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Chapter 1: Problem 20

Three dice are thrown. What is the probability the same number appears on exactly two of the three dice?

Short Answer

Expert verified
The probability that exactly two of the three dice show the same number is \(\frac{5}{12}\).

Step by step solution

01

Calculate the Total Possibilities

First, we need to find the number of possible outcomes. In each die, there are 6 possibilities (the numbers 1 to 6). Since there are three dice, the total number of outcomes is: Total Possible Outcomes = 6 (possible numbers on die 1) * 6 (possible numbers on die 2) * 6 (possible numbers on die 3) = \(6^3\)
02

Calculate the Favorable Outcomes

To have exactly two dice with the same number, we must ensure that the third die shows a different number. We will calculate the favorable outcomes in the following manner: 1. Choose one pair of dice to have the same number. 2. Choose a number to appear on the chosen pair. 3. Choose a number for the third die which is different from the chosen pair. We can calculate the number of ways we can do each step: 1. Choose one pair of dice: There are 3 pairs possible {Die1, Die2}, {Die1, Die3}, {Die2, Die3}. 2. Choose a number to appear on the chosen pair: There are 6 possible numbers (1 to 6). 3. Choose a number for the third die which is different from the chosen pair: As the chosen pair has only one number, we have 5 possibilities for the third die. Now, to calculate the total favorable outcomes, we multiply the number of ways for each step: Favorable outcomes = 3 (possible pairs) * 6 (possible numbers in the pair) * 5 (possible numbers for the third die)
03

Calculate the Probability

Now that we have the Total Possible Outcomes and Favorable Outcomes, we can calculate the probability as follows: Probability = Favorable Outcomes / Total Possible Outcomes Substituting the values we calculated: Probability = (3 * 6 * 5) / (\(6^3\))
04

Simplify the Probability

Now, let's simplify the probability: Probability = (3 * 6 * 5) / (\(6^3\)) = 90 / 216 The probability can be further simplified: Probability = 90/216 = 5/12 So, the probability that exactly two of the three dice show the same number is 5/12.

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Key Concepts

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

Combinatorics
Combinatorics is a key area in mathematics that involves counting, arrangement, and combination of elements within a set. When dealing with problems like the one involving dice, combinatorics helps to determine the number of possible outcomes and favorable outcomes.

In this problem, we are particularly interested in counting the number of ways we can get two dice to show the same number and the third one to show a different number.
  • We start by choosing a pair of dice that will show the same number. There are 3 different pairs possible: \({D1, D2}, {D1, D3}, {D2, D3}\).
  • Next, we choose a number for this pair. Since a die has 6 sides, there are 6 choices (numbers 1 through 6).
  • Finally, for the remaining die, which should show a different number from the chosen pair, we have just 5 choices left for numbers.

So, by multiplying the number of ways to choose each of these steps, combinatorics helps us find the total number of favorable outcomes.
Discrete Probability
Discrete probability is a branch of probability theory that deals with discrete random variables. These variables take finite or countable values, like the numbers shown on a die.

In our dice problem, we calculate discrete probability by examining all possible outcomes and identifying those that meet our criteria.
  • Total possible outcomes refer to all combinations that can occur when three dice are thrown. Since each die has 6 sides, there are \(6^3\) total combinations, which gives us 216 outcomes.
  • Favorable outcomes are the scenarios that meet our criteria: exactly two of the three dice show the same number. From combinatorics, we calculated that this number is 90.
  • To find the probability, we use the formula: \(\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Possible Outcomes}}\), yielding a simplified probability of \(\frac{5}{12}\).

Discrete probability, therefore, quantifies the likelihood of a specific event within a finite sample space.
Random Experiments
Random experiments are processes or actions that lead to one of several possible outcomes, where the results cannot be predicted with certainty in advance. Rolling dice is a classic example of a random experiment.

When we roll three dice, we perform a random experiment with multiple trials, each resulting in a number between 1 and 6. Each die roll is independent of the others, meaning the result of one does not affect the others.
  • These random experiments generate a sample space of outcomes, which is vital for counting possibilities and determining probabilities.
  • In this scenario, we're interested in the likelihood of a specific outcome: two dice showing the same number and the third showing a different number, among many other potential outcomes.
  • The uncertainty inherent in each die roll, and the independence of each trial, add complexity to calculating probabilities, helping us understand the nature of chance in random experiments.

Thus, random experiments underlie many probability problems, illustrating how chance works in uncertain scenarios.

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Most popular questions from this chapter

In an election, candidate \(A\) receives \(n\) votes and candidate \(B\) receives \(m\) votes, where \(n>m\). Assume that in the count of the votes all possible orderings of the \(n+m\) votes are equally likely. Let \(P_{n, m}\) denote the probability that from the first vote on \(A\) is always in the lead, Find (a) \(P_{2,1}\) (b) \(P_{3,1}\) (c) \(P_{n, 1}\) (d) \(P_{3,2}\) (e) \(P_{4,2}\) (f) \(P_{n, 2}\) (g) \(P_{4,3}\) (h) \(P_{5,3}\) (i) \(P_{5,4}\) (j) Make a conjecture as to the value of \(P_{n, m}\).

In a class there are four freshman boys, six freshman girls, and six sophomore boys. How many sophomore girls must be present if sex and class are to be independent when a student is selected at random?

Argue that \(E=E F \cup E F^{c}, E \cup F=E \cup F E^{c}\)

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. Now what is the probability that it is the fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?

Urn 1 has five white and seven black balls. Urn 2 has three white and twelve black balls. We flip a fair coin. If the outcome is heads, then a ball from urn 1 is selected, while if the outcome is tails, then a ball from urn 2 is selected. Suppose that a white ball is selected. What is the probability that the coin landed tails?
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