/*! 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 25 Find the probability for the exp... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the probability for the experiment of tossing a six-sided die twice. The sum is 6.

Short Answer

Expert verified
The probability of the sum being 6 when rolling a die twice is \( \frac{5}{36}\)

Step by step solution

01

Determine all possibilities

This exercise involves throwing a six-sided die. There are 6 possibilities for each die, so there are \(6 * 6 = 36\) total outcomes when throwing both dice.
02

Find the outcomes that sum to 6

We need to identify all the outcomes that result in a sum of 6. They are (1,5), (2,4), (3,3), (4,2) and (5,1) . So, there are 5 outcomes that result in a sum of 6.
03

Calculate the probability

The probability is the favorable outcomes divided by the total number of outcomes. Therefore, the probability of the sum being 6 when rolling a die twice is \(P(6) = \frac{numberOfFavorableOutcomes}{totalNumberofOutcomes}\) . Substituting values, we get \( P(6) = \frac{5}{36}\).

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Key Concepts

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

Favorable Outcomes
Favorable outcomes refer to the specific results that satisfy the condition we're interested in, in a probability experiment. Here, we are rolling two dice and are interested in the outcomes where their sum equals 6. Each possible pair that sums to 6 is a favorable outcome. Let's look at these pairs:
  • (1,5)
  • (2,4)
  • (3,3)
  • (4,2)
  • (5,1)
These five pairs are the only combinations, with numbers from 1 to 6 on each die, that yield a total of 6 when added together. Therefore, there are 5 favorable outcomes in this scenario.
Total Outcomes
Total outcomes encompass all the possible results that can occur when an experiment is conducted. In the case of rolling two six-sided dice, each die has 6 possible faces. When working out the total number of outcomes, you multiply the possibilities for the first die with the possibilities for the second. Thus, there are:
  • 6 possibilities from the first die
  • 6 possibilities from the second die
This multiplication results in a total of 36 outcomes, because each of the 6 faces can pair with any of the 6 faces of the second die. Understanding the total number of outcomes is key when calculating probabilities, as it serves as the denominator in our probability formula.
Sum of Two Dice
The sum of two dice is simply the combined total of the numbers showing on each die after they are rolled. When calculating probability based on the sum of two dice, you're focusing on specific numerical results that can be obtained. For example, when interested in the sum being 6, you need to consider which pairs of numbers from the first and second die add up to 6. With six possible numbers on each die, the sums can vary from 2 (1+1) to 12 (6+6). In our exercise, the sum of 6 is created by the combinations (1,5), (2,4), (3,3), (4,2), and (5,1). Each pairing aligns with our favorable outcomes, making these combinations critical in determining the probability of rolling a sum of 6. The clarity in knowing which pairs result in the desired sum facilitates more straightforward calculation of probabilities.

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Most popular questions from this chapter

Consider a group of \(n\) people. (a) Explain why the following pattern gives the probabilities that the \(n\) people have distinct birthdays. $$\begin{array}{l}{n=2 : \frac{365}{365} \cdot \frac{364}{365}=\frac{365 \cdot 364}{365^{2}}} \\ {n=3 : \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}=\frac{365 \cdot 364 \cdot 363}{365^{3}}}\end{array}$$ (b) Use the pattern in part (a) to write an expression for the probability that \(n=4\) people have distinct birthdays. (c) Let \(P_{n}\) be the probability that the \(n\) people have distinct birthdays. Verify that this probability can be obtained recursively by $$P_{1}=1\( and \)P_{n}=\frac{365-(n-1)}{365} P_{n-1}$$ (d) Explain why \(Q_{n}=1-P_{n}\) gives the probability that at least two people in a group of \(n\) people have the same birthday. (e) Use the results of parts (c) and (d) to complete the table. (f) How many people must be in a group so that the probability of at least two of them having the same birthday is greater than \(\frac{1}{2} ?\) Explain.

Graphical Reasoning In Exercises 83 and \(84,\) use a graphing utility to graph \(f\) and \(g\) in the same viewing window. What is the relationship between the two graphs? Use the Binomial Theorem to write the polynomial function \(g\) in standard form. $$f(x)=x^{3}-4 x, \quad g(x)=f(x+4)$$

Linear Model, Quadratic Model, or Neither? In Exercises \(61-68\) , write the first six terms of the sequence beginning with the given term. Then calculate the first and second differences of the sequence. State whether the sequence has a perfect linear model, a perfect quadratic model, or neither. $$a_{1}=2$$ $$a_{n}=a_{n-1}+2$$

Simplifying a Difference Quotient In Exercises \(67-72\) , simplify the difference quotient, using the Binomial Theorem if necessary. $$\frac{f(x+h)-f(x)}{b} \quad$$ Difference quotient $$f(x)=x^{8}$$

Forming Rows of Pascal's Triangle Form rows \(8-10\) of Pascal's Triangle.

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