Problem 21
Question
Two dice are thrown. The probability that the sum of numbers appearing is more than 10 is (a) \(1 / 18\) (b) \(1 / 12\) (c) \(1 / 6\) (d) None of these
Step-by-Step Solution
Verified Answer
The probability that the sum is more than 10 is \(\frac{1}{12}\), so the answer is (b).
1Step 1: Understand Total Outcomes
When two six-sided dice are thrown, each die has 6 faces. Therefore, the total number of possible outcomes is \(6 \times 6 = 36\).
2Step 2: Sum Conditions
We are interested in the outcomes where the sum of the numbers is more than 10. The possible sums greater than 10 are 11 and 12.
3Step 3: Find Favorable Outcomes for Sum 11
To find outcomes where the sum is 11, consider the pairs: \((5, 6)\) and \((6, 5)\). Thus, there are 2 favorable outcomes for a sum of 11.
4Step 4: Find Favorable Outcomes for Sum 12
To achieve a sum of 12, the only possible pair is \((6, 6)\). Thus, there is 1 favorable outcome for a sum of 12.
5Step 5: Calculate Total Favorable Outcomes
Adding the favorable outcomes for sums 11 and 12 gives a total of \(2 + 1 = 3\) favorable outcomes.
6Step 6: Calculate Probability
The probability is calculated by dividing the number of favorable outcomes by the total number of outcomes. Therefore, the probability that the sum is more than 10 is \(\frac{3}{36} = \frac{1}{12}\).
Key Concepts
Dice ProbabilitiesOutcome AnalysisSum of DiceFavorable Outcomes
Dice Probabilities
Dice probabilities help us to determine the likelihood of a particular outcome when rolling dice. When you roll two dice, each die has 6 faces with numbers ranging from 1 to 6. Thus, for a single die, the probability of any specific number is one in six or \(\frac{1}{6}\). For two dice, the probabilities evolve because there are 36 possible combinations.
Understanding dice probabilities involves knowing:
Understanding dice probabilities involves knowing:
- The total number of possible outcomes (which is 36, from \(6 \times 6\))
- The desired sum or outcome you are calculating probability for
- The number of favorable combinations for that sum or result
Outcome Analysis
Outcome analysis refers to the process of examining all possible results of throwing dice to understand how often a particular event can occur. Whenever we analyze a problem involving dice, we list all potential outcomes and see which ones meet our conditions.
Example: Rolling Two Dice
When two dice are thrown, they can produce numerous total scores ranging from 2 to 12. To conduct an outcome analysis:- Consider all pairs, such as (1,1), (1,2), ..., (6,6).
- Calculate the sum for each combination.
- Count the number of combinations that meet the criteria, like achieving a sum greater than 10.
Sum of Dice
The sum of dice refers to the combined total when the numbers on the top faces of multiple dice are added together. This concept is central to games and probability problems involving dice. Understanding which sums can arise and how frequently they appear involves a straightforward method.
Key Points: Understanding Dice Sums
- A single die can result in numbers from 1 to 6.
- When two dice are rolled, possible sums range from 2 (1+1) to 12 (6+6).
- Each sum between 2 and 12 has a different probability based on how many combinations can achieve it.
Favorable Outcomes
Favorable outcomes are the specific results that align with the conditions set by a problem. In probability, determining these outcomes is crucial for calculating how likely an event is to occur.
Example of Favorable Outcomes
Consider rolling two dice and wanting to find sums greater than 10. You identify and count the rolls yielding such sums:- For a sum of 11, possible rolls: (5,6), (6,5).
- For a sum of 12, possible roll: (6,6).
Other exercises in this chapter
Problem 19
Two dice are thrown together. The probability that sum of the two numbers will be a multiple of 4 is (a) \(1 / 9\) (b) \(1 / 3\) (c) \(1 / 4\) (d) \(5 / 9\)
View solution Problem 20
One of the two events must occur. If the chance of one is \(2 / 3\) of the other, then odds in favour of the other are (a) \(2: 3\) (b) \(1: 3\) (c) \(3: 1\) (d
View solution Problem 22
A coin is tossed twice. The probability of getting head both the times is (a) \(1 / 2\) (b) \(1 / 4\) (c) \(3 / 4\) (d) 1
View solution Problem 23
If two balanced dice are tossed once, the probability of the event that the sum of the integers coming on the upper sides of the two dice is 9 is (a) \(7 / 18\)
View solution