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

Step by step solution

01

Identify the sample space

The sample space contains all the possible outcomes that can occur when rolling two dice. There are 6_possible_outcomes for the first die and 6_possible_outcomes for the second die. Therefore, the total sample space has 6 * 6 = 36 possible outcomes. Some examples of possible outcomes are (1,1), (1,2), (1,3),...(6,6).
02

Count the number of ways the event can occur

Now, we need to count the number of ways the event we are interested in can occur - one die shows a 6, and the other die shows a number_less_than_3. There are two cases for this event: 1. The first die shows a 6, and the second die shows a number_less_than_3. 2. The second die shows a 6, and the first die shows a number_less_than_3. For the first case, there are 2_possible_outcomes for the second die: 1 and 2. For the second case, there are also 2_possible_outcomes for the first die: 1 and 2. So, there are 2_possible_outcomes + 2_possible_outcomes = 4_possible_outcomes for the event.
03

Calculate the probability of the event

The probability of the event can be calculated by dividing the number of ways the event can occur (found in Step 2) by the total number of_possible_outcomes in the sample space (found in Step 1): \(P(event) = \frac{number\_of\_ways\_event\_can\_occur}{total\_possible\_outcomes} = \frac{4}{36}\)
04

Simplify the probability

To simplify the probability, we divide both numbers by their greatest common divisor (GCD). The GCD of 4 and 36 is 4. Therefore, we simplify the fraction as follows: \(P(event) = \frac{4 \div 4}{36 \div 4} = \frac{1}{9}\) The probability that one die shows a 6 and the other die shows 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.

Sample Space in Dice Rolling
Understanding the sample space is the first step in solving any probability problem. It refers to the set of all possible outcomes that can occur in a given random experiment. In the exercise, the experiment involves rolling a pair of dice. Since each die has 6 faces, when you roll two dice together, each die operates independently of the other, creating a scenario where multiple outcomes can be combined to form pairs. This leads to a sample space with 36 possible outcomes (6 possible outcomes for the first die multiplied by 6 possible outcomes for the second).

These outcomes can be represented as ordered pairs, like (1,1), (1,2), (2,1), and so on, up to (6,6). It's essential to consider all the possible combinations to accurately determine the sample space. Without a complete sample space, probability calculations might be off, resulting in incorrect conclusions.
Probability Calculations
Probability calculations help us determine how likely an event is to occur within the context of all possible outcomes. The basic probability formula is simple:
\[P(event) = \frac{number\_of\_ways\_event\_can\_occur}{total\_possible\_outcomes}\]
The exercise requires calculating the probability of rolling a pair of dice where one die shows a 6 and the other shows a number less than 3. This event can occur in 4 ways (6,1), (6,2), (1,6), and (2,6). Since the total number of possible outcomes is 36, the probability of this specific event is calculated by placing the 4 favorable outcomes over the 36 possible outcomes, which simplifies to a probability of \(\frac{1}{9}\).

This calculation enables students to quantify the likelihood of an event in a dice rolling scenario. By breaking down complex events into simpler, countable outcomes, probability becomes a more straightforward, manageable concept.
Combinatorics in Probability
Combinatorics is a mathematics field that studies the counting, arrangement, and combination of objects. In the context of probability, combinatorics is fundamental for calculating various possibilities used in our probability assessment.

For example, when dealing with two dice, combinatorics can be applied to count the number of outcomes where at least one die shows a specific number, like in our exercise. The exercise improvement advice suggests clearly establishing the favorable outcomes of the dice roll and then using combinatorics - the systematic counting of those outcomes - to ensure accuracy in the probability calculations.

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Most popular questions from this chapter

According to Mediamark Research, 84 million out of 179 million adults in the United States correct their vision by using prescription eyeglasses, bifocals, or contact lenses. (Some respondents use more than one type.) What is the probability that an adult selected at random from the adult population uses corrective lenses?

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) is a subset of \(B\), then \(P(A) \leq P(B)\).

Human blood is classified by the presence or absence of three main antigens (A, B, and Rh). When a blood specimen is typed, the presence of the \(\mathrm{A}\) and/or \(\mathrm{B}\) antigen is indicated by listing the letter \(A\) and/or the letter \(B\). If neither the A nor B antigen is present, the letter \(\mathrm{O}\) is used. The presence or absence of the \(\mathrm{Rh}\) antigen is indicated by the symbols \(+\) or \(-\), respectively. Thus, if a blood specimen is classified as \(\mathrm{AB}^{+}\), it contains the \(\mathrm{A}\) and the \(\mathrm{B}\) antigens as well as the \(\mathrm{Rh}\) antigen. Similarly, \(\mathrm{O}^{-}\) blood contains none of the three antigens. Using this information, determine the sample space corresponding to the different blood groups.

In an online survey for Talbots of 1095 women ages \(35 \mathrm{yr}\) and older, the participants were asked what article of clothing women most want to fit perfectly. A summary of the results of the survey follows: $$ \begin{array}{lc} \hline \text { Article of Clothing } & \text { Respondents } \\ \hline \text { Jeans } & 470 \\ \hline \text { Black Pantsuit } & 307 \\ \hline \text { Cocktail Dress } & 230 \\ \hline \text { White Shirt } & 22 \\ \hline \text { Gown } & 11 \\ \hline \text { Other } & 55 \\ \hline \end{array} $$ If a woman who participated in the survey is chosen at random, what is the probability that she most wants a. Jeans to fit perfectly? b. A black pantsuit or a cocktail dress to fit perfectly?

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}\right\\}\) be the sample space associated with an experiment having the following probability distribution: $$ \begin{array}{lccccc} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} \\ \hline \text { Probability } & \frac{1}{14} & \frac{3}{14} & \frac{6}{14} & \frac{2}{14} & \frac{2}{14} \\ \hline \end{array} $$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{2}, s_{4}\right\\}\) b. \(B=\left\\{s_{1}, s_{5}\right\\}\) c. \(C=S\)

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