/*! 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 15 A die is rolled. Find the probab... [FREE SOLUTION] | 91Ó°ÊÓ

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A die is rolled. Find the probability of getting a number greater than 4

Short Answer

Expert verified
The probability of getting a number greater than 4 when a die is rolled is \( \frac{2}{6} \) or reduced to \( \frac{1}{3} \)

Step by step solution

01

- Identify possible outcomes

Analyzing the die, there are 6 possible outcomes, (assuming it's a fair six-sided die) with the numbers 1, 2, 3, 4, 5, and 6.
02

- Identify favorable outcomes

The problem asks for the probability of getting a number greater than 4. The favorable outcomes in this case would be getting a 5 or a 6. Thus, there are 2 favorable outcomes.
03

- Calculate the Probability

Probability is calculated as the ratio of the number of favorable outcomes to the total number of outcomes. Therefore, the probability of getting a number greater than 4, denoted as P(A), is calculated as follows: \( P(A) = \frac{Number of Favorable Outcomes}{Total Number of Outcomes} = \frac{2}{6} \)

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Key Concepts

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

Probability of an Event
Understanding the probability of an event is essential not just in algebra, but also in making daily decisions. Probability refers to the likelihood that a specific event will occur. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 denotes certainty. When you roll a die, for instance, what is the chance that you'll get a number greater than 4? To find this, you evaluate all possible outcomes and the favorable ones to that event.

In calculating the probability of an event, divide the number of ways the event can happen by the total number of possible outcomes. For our die-rolling example, we figured out the probability following these exact principles. If you remember the numbers on the die (1 through 6), any single roll has six possible outcomes. To visualize these concepts better, it often helps to write down a list or create a simple diagram that depicts all possible outcomes. This visual aid can make the abstract concept of probability more concrete and easier to grasp.
Favorable Outcomes
Favorable outcomes are the heart of probability calculations. These are the outcomes which align with the event we're interested in. If you're determining the chance of rolling a number greater than 4 on a die, you're not concerned with all six faces—just the ones with 5 and 6.

It's important to accurately determine which outcomes are considered favorable, as this affects the final probability. For example, if the event was getting an even number, the favorable outcomes would be different (2, 4, and 6). So always ensure you've identified the right favorable outcomes based on the question asked. In some scenarios, it might also be useful to consider if the favorable outcomes are equally likely, as this is presumed in basic probability calculations.
Total Number of Outcomes
The total number of outcomes in a probability scenario is the complete set of all possible results that can occur. For a standard six-sided die, the total number of outcomes is six, corresponding to the six faces of the die, each showing a different number from 1 to 6. This number is crucial because it forms the denominator of the probability fraction, against which the number of favorable outcomes is compared.

To solidify the understanding of total outcomes, one might practice with various objects or scenarios, such as flipping a coin (two outcomes) or drawing a card from a standard deck (52 outcomes). It's critical to distinguish between ‘possible’ outcomes, which may occur, and ‘favorable’ outcomes, which are the ones that contribute to the event in question. Mixing up these two can lead to incorrect calculations and misunderstandings of probability.

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Most popular questions from this chapter

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Assuming the next U.S. president will be a Democrat or a Republican, the probability of a Republican president is 0.5.

Determine whether each statement makes sense or does not make sense, and explain your reasoning. The probability that I will go to graduate school is 1.5.

Follow the outline below and use mathematical induction to prove the Binomial Theorem: $$\begin{aligned}(a+b)^{n} &-\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\&+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned}$$ a. Verify the formula for \(n-1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$\begin{array}{l}(a+b)^{k+1}-\left(\begin{array}{c}k \\\0\end{array}\right)a^{k+1}+\left[\left(\begin{array}{c}k \\\0\end{array}\right)+\left(\begin{array}{c}k \\\1\end{array}\right)\right] a^{k} b \\\\+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1}\end{array}$$ e. Use the result of Exercise 84 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) $$\begin{aligned}&-\left(\begin{array}{l}n+1 \\\r+1\end{array}\right), \text { then }\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)-\left(\begin{array}{c}k+1 \\\1\end{array}\right) \text { and }\\\&\left(\begin{array}{l}k \\\1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)-\left(\begin{array}{c}k+1 \\\2\end{array}\right)\end{aligned}$$ f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)-\left(\begin{array}{c}k+1 \\ 0\end{array}\right)(\text { why? })\) and \(\left(\begin{array}{l}k \\\ k\end{array}\right)-\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

Show that the sum of the first \(n\) positive odd integers, $$1+3+5+\dots+(2 n-1)$$ is \(n^{2}\)

Solve by the method of your choice. Nine comedy acts will perform over two evenings. Five of the acts will perform on the first evening and the order in which the acts perform is important. How many ways can the schedule for the first evening be made?

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