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Chapter 2: Problem 24

If two dice are rolled, what is the probability that the sum of the upturned faces equals \(i ?\) Find it for \(i=\) \(2,3, \ldots, 11,12\)

Short Answer

Expert verified
For each \(i \ (2 \leq i \leq 12) \), the probabilities are: \(P(2) = \frac{1}{36}\), \(P(3) = \frac{2}{36}\), \(P(4) = \frac{3}{36}\), \(P(5) = \frac{4}{36}\), \(P(6) = \frac{5}{36}\), \(P(7) = \frac{6}{36}\), \(P(8) = \frac{5}{36}\), \(P(9) = \frac{4}{36}\), \(P(10) = \frac{3}{36}\), \(P(11) = \frac{2}{36}\), \(P(12) = \frac{1}{36}\).

Step by step solution

01

Identify the sample space

We know that a regular dice has 6 faces with numbers 1 to 6. When rolling two dice, there are a total of 6 * 6 = 36 possible outcomes since each outcome on the first die can be paired with each outcome on the second die. These outcomes can be represented as ordered pairs (a, b), where a is the number on the first die and b is the number on the second die.
02

Determine favorable outcomes for each value

Now, for each given value of \(i \ (2 \leq i \leq 12) \), we will list all the pairs of numbers on the two dice whose sum equals i. These pairs, or favorable outcomes, will help us calculate the probability for each i.
03

Calculate the probability for each value

To find the probability of each given value i, we will divide the number of favorable outcomes (pairs whose sum equals i) by the total possible outcomes in the sample space (36). The probabilities for each value are: i = 2: Favorable outcomes: (1, 1) Probability = Number of favorable outcomes / Total possible outcomes = 1/36 i = 3: Favorable outcomes: (1, 2), (2, 1) Probability = 2/36 i = 4: Favorable outcomes: (1, 3), (2, 2), (3, 1) Probability = 3/36 i = 5: Favorable outcomes: (1, 4), (2, 3), (3, 2), (4, 1) Probability = 4/36 i = 6: Favorable outcomes: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) Probability = 5/36 i = 7: Favorable outcomes: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) Probability = 6/36 i = 8: Favorable outcomes: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) Probability = 5/36 i = 9: Favorable outcomes: (3, 6), (4, 5), (5, 4), (6, 3) Probability = 4/36 i = 10: Favorable outcomes: (4, 6), (5, 5), (6, 4) Probability = 3/36 i = 11: Favorable outcomes: (5, 6), (6, 5) Probability = 2/36 i = 12: Favorable outcomes: (6, 6) Probability = 1/36 It is important to note that you can simplify these probabilities to their lowest terms, but for consistency, we have kept the denominator as 36.

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Key Concepts

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

Sample Space Calculation
Understanding the sample space is crucial when calculating probabilities in dice games. It refers to the set of all possible outcomes that can occur. For a single six-sided dice, there are 6 outcomes, namely 1 through 6. But when rolling two dice, each die operates independently, meaning the outcomes of one don't affect the other. So, for two dice, you calculate the sample space by multiplying the number of outcomes for the first die by the number of outcomes for the second one.

Thus, the sample space, which can be visualized as a 6x6 grid, comprises 36 possibilities (6 for the first die multiplied by 6 for the second), represented by ordered pairs \( (a, b) \), where \( a \) and \( b \) are the numbers from the upturned faces of the first and second dice, respectively. These pairs can range from \( (1, 1) \) to \( (6, 6) \) and cover every combination in between.
Favorable Outcomes
When it comes to finding the probability of a specific sum with two dice, we need to consider the favorable outcomes. These are the outcomes that meet our criteria 鈥� in this case, the sum of the two dice equaling a certain number, \( i \).

Determining Favorable Outcomes

To determine these, we list all the pairs that add up to the target sum. For instance, if we're looking for a sum of seven, the favorable outcomes are all the dice pairs that add up to seven: \( (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \). By identifying all such pairs for each sum, we can quantify our chances of rolling that sum.
Combinatorial Probability
Combinatorial probability involves counting and calculating the likelihood of events that involve combining items or numbers. In dice probability, we use this concept to find how likely we are to get a particular sum of the numbers on the dice faces.

To figure out the probability, we divide the number of favorable outcomes 鈥� combinations that yield our target sum 鈥� by the total number of potential outcomes in our sample space. For example, a sum of 7 has 6 favorable outcomes \( (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \), so the probability of rolling a sum of 7 is \( \frac{6}{36} \), which simplifies to \( \frac{1}{6} \). It's this approach that allows us to calculate the probabilities for each possible sum when rolling two dice, laying out the essentials of how likely certain rolls are.

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Most popular questions from this chapter

A hospital administrator codes incoming patients suffering gunshot wounds according to whether they have insurance (coding 1 if they do and 0 if they do not) and according to their condition, which is rated as good (g), fair (f), or serious (s). Consider an experiment that consists of the coding of such a patient. (a) Give the sample space of this experiment. (b) Let \(A\) be the event that the patient is in serious condition. Specify the outcomes in \(A\) (c) Let \(B\) be the event that the patient is uninsured. Specify the outcomes in \(B\). (d) Give all the outcomes in the event \(B^{c} \cup A\).

Five people, designated as \(A, B, C, D, E,\) are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that (a) there is exactly one person between \(A\) and \(B ?\) (b) there are exactly two people between \(A\) and \(B ?\) (c) there are three people between \(A\) and \(B ?\)

An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. The classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 16 in the German class. There are 12 students who are in both Spanish and French, 4 who are in both Spanish and German, and 6 who are in both French and German. In addition, there are 2 students taking all 3 classes. (a) If a student is chosen randomly, what is the probability that he or she is not in any of the language classes? (b) If a student is chosen randomly, what is the probability that he or she is taking exactly one language class? (c) If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?

How many people have to be in a room in order that the probability that at least two of them celebrate their birthday in the same month is at least \(\frac{1}{2} ?\) Assume that all possible monthly outcomes are equally likely.

If a die is rolled 4 times, what is the probability that 6 comes up at least once?
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