We have: P(C) = P(3) + P(4) + P(5) + P(6) = \(\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{4}{6}\).
05

Simplify the probability of the event \(C\)

Simplify the fraction \(\frac{4}{6}\) : P(C) = \(\frac{2}{3}\). The probability of observing a number greater than 2 when tossing a single fair die is \(\frac{2}{3}\).

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Key Concepts

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

Understanding Fair Die Probability
When we talk about a 'fair die', we refer to an ideal dice that has equal chances of landing on any of its faces when thrown. It's essential in understanding probability, as it provides a basis for an unbiased outcome.

A standard fair die has six faces, each marked with a different number of dots from 1 to 6. Since it is fair, each of these numbers has an equal chance of coming up, which is the core concept behind the 'fair die probability'.

In mathematical terms, this equal probability is represented as \( \frac{1}{6} \) for each face, because there are six possible outcomes and each is equally likely. Thus, when you roll a fair die, the probability of rolling any single number is always one-sixth.
Calculating Event Probability
The likelihood of an event happening in a probability experiment is referred to as the 'event probability'. When we calculate this probability, we're essentially asking, 'What are the chances of a certain event occurring?'

Event probability is calculated by dividing the number of ways an event can occur by the total number of possible outcomes. In the case of a fair die, if we want to calculate an event probability, we list all the outcomes that fulfill our event condition and then add their probabilities together.

For example, as in the given exercise, if we are looking for the probability of rolling a number greater than 2 (event C), we only consider the favorable outcomes: 3, 4, 5, and 6. We then sum the probabilities for these outcomes: \( P(C) = P(3) + P(4) + P(5) + P(6) \) which gives us a result of \( \frac{2}{3} \) after simplification.
Simple Events in Probability
A 'simple event' in probability is an event that cannot be broken down into smaller parts. In other words, it's an outcome that's completely independent from any other outcome and has a single possibility.

In a fair die throw, each number rolling out is considered a simple event because it stands alone and does not depend on other rolls. The outcomes {1}, {2}, {3}, {4}, {5}, or {6} are all examples of simple events - each with a probability of \( \frac{1}{6} \) as provided in the exercise.

Understanding simple events is crucial since more complex events can often be expressed as combinations of simple events. This idea helps when calculating the overall event probability, making it easier to manage and understand the probability of various outcomes, especially when dealing with compound events that encompass multiple simple events.

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Most popular questions from this chapter

Professional basketball is now a reality for women basketball players in the United States. There are two conferences in the WNBA, each with six teams, as shown in the following table. \(^{3}\) $$ \begin{array}{ll} \hline \text { Western Conference } & \text { Eastern Conference } \\ \hline \text { Minnesota Lynx } & \text { Atlanta Dream } \\ \text { Phoenix Mercury } & \text { Indiana Fever } \\ \text { Dallas Wings } & \text { New York Liberty } \\ \text { Los Angeles Sparks } & \text { Washington Mystics } \\ \text { Seattle Storm } & \text { Connecticut Sun } \\ \text { San Antonio Stars } & \text { Chicago Sky } \end{array} $$ Two teams, one from each conference, are randomly selected to play an exhibition game. a. How many pairs of teams can be chosen? b. What is the probability that the two teams are Los Angeles and New York? c. What is the probability that the Western Conference team is not from California?

City crime records show that \(20 \%\) of all crimes are violent and \(80 \%\) are nonviolent, involving theft, forgery, and so on. Ninety percent of violent crimes are reported versus \(70 \%\) of nonviolent crimes. a. What is the overall reporting rate for crimes in the city? b. If a crime in progress is reported to the police, what is the probability that the crime is violent? What is the probability that it is nonviolent? c. Refer to part b. If a crime in progress is reported to the police, why is it more likely that it is a nonviolent crime? Wouldn't violent crimes be more likely to be reported? Can you explain these results?

Two cold tablets are unintentionally put in a box containing two aspirin tablets, that appear to be identical. One tablet is selected at random from the box and swallowed by the first patient. The second patient selects another tablet at random and swallows it. a. List the simple events in the sample space \(S\). b. Find the probability of event \(A\), that the first patient swallowed a cold tablet. c. Find the probability of event \(B\), that exactly one of the two patients swallowed a cold tablet. d. Find the probability of event \(C,\) that neither patient swallowed a cold tablet.

A large number of adults are classified according to whether they were judged to need eyeglasses for reading and whether they actually used eyeglasses when reading. The proportions falling into the four categories are shown in the table. A single adult is selected from this group. Find the probabilities given here. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Used Eyeglasses } \\ \text { for Reading } \end{array} & \\ \hline \text { Judged to Need Eyeglasses } & \text { Yes } & \text { No } \\ \hline \text { Yes } & .44 & .14 \\ \text { No } & .02 & .40 \end{array} $$ a. The adult is judged to need eyeglasses. b. The adult needs eyeglasses for reading but does not use them. c. The adult uses eyeglasses for reading whether he or she needs them or not. d. An adult used glasses when they didn't need them.

A certain manufactured item is visually inspected by two different inspectors. When a defective item comes through the line, the probability that it gets by the first inspector is \(.1 .\) Of those that get past the first inspector, the second inspector will "miss" 5 out of \(10 .\) What fraction of the defective items get by both inspectors?

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