/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 A pair of dice is rolled, and th... [FREE SOLUTION] | 91Ó°ÊÓ

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A pair of dice is rolled, and the number that appears uppermost on each die is observed.refer to this experiment and find the probability of the given event. A pair of 1 s is thrown.

Short Answer

Expert verified
The probability of rolling a pair of 1s on a pair of dice is \(\frac{1}{36}\).

Step by step solution

01

Identify the total possible outcomes

When rolling a pair of dice, there are 6 sides on each die, so there are 6 x 6 = 36 possible outcomes.
02

Determine the occurrence of the desired outcome

The desired outcome is a pair of 1s, which can occur only once (1 on the first die and 1 on the second die).
03

Calculate the probability

To find the probability of rolling a pair of 1s, we need to divide the occurrence of the desired outcome by the total possible outcomes: Probability = (Desired Outcome) / (Total Possible Outcomes) In this case, the probability of rolling a pair of 1s is: Probability = 1 / 36

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Key Concepts

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

Dice Rolling
In dice games, rolling dice is one of the most fundamental actions. A standard die has six sides, numbered from 1 to 6. When you roll a die, any of these six outcomes is equally likely to occur. This means each number has a probability of \( \frac{1}{6} \) of being rolled. With multiple dice, like a pair in this exercise, the number of outcomes multiplies.
Each die acts independently of the other, which means the outcome of one die doesn't affect the outcome of the other. Thus, when rolling two dice, there are 36 combinations (6 possible outcomes for each die). These combinations include outcomes like getting a 1 on the first die and a 3 on the second, or a 4 on the first die and a 6 on the second. Hence, understanding how dice rolling works is the first step to computing probabilities in dice games.
Probability Calculation
Calculating probability involves considering both the desired outcomes and all possible outcomes of the event. Probability is often expressed as a fraction or ratio showing the likelihood of an event.
To compute the probability of a specific outcome when rolling a pair of dice, you can use the formula: \[ \text{Probability} = \frac{\text{Number of Desired Outcomes}}{\text{Total Number of Outcomes}} \]
For our exercise, if you’re calculating the probability of rolling two 1s, you count how frequently you can achieve this; there is only one such outcome (rolling a 1 on both dice). This singular event remains the numerator in our calculation, resulting in a formula: \( \frac{1}{36} \).
Thus, the skill in probability calculation lies in correctly determining your desired outcome and the total possible outcomes of your event, which is essential for answering any probability-based query.
Probability of Outcomes
The concept of the probability of outcomes is central to understanding any probabilistic event, such as dice rolling. Here, each outcome of rolling dice is part of a sample space, which is the set of all possible outcomes.
The probability of an outcome or event helps us predict how likely it is to occur. For instance, the probability of rolling a pair of 1s with two dice is calculated as \( \frac{1}{36} \), meaning this outcome is relatively rare.
Understanding probability allows you to anticipate results over multiple trials. Since probabilities need to add up to 1, each possible outcome can thereafter be quantified and anticipated. Hence, while the probability of rolling a specific pair is low, understanding it can help strategize for games or grasp the principles of randomness and chance that govern dice games and other probabilistic circumstances.

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Most popular questions from this chapter

Data compiled by the Highway Patrol Department regarding the use of seat belts by drivers in a certain area after the passage of a compulsory seat-belt law are shown in the accompanying table. $$\begin{array}{lcc} \hline & & \text { Percent of } \\ \text { Drivers } & \begin{array}{c} \text { Percent } \\ \text { of Drivers } \\ \text { in Group } \end{array} & \begin{array}{c} \text { Group Stopped } \\ \text { for Moving } \end{array} \\ \hline \text { Group I } & \text { Violation } \\ \text { (using seat belts) } & 64 & .2 \\ \hline \text { Group II } & & \\ \text { (not using seat belts) } & 36 & .5 \\ \hline \end{array}$$ If a driver in that area is stopped for a moving violation, what is the probability that he or she a. Will have a seat belt on? b. Will not have a seat belt on?

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Let \(E\) and \(F\) be mutually exclusive events and suppose \(P(F) \neq 0\). Find \(P(E \mid F)\) and interpret your result.

In a survey of 106 senior information technology and data security professionals at major U.S. companies regarding their confidence that they had detected all significant security breaches in the past year, the following responses were obtained. $$\begin{array}{lcccc} \hline & \begin{array}{c} \text { Very } \\ \text { confident } \end{array} & \begin{array}{c} \text { Moderately } \\ \text { confident } \end{array} & \begin{array}{l} \text { Not very } \\ \text { confident } \end{array} & \begin{array}{l} \text { Not at all } \\ \text { confident } \end{array} \\ \hline \text { Respondents } & 21 & 56 & 22 & 7 \\ \hline \end{array}$$ What is the probability that a respondent in the survey selected at random a. Had little or no confidence that he or she had detected all significant security breaches in the past year? b. Was very confident that he or she had detected all significant security breaches in the past year?

The estimated probability that a brand-A, a brand-B, and a brand-C plasma TV will last at least \(30,000 \mathrm{hr}\) is \(.90, .85\), and \(.80\), respectively. Of the 4500 plasma TVs that Ace TV sold in a certain year, 1000 were brand A, 1500 were brand \(\mathrm{B}\), and 2000 were brand \(\mathrm{C}\). If a plasma TV set sold by Ace TV that year is selected at random and is still working after \(30,000 \mathrm{hr}\) of use a. What is the probability that it was a brand-A TV? b. What is the probability that it was not a brand-A TV?

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