Apps My Blue Hen Home Carvas Math QuizletHandshake, Pat Math 114 College Math and Statistics – Thompson – Fall 2020 Homework: Math 114 Week 12 Homework Score: 0 of 1 pt 8.5.1 Suppose a single fair die is rolled. Find the probability that it is a 1, given that it is an even number. The probability is (Type an integer or a simplified fraction.) Enter your answer in the answer box and then click Check Answer, All parts showing
The Correct Answer and Explanation is:
To solve this probability problem, we need to apply conditional probability.
Given:
- A single fair die is rolled.
- We are told the outcome is an even number.
- We need to find the probability that it is a 1, given this condition.
Step 1: Understand the Sample Space
A fair six-sided die has the numbers:
1, 2, 3, 4, 5, 6
The even numbers among these are:
2, 4, 6
So, the new sample space (since the condition is that the result is even) is:
{2, 4, 6}
There are 3 outcomes in this conditional sample space.
Step 2: Define the Event of Interest
We are asked to find the probability that the number is 1, given that the number is even.
But looking at the even numbers (2, 4, 6), we see:
1 is not an even number, and therefore it is not in the new sample space.
That means:
It is impossible to roll a 1 if we already know the number is even.
So, the event “the number is 1” cannot occur under the condition that the number is even.
Step 3: Write the Conditional Probability
The formula for conditional probability is: P(A∣B)=P(A∩B)P(B)P(A \mid B) = \frac{P(A \cap B)}{P(B)}
Let:
- A = “number is 1”
- B = “number is even”
Since 1 is not even, A∩B=∅A \cap B = \emptyset, so: P(A∩B)=0P(A \cap B) = 0 P(B)=36=12P(B) = \frac{3}{6} = \frac{1}{2}
So: P(1∣even)=01/2=0P(\text{1} \mid \text{even}) = \frac{0}{1/2} = 0
Final Answer:
0\boxed{0}
300-Word Explanation:
This problem is an example of conditional probability, where we are given that one event has occurred and asked to find the probability of another event happening under that condition. In this case, we are told that a fair die is rolled and the outcome is an even number. The even numbers on a standard six-sided die are 2, 4, and 6. This means we now only consider these numbers when calculating probabilities.
We are then asked to find the probability that the number rolled is a 1, given that we already know it is an even number. Since the number 1 is not even, and is not part of the subset {2, 4, 6}, it is not possible for the outcome to be a 1 under the given condition. When an event cannot happen within the restricted sample space, the probability of that event is 0.
This situation shows why understanding the sample space is critical when dealing with conditional probabilities. The conditional probability formula is: P(A∣B)=P(A∩B)P(B)P(A \mid B) = \frac{P(A \cap B)}{P(B)}
Here, P(A∩B)P(A \cap B) represents the probability that both events A and B occur simultaneously. In our case, “rolling a 1” and “rolling an even number” cannot both happen at the same time, so their intersection is empty and the probability is 0. Dividing 0 by any non-zero probability (such as P(B)=0.5P(B) = 0.5) still gives 0.
Therefore, the final answer is 0, meaning it is impossible to roll a 1 if we already know the outcome is even.