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Five fair dice are rolled. What is the probability that the faces showing constitute a "full house"-that is, three faces show one number and two faces show a second number?

Short Answer

Expert verified
The probability of getting a 'full house' when five fair dice are rolled is \(\frac{15 * 10}{6^5}\)

Step by step solution

01

Determine total outcomes

First, determine the total number of outcomes when five dice are rolled. All the dice are independent of each other and each dice can result in one of six outcomes (1,2,3,4,5,6). Thus, the total number of outcomes is \(6^5\).
02

Calculate 'full house' outcomes

Next, calculate the number of ways you can get a 'full house'. This can be broken into parts: 1) the number of ways you can choose two distinct numbers from six (which is \( \binom{6}{2} = 15\)), 2) the number of ways to distribute the five dice into two groups of three and two (which is \( \binom{5}{3} = 10 \)). Multiply these together to get the total number of 'full house' outcomes.
03

Compute the probability

Now, the probability is the ratio of the 'full house' outcomes to the total outcomes. Using formulas from steps 1 and 2, divide the 'full house' outcomes by the total outcomes.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arranging, and combining objects. In the context of rolling five dice to get a full house, combinatorics helps to determine the number of possible arrangements of dice faces. Understanding the basics of combinatorics is crucial to solve this problem because it allows us to count the different ways in which the dice can show a full house.

For instance, when we roll five dice, the number of total possible outcomes is dependent on the principle of counting which in combinatorial terms is expressed as raising the number of options (6 faces per die) to the power of the number of dice (5 dice), giving us a total of \(6^5\) outcomes.
Binomial Coefficient
The binomial coefficient, often read as 'n choose k', is a fundamental concept in combinatorics represented by \(\binom{n}{k}\). It is used to calculate the number of ways to choose k items from a set of n distinct items without considering the order. In the full house problem, this concept is applied to select the two distinct numbers from the six possible numbers on a die.Calculating the binomial coefficient for our textbook problem involves two parts: choosing the two numbers that will appear in our full house (\(\binom{6}{2}\) ways) and distributing the dice to show these numbers (\(\binom{5}{3}\) ways for selecting three dice for one number).
Probability Outcomes
Probability outcomes measure the likelihood of a particular event occurring. This is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. To find the probability of a full house in our five dice roll, we must first identify the favorable outcomes—those that show three of one number and two of another.

By using combinatorics and the binomial coefficient, we determine the number of favorable outcomes and then compute the probability by dividing the favorable outcomes by the total possible outcomes, resulting in a specific chance to achieve a full house when rolling five dice.
Independent Events
Independent events are those whose outcomes do not affect each other. In the case of rolling dice, each roll is independent; the outcome of one die does not impact the outcomes of the other dice. This is why we are able to raise the number of possible outcomes for one die to the power of the number of dice being rolled to determine the total number of outcomes for all five dice.

Understanding the principle of independent events assures us that the probability of rolling any number on one die is always \(\frac{1}{6}\), regardless of what has been rolled on the other dice. This independence is a key assumption in calculating probability outcomes in many statistical analyses and is particularly important in our full house example.

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Most popular questions from this chapter

An automobile insurance company has compiled the information summarized below on its policy-holders. Suppose someone calls to file a claim. To which age group does he or she most likely belong? $$ \begin{array}{lcc} \hline \text { Age Group } & \begin{array}{c} \% \text { of } \\ \text { Policyholders } \end{array} & \begin{array}{c} \% \text { Involved in } \\ \text { Accidents Last Year } \end{array} \\ \hline \text { Young }(<30) & 20 & 35 \\ \text { Middle-aged }(30-64) & 50 & 15 \\ \text { Elderly }(65+) & 30 & 25 \\ \hline \end{array} $$

A man has \(n\) keys on a key ring, one of which opens the door to his apartment. Having celebrated a bit too much one evening, he returns home only to find himself unable to distinguish one key from another. Resourceful, he works out a fiendishly clever plan: He will choose a key at random and try it. If it fails to open the door, he will discard it and choose at random one of the remaining \(n-1\) keys, and so on. Clearly, the probability that he gains entrance with the first key he selects is \(1 / n\). Show that the probability the door opens with the third key he tries is also \(1 / n\). (Hint: What has to happen before he even gets to the third key?)

A toy manufacturer buys ball bearings from three different suppliers \(-50 \%\) of her total order comes from supplier \(1,30 \%\) from supplier 2, and the rest from supplier \(3 .\) Past experience has shown that the quality-control standards of the three suppliers are not all the same. Two percent of the ball bearings produced by supplier 1 are defective, while suppliers 2 and 3 produce defective bearings \(3 \%\) and \(4 \%\) of the time, respectively. What proportion of the ball bearings in the toy manufacturer's inventory are defective?

Your statistics teacher announces a twenty-page reading assignment on Monday that is to be finished by Thursday morning. You intend to read the first \(x_{1}\) pages Monday, the next \(x_{2}\) pages Tuesday, and the final \(x_{3}\) pages Wednesday, where \(x_{1}+x_{2}+x_{3}=20\), and each \(x_{i} \geq 1\). In how many ways can you complete the assignment? That is, how many different sets of values can be chosen for \(x_{1}\), \(x_{2}\), and \(x_{3}\) ?

A pair of fair dice are rolled until the first sum of 8 appears. What is the probability that a sum of 7 does not precede that first sum of 8 ?

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