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Three dice are tossed. How many simple events are in the sample space?

Short Answer

Expert verified
Answer: There are 216 simple events in the sample space when three dice are tossed.

Step by step solution

01

Determine the number of outcomes for each die

A single die has 6 faces, each with a unique number from 1 to 6. Therefore, each die can have 6 different outcomes for each toss.
02

Determine the total number of outcomes for tossing three dice

Since each die has 6 possible outcomes and there are 3 dice, we can find the total number of outcomes for all three dice by calculating the Cartesian product of the outcomes of each die. In other words, we can find all the possible combinations of outcomes by multiplying the number of outcomes for each die. So, the total number of outcomes for three dice can be found using the formula: Total outcomes = 6 (for first die) * 6 (for second die) * 6 (for third die)
03

Calculate the total number of outcomes for all three dice

Using the formula from Step 2, we can calculate the total number of outcomes for all three dice: Total outcomes = 6 * 6 * 6 Total outcomes = 216 The sample space consists of 216 simple events when three dice are tossed.

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Key Concepts

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

Sample Space
When discussing probability, the term 'sample space' refers to the set of all possible outcomes or events that can occur during an experiment. For example, when considering three dice being tossed, each die can land on any number from 1 to 6. Therefore, the sample space for one die includes six possible outcomes.

To determine the sample space for three dice, we must consider all possible outcomes for each individual die. Because each die has six sides, we use the multiplication principle to calculate the overall sample space. This principle tells us to multiply the number of possible outcomes for each individual element in the experiment.

Thus, for three dice, the sample space is the product of the outcomes for each die:
  • The first die has 6 possible outcomes.
  • The second die also has 6 possible outcomes.
  • The third die similarly has 6 possible outcomes.
This results in a sample space of 6 * 6 * 6, equaling 216 possible outcomes or simple events.
Combinatorics
Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects. It's particularly useful in understanding probability by determining the total number of possible outcomes in a given scenario.

In the context of rolling dice, combinatorics helps us calculate the number of ways we can observe different outcomes across multiple dice. Each die tossed is an independent event, with its own unique set of potential results. Using the principles of combinatorics, we multiply the number of possible outcomes for each independent event, as demonstrated with three dice:
  • First die options: 6
  • Second die options: 6
  • Third die options: 6
By applying combinatoric principles, specifically through multiplication, we obtain the total number of combinations possible — which, for three dice, is 216.
Outcomes
An outcome in probability refers to the result of a single experiment or event. When tossing dice, an outcome is the specific number that appears on the top face of a die after it lands. Each die has 6 possible outcomes, corresponding to the numbers 1 through 6.

When multiple dice are used, the total number of outcomes increases exponentially, as shown by multiplying the possibilities for each die together. This means for three dice:
  • Outcome for die one could be any of 1, 2, 3, 4, 5, 6.
  • Outcome for die two can independently be any of those numbers.
  • Similarly, outcome for die three can be any of those numbers as well.
Each combination of these numbers represents a unique outcome, and when considering three dice, the total outcome set includes 216 possibilities. This allows us to evaluate the likelihood of any specific combination occurring during a toss of these dice.

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Most popular questions from this chapter

The maximum patent life for a new drug is 17 years. Subtracting the length of time required by the FDA for testing and approval of the drug provides the actual patent life of the drug \(-\) that is, the length of time that a company has to recover research and development costs and make a profit. Suppose the distribution of the lengths of patent life for new drugs is as shown here: $$\begin{array}{l|cccccc}\text { Years, } x & 3 & 4 & 5 & 6 & 7 & 8 \\\\\hline p(x) & .03 & .05 & .07 & .10 & .14 & .20 \\\\\text { Years, } x & 9 & 10 & 11 & 12 & 13 & \\\\\hline p(x) & .18 & .12 & .07 & .03 & .01 &\end{array}$$ a. Find the expected number of years of patent life for a new drug. b. Find the standard deviation of \(x\). c. Find the probability that \(x\) falls into the interval \(\mu \pm 2 \sigma\).

Player \(A\) has entered a golf tournament but it is not certain whether player \(B\) will enter. Player \(A\) has probability \(1 / 6\) of winning the tournament if player \(B\) enters and probability \(3 / 4\) of winning if player \(B\) does not enter the tournament. If the probability that player \(B\) enters is \(1 / 3,\) find the probability that player \(A\) wins the tournament.

In Exercise 4.92 you found the probability distribution for \(x\), the number of sets required to play a best-of-five-sets match, given that the probability that \(A\) wins any one set \(-\) call this \(P(A)-\) is .6 a. Find the expected number of sets required to complete the match for \(P(A)=.6\). b. Find the expected number of sets required to complete the match when the players are of equal ability- that is, \(P(A)=.5\). c. Find the expected number of sets required to complete the match when the players differ greatly in ability - that is, say, \(P(A)=.9\). d. What is the relationship between \(P(A)\) and \(E(x),\) the expected number of sets required to complete the match?

In how many ways can you select five people from a group of eight if the order of selection is important?

A certain virus afflicted the families in three adjacent houses in a row of 12 houses. If houses were randomly chosen from a row of 12 houses, what is the probability that the three houses would be adjacent? Is there reason to believe that this virus is contagious?

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