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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. One die shows a 6 , and the other is a number less than 3 .

Short Answer

Expert verified
The probability of one die showing a 6 and the other die showing a number less than 3 when rolling a pair of dice is \(\frac{1}{9}\).

Step by step solution

01

Define the sample space

For this experiment, the sample space consists of all possible outcomes of rolling two dice. Each die has 6 sides, so the sample space will contain 6 x 6 = 36 possible outcomes, represented as ordered pairs (x, y), where x is the result of the first die roll and y is the result of the second die roll.
02

Identify the desired outcomes

The desired outcomes are the situations where one die shows a 6 and the other die shows a number less than 3. There are two possible scenarios: 1) The first die shows a 6 and the second die shows a number less than 3, i.e., (6,1) or (6,2). 2) The second die shows a 6 and the first die shows a number less than 3, i.e., (1,6) or (2,6).
03

Determine the probability

The total number of outcomes in this experiment is 36 (from Step 1). The number of desired outcomes is 4, as we have the following possibilities: (6,1), (6,2), (1,6), and (2,6). To find the probability of the given event, we can use the formula: Probability = (Number of desired outcomes) / (Total number of possible outcomes) In this case, the probability is: \(P(E) = \frac{4}{36}\)
04

Simplify the fraction

To simplify the fraction, we can divide the numerator and denominator by their greatest common divisor (GCD). In this case, the GCD of 4 and 36 is 4. \(P(E) = \frac{4}{36} = \frac{4 \div 4}{36 \div 4} = \frac{1}{9}\)
05

Conclusion

The probability of one die showing a 6 and the other die showing a number less than 3 when rolling a pair of dice is \(\frac{1}{9}\).

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Key Concepts

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

Dice Probability
Dice probability involves understanding how likely different outcomes are when rolling dice. When you roll two six-sided dice, each die has possibilities numbered 1 through 6. This means, every roll of the dice is an independent event, and each number has an equal chance of appearing. Therefore, the probability for each possible result depends only on the number of ways we can achieve that result, divided by the total number of possibilities.

In this exercise, we are interested in the specific event where one die shows a 6, and the other shows a number less than 3. Understanding dice probability helps us determine the frequency of these specific outcomes. Exploring dice probability gives us insight into how random events behave and can be applied in games and real-life decision-making.
Sample Space
The concept of a sample space is crucial when dealing with probability, as it represents all possible outcomes of an experiment.

For this exercise, since we are rolling two dice, the sample space refers to all possible pairs of numbers we might see. Each die has 6 faces, so there are 6 possibilities for the first die and 6 for the second, leading to a total of 36 possible outcomes. These outcomes can be written as ordered pairs (x, y), where x and y are the numbers on each die. For example, (1,1), (1,2), ..., (6,6).

Having a clear understanding of the sample space ensures that we don't overlook any possible outcomes, allowing us to calculate probabilities accurately. Knowing how to enumerate the sample space is key when solving probability problems involving dice.
Desired Outcomes
Desired outcomes are the specific results we want to observe in a probability experiment. These are the cases that we are interested in when calculating the probability of a particular event.

For this setup with two dice, we are looking for scenarios where one die shows a 6, and the other shows a number less than 3. There are only four such possible outcomes: (6, 1), (6, 2), (1, 6), and (2, 6). These outcomes match our condition perfectly. It’s essential to correctly identify these desired outcomes as they directly influence the probability calculation.

Highlighting the importance of desired outcomes is necessary because making any mistakes here can lead to incorrect probability calculations. They form the basis for any further probability computations.
Probability Calculation
Probability calculation involves determining the likelihood of a specific event occurring within the sample space. The probability of an event can be found using the formula:
  • Probability = (Number of desired outcomes) / (Total number of possible outcomes)
Applying this formula, we first identified that there are 4 desired outcomes (as determined in the previous section) and 36 total possible outcomes (from the sample space). Thus, the initial probability is given by:
  • \(P(E) = \frac{4}{36}\)
To simplify the probability, we need to divide the numerator and the denominator by their greatest common divisor (GCD), which is 4 in this case. This reduces the probability to:
  • \(P(E) = \frac{1}{9}\)
This means, there is a 1 in 9 chance for one die to show a 6 and the other to show a number less than 3 when rolling a pair of dice. This straightforward method of calculating probabilities provides a foundational understanding that can be applied to a variety of situations involving random events.

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Most popular questions from this chapter

In a poll conducted in 2007,2000 adults ages 18 yr and older were asked how frequently they are in touch with their parents by phone. The results of the poll are as follows: $$\begin{array}{lccccc} \hline \text { Answer } & \text { Monthly } & \text { Weekly } & \text { Daily } & \text { Don't know } & \text { Less } \\ \hline \text { Respondents, } \% & 11 & 47 & 32 & 2 & 8 \\ \hline \end{array}$$ If a person who participated in the poll is selected at random, what is the probability that the person said he or she kept in touch with his or her parents a. Once a week? b. At least once a week?

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