Problem 15
Question
A die is rolled twice. Find each probability. \(P(1, \text { then any number) }\)
Step-by-Step Solution
Verified Answer
The probability is \(\frac{1}{6}.\)
1Step 1: Understand the Event
To solve for the probability of the event where a die shows a 1 on the first roll and any number on the second roll, identify the sample space for each roll of the die. A single die has 6 sides and may show any number from 1 to 6.
2Step 2: Determine the Events for Each Roll
The first roll needs to show a 1, which is only 1 possible outcome (
1
). The second roll can show any number from 1 to 6, so there are 6 possible outcomes for the second roll.
3Step 3: Calculate the Probability for Each Roll
The probability of rolling a 1 on the first roll is: \[ P( ext{roll} = 1) = \frac{1}{6} \]The probability of rolling any number on the second roll is: \[ P( ext{any number}) = 1 \] since every roll of the die results in any number from 1 to 6.
4Step 4: Calculate the Combined Probability
Since these are independent events, the combined probability is the product of both probabilities:\[ P(1, \text{ then any number}) = \frac{1}{6} \times 1 = \frac{1}{6} \]
Key Concepts
Independent EventsSample SpaceRolling DiceCombined Probability Calculation
Independent Events
When we talk about independent events in probability, we refer to events where the outcome of one event does not affect the outcome of another. For example, rolling a die two times involves independent events because the result of the first roll has no impact on the result of the second roll.
In technical terms, two events, A and B, are considered independent if the probability of both events occurring is the product of the probabilities of each event happening separately. Mathematically, this can be expressed as:
In technical terms, two events, A and B, are considered independent if the probability of both events occurring is the product of the probabilities of each event happening separately. Mathematically, this can be expressed as:
- \(P(A \text{ and } B) = P(A) \times P(B) \)
Sample Space
Sample space is a fundamental concept in probability that refers to the set of all possible outcomes of a random experiment. When you roll a single die, the sample space is all the numbers that the die can show, which are 1, 2, 3, 4, 5, and 6.
If you're doing more complex experiments, say rolling a die twice, the sample space becomes all possible combinations of results from both rolls. For each roll of the die, the sample space remains the same with six possible outcomes.
To illustrate further:
If you're doing more complex experiments, say rolling a die twice, the sample space becomes all possible combinations of results from both rolls. For each roll of the die, the sample space remains the same with six possible outcomes.
To illustrate further:
- First roll sample space: \(\{1, 2, 3, 4, 5, 6\}\)
- Second roll sample space: \(\{1, 2, 3, 4, 5, 6\}\)
Rolling Dice
Rolling dice is a common activity used to teach and understand basic probability. Each roll of a standard die has an equal likelihood of landing on any number from 1 to 6. This is what makes rolling dice a fair random event.
The probability of any single outcome, such as rolling a 1, is given by dividing the number of favorable outcomes by the total number of possible outcomes. For a single die:
The probability of any single outcome, such as rolling a 1, is given by dividing the number of favorable outcomes by the total number of possible outcomes. For a single die:
- Probability of rolling a specific number (e.g., 1): \( P(\text{specific number}) = \frac{1}{6} \)
Combined Probability Calculation
Calculating combined probabilities involves determining the likelihood of two or more events happening together. If the events are independent, as with rolling a die twice, their combined probability is found by multiplying their individual probabilities.
Using the example of rolling a die twice:
Using the example of rolling a die twice:
- The probability of rolling a 1 on the first roll is \(\frac{1}{6}\)
- The probability of rolling any number on the second roll is 1, since the second roll can result in any number from the sample space
- \(P(1, \text{ then any number}) = \frac{1}{6} \times 1 = \frac{1}{6}\)
Other exercises in this chapter
Problem 15
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