Problem 17
Question
Two dice are tossed. Find the proba- bility of rolling each of the following events: A sum of 6 \(\frac{5}{36}\)
Step-by-Step Solution
Verified Answer
The probability of rolling a sum of 6 is \( \frac{5}{36} \).
1Step 1: Understand the problem
We need to find the probability of rolling a sum of 6 when two dice are tossed. Each die has six faces, numbered from 1 to 6.
2Step 2: Identify possible outcomes for a single die
A single die has 6 outcomes: 1, 2, 3, 4, 5, and 6. Thus, when two dice are thrown, there are a total of 6 × 6 = 36 possible combinations of outcomes.
3Step 3: List combinations resulting in a sum of 6
We list all pairs of dice rolls that sum to 6: (1,5), (2,4), (3,3), (4,2), and (5,1). These are the combinations that give a sum of 6.
4Step 4: Count the favorable outcomes
There are 5 combinations (1,5), (2,4), (3,3), (4,2), and (5,1) that result in a sum of 6. Thus, there are 5 favorable outcomes.
5Step 5: Calculate the probability
The probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. So, the probability is given by \( \frac{5}{36} \).
Key Concepts
Dice combinationsFavorable outcomesOutcome analysis
Dice combinations
When rolling two dice, each die can land on any of its six faces. Each face is marked with a number from 1 to 6.
This results in 6 outcomes per die. Thus, when rolling two dice, the total number of possible combinations can be calculated by multiplying the number of outcomes from each die:
\( 6 \times 6 = 36 \).
These 36 combinations represent all the possible outcomes, ranging from a roll of (1,1) to (6,6).
It's important to list these combinations because they form the basis for calculating probabilities.
This results in 6 outcomes per die. Thus, when rolling two dice, the total number of possible combinations can be calculated by multiplying the number of outcomes from each die:
\( 6 \times 6 = 36 \).
These 36 combinations represent all the possible outcomes, ranging from a roll of (1,1) to (6,6).
It's important to list these combinations because they form the basis for calculating probabilities.
- The pairs (1,1), (1,2), ..., (6,6) are simply ordered outcomes.
- Each combination should be considered unique in determining specific events such as sums, product, or matching numbers.
Favorable outcomes
Favorable outcomes are the specific event occurrences that we are interested in when analyzing probability. In this exercise, we want to find combinations of two dice that sum up to 6.
Such combinations, called favorable outcomes, are crucial as they directly impact the probability calculation.
To find these outcomes:
Such combinations, called favorable outcomes, are crucial as they directly impact the probability calculation.
To find these outcomes:
- Carefully list all possible tuples – in this case, dice rolls – that meet the condition of summing to 6.
- Each pair counts as one favorable outcome.
- For the sum of 6, these pairs are (1,5), (2,4), (3,3), (4,2), and (5,1).
Outcome analysis
Outcome analysis involves evaluating both the total and the favorable outcomes to determine the likelihood of an event. For this probability exercise, we apply this concept by:
By breaking down the process and understanding each part of the analysis, we can ensure accuracy in our probability calculations. This method allows us to see how often the desired event happens compared to all possible events.
- Counting the number of combinations that meet the event condition (a sum of 6).
- Dividing the count of favorable outcomes by the total number of combinations possible from the dice rolls.
- Total combinations = 36
- Favorable combinations = 5
- Probability = \( \frac{5}{36} \)
By breaking down the process and understanding each part of the analysis, we can ensure accuracy in our probability calculations. This method allows us to see how often the desired event happens compared to all possible events.
Other exercises in this chapter
Problem 17
What is the probability of getting a sum of 8 with one roll of a pair of dice? \(\frac{5}{36}\)
View solution Problem 17
Toss a pair of dice. What is the probability of not getting a double? \(\frac{5}{6}\)
View solution Problem 17
How many three-letter permutations can be formed from the first eight letters of the alphabet (a) if repetitions are not allowed? (b) if repetitions are allowed
View solution Problem 17
In how many ways can Amy, Bob, Cindy, Dan, and Elmer be seated in a row of five seats so that neither Amy nor Bob occupies an end seat? 36
View solution