Problem 20
Question
Work each problem. Two dice are rolled. Find the probability of each event. (a) The sum is at least \(10 .\) (b) The sum is either 7 or at least \(10 .\) (c) The sum is 2 or the dice both show the same number.
Step-by-Step Solution
Verified Answer
(a) \(\frac{1}{6}\), (b) \(\frac{1}{3}\), (c) \(\frac{1}{6}\).
1Step 1: Total Possible Outcomes
When rolling two dice, each die has 6 faces, so the total number of possible outcomes is calculated by multiplying the number of choices for each die: 6 x 6 = 36.
2Step 2: Outcomes for Sum at Least 10
To find outcomes where the sum is at least 10, enumerate pairs (x, y) such that x + y ≥ 10. The outcomes are (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6). There are 6 outcomes.
3Step 3: Calculate Probability for Part (a)
The probability of the sum being at least 10 is given by the number of favorable outcomes divided by the total outcomes: \( \frac{6}{36} = \frac{1}{6} \).
4Step 4: Outcomes for Sum Equal to 7
List outcomes for the sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 outcomes.
5Step 5: Outcomes for Sum at Least 10 (Repeated)
Recall there are 6 outcomes: (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6) as found before.
6Step 6: Combine Outcomes for Part (b)
Combine unique outcomes from sums equal to 7 or at least 10: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (4,6), (5,5), (5,6), (6,4), (6,5), (6,6). There are 12 unique outcomes.
7Step 7: Calculate Probability for Part (b)
The probability of the sum being 7 or at least 10 is \( \frac{12}{36} = \frac{1}{3} \).
8Step 8: Outcomes for Sum Equal to 2
For the sum to be 2, the only outcome is (1,1). There is 1 outcome.
9Step 9: Outcomes for Dice Showing the Same Number
List outcomes where both dice show the same number: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). There are 6 outcomes.
10Step 10: Combine Outcomes for Part (c)
Since (1,1) is a common outcome, combine lists: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). There are 6 unique outcomes.
11Step 11: Calculate Probability for Part (c)
The probability for the sum to be 2 or both dice showing the same number is \( \frac{6}{36} = \frac{1}{6} \).
Key Concepts
dice probabilitysum of diceequally likely outcomescombinatorial probability
dice probability
Dice probability involves calculating the chances of specific outcomes when rolling one or more dice. A standard die has six faces, numbered from 1 to 6. When we roll two such dice, they lead us to a total of 36 possible outcomes, because each die has 6 potential results. Doing the math: we multiply the possibilities of each die: \(6 \times 6 = 36\).
Calculating probabilities with dice relies on understanding the ratio of favorable outcomes to the total possible outcomes. Probability can be expressed as a fraction: if there are 6 ways to achieve a desired outcome, with 36 possible outcomes in total, probability becomes \(\frac{6}{36} = \frac{1}{6}\). This basic concept lays the foundation for more complex probability calculations.
Calculating probabilities with dice relies on understanding the ratio of favorable outcomes to the total possible outcomes. Probability can be expressed as a fraction: if there are 6 ways to achieve a desired outcome, with 36 possible outcomes in total, probability becomes \(\frac{6}{36} = \frac{1}{6}\). This basic concept lays the foundation for more complex probability calculations.
sum of dice
The sum of dice is a common problem where we calculate the total of numbers shown on two or more dice. Each sum from rolling two dice can range from 2 (1+1) to 12 (6+6). Every sum has a different number of combinations to be formed.
For instance, if we're looking for a sum of 7, there are several combinations: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 ways in total to achieve this sum. Notice how each pair adds exactly to 7. Similarly, finding outcomes where the sum is at least 10 (such as 10, 11, or 12) requires listing all pairs that yield these results, which in this case are (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6).
Using sums in probability calculations allows us to explore various interesting outcomes from dice rolls.
For instance, if we're looking for a sum of 7, there are several combinations: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 ways in total to achieve this sum. Notice how each pair adds exactly to 7. Similarly, finding outcomes where the sum is at least 10 (such as 10, 11, or 12) requires listing all pairs that yield these results, which in this case are (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6).
Using sums in probability calculations allows us to explore various interesting outcomes from dice rolls.
equally likely outcomes
In probability theory, equally likely outcomes mean every single possible result has the same chance of happening. When rolling a single fair die, each face (1 through 6) is equally likely to show up, making each outcome have a probability of \(\frac{1}{6}\).
Extending this to two dice, the principle remains the same. Each of the 36 combinations from a pair of dice has an equal probability of occurrence. This is because each die does not influence the result of the other. For example, getting any specific pair like (2,3) or (6,6) is as likely as any other of the 36 outcomes due to the fairness of the dice.
Understanding equally likely outcomes is essential to accurately calculate probabilities in dice-based games and real-world scenarios.
Extending this to two dice, the principle remains the same. Each of the 36 combinations from a pair of dice has an equal probability of occurrence. This is because each die does not influence the result of the other. For example, getting any specific pair like (2,3) or (6,6) is as likely as any other of the 36 outcomes due to the fairness of the dice.
Understanding equally likely outcomes is essential to accurately calculate probabilities in dice-based games and real-world scenarios.
combinatorial probability
Combinatorial probability involves using counting methods to determine the likelihood of different outcomes. Here, we use combinations and permutations.
Let's consider finding the probability of two dice yielding results that sum to a specific number. To do this, list all possible combinations that satisfy the condition. For example, the combinations for a sum of 10 include (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6). There are 6 combinations that work.
But what if we're interested in the probability of either a sum of 2 or both dice showing the same number? In this case, outcomes like (1,1), (2,2), (3,3), etc., have to be considered. Analyze and list these combinations, allowing for quick calculation of probability: for example, \(\frac{6}{36} = \frac{1}{6}\).
Let's consider finding the probability of two dice yielding results that sum to a specific number. To do this, list all possible combinations that satisfy the condition. For example, the combinations for a sum of 10 include (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6). There are 6 combinations that work.
But what if we're interested in the probability of either a sum of 2 or both dice showing the same number? In this case, outcomes like (1,1), (2,2), (3,3), etc., have to be considered. Analyze and list these combinations, allowing for quick calculation of probability: for example, \(\frac{6}{36} = \frac{1}{6}\).
- List all possible outcomes for the desired condition
- Count them carefully to avoid duplication or missing out any
- Compare with the total number of possibilities for probability
Other exercises in this chapter
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