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Chapter 9: Problem 10

Imagine rolling five regular dice and looking for outcomes when all five dice match. a. How many different outcomes are possible on a roll of five dice? Explain. b. In how many of the possible outcomes do all five dice match? c. What is the probability of getting all five dice to match on a single roll? d. Suppose Tamika is given three rolls to get five matching dice. On the second and third rolls, she may roll some or all of the five dice again. On her first roll, Tamika gets three \(3 \mathrm{s}, \mathrm{a} 2\) and a \(6 .\) She picks up the dice showing 2 and 6 and rolls them again. What is the probability that she will get two more 3 \(\mathrm{s}\) on this roll?

Short Answer

Expert verified
a) 7776b) 6c) \( \frac{1}{1296} \)d) \( \frac{1}{36} \)

Step by step solution

01

Determine total outcomes for rolling five dice

Each die has 6 faces, and there are 5 dice. The total number of outcomes when rolling five dice is calculated as: \[ 6^5 = 7776 \] Thus, there are 7776 possible outcomes for rolling five dice.
02

Outcomes where all five dice match

For all five dice to match, each die must show the same number. There are 6 possible matching numbers (1 through 6). So, there are 6 outcomes where all five dice match.
03

Calculate the probability of all five dice matching

Probability is calculated by dividing the number of outcomes where all five dice match by the total number of outcomes: \[ \text{Probability} = \frac{6}{7776} = \frac{1}{1296} \] Therefore, the probability of all five dice matching is \( \frac{1}{1296} \).
04

Analyze Tamika's second roll

On her first roll, Tamika has three dice showing 3, one showing 2, and one showing 6. She rolls the dice showing 2 and 6 again. The probability that each new roll shows a 3 is independent, and the probability of rolling a 3 on a single die is \( \frac{1}{6} \). The probability that both dice show a 3 is: \[ \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \left(\frac{1}{6}\right)^2 = \frac{1}{36} \] Thus, the probability that she will get two more 3's on this roll is \( \frac{1}{36} \).

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Key Concepts

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

rolling dice probability
When we talk about rolling dice, each die has six faces, each showing a number from 1 to 6. Every time you roll a die, the result is one of these six numbers. This means each face has an equal chance of landing face up.
Thus, the probability of rolling a specific number on one die is 1 out of 6, or \( \frac{1}{6} \). When rolling multiple dice at once, each die鈥檚 result is independent of the others' results. This means the outcome of one die doesn't affect the others, making the probabilities straightforward to calculate by multiplying the probabilities of individual outcomes together.
outcome calculations
To determine the number of possible outcomes when rolling multiple dice, we expand on the concept of single-die outcomes. For example, rolling one die gives six possible outcomes. For two dice, we calculate the outcomes as \( 6 \times 6 = 36 \).
Extending this to five dice, the total outcome calculation becomes \[ 6^5 = 7776 \].
This equation shows that each die's outcome multiplies the total possible outcomes, resulting in 7776 different outcomes when rolling five dice.
probability of matching outcomes
In scenarios where we seek matching outcomes, such as all dice showing the same number, we calculate the probability by focusing on specifics. For five dice to match, say all showing 3s, only six possibilities exist (one for each face of the die).
This is a much smaller subset compared to the total possible outcomes. To find this probability, we divide the number of successful outcomes by the total possible outcomes:\[ \text{Probability} = \frac{6}{7776} = \frac{1}{1296} \]
Hence, the probability of getting all five dice to match in one roll is \frac{1}{1296}\.
independent events in probability
Independent events in probability means the outcome of one event doesn't influence the outcome of another. This principle applies well to dice rolls.
For Tamika's problem, despite her earlier rolls, each new roll's probability remains \frac{1}{6} \ for any specific number, including 3. If she needs both dice to show 3, we calculate the combined probability using multiplication for independent events:\[ \text{Probability} = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \]
Thus, the probability of both dice showing 3 on this roll is \frac{1}{36}\.

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