Question
Suppose you roll a standard number cube. What is the probability of rolling a number less than 3 or an odd number?(1 point) Responses 12 Start Fraction 1 over 2 End Fraction 23 Start Fraction 2 over 3 End Fraction 13 Start Fraction 1 over 3 End Fraction 56
Answers
GPT-4o mini
To find the probability of rolling a number less than 3 or an odd number on a standard six-sided die, we first identify the relevant outcomes.
The numbers on a standard number cube (1 through 6) are as follows:
- **Numbers less than 3:** 1, 2 (2 outcomes)
- **Odd numbers:** 1, 3, 5 (3 outcomes)
Next, we need to find the total outcomes for the event "rolling a number less than 3 or an odd number." We will use the principle of inclusion-exclusion to avoid double-counting.
1. **Count of numbers less than 3:**
- There are 2 outcomes: {1, 2}.
2. **Count of odd numbers:**
- There are 3 outcomes: {1, 3, 5}.
3. **Count of outcomes that are both less than 3 and odd:**
- The only outcome that fits both criteria is 1, so there is 1 outcome: {1}.
Now we apply the inclusion-exclusion principle:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Where:
- \(P(A)\) is the probability of rolling a number less than 3,
- \(P(B)\) is the probability of rolling an odd number, and
- \(P(A \cap B)\) is the probability of rolling a number that is both less than 3 and odd.
Calculating the probabilities:
\[
P(A) = \frac{2}{6} = \frac{1}{3}
\]
\[
P(B) = \frac{3}{6} = \frac{1}{2}
\]
\[
P(A \cap B) = \frac{1}{6}
\]
Now substituting back into our formula:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{6}
\]
To add these fractions, we need a common denominator. The least common multiple of 3, 2, and 6 is 6.
Converting each fraction:
\[
P(A) = \frac{1}{3} = \frac{2}{6}
\]
\[
P(B) = \frac{1}{2} = \frac{3}{6}
\]
\[
P(A \cap B) = \frac{1}{6}
\]
Now, substituting:
\[
P(A \cup B) = \frac{2}{6} + \frac{3}{6} - \frac{1}{6} = \frac{2 + 3 - 1}{6} = \frac{4}{6} = \frac{2}{3}
\]
Thus, the probability of rolling a number less than 3 or an odd number is
\[
\frac{2}{3}
\]
So the final answer is:
**Response: \(\frac{2}{3}\)**
The numbers on a standard number cube (1 through 6) are as follows:
- **Numbers less than 3:** 1, 2 (2 outcomes)
- **Odd numbers:** 1, 3, 5 (3 outcomes)
Next, we need to find the total outcomes for the event "rolling a number less than 3 or an odd number." We will use the principle of inclusion-exclusion to avoid double-counting.
1. **Count of numbers less than 3:**
- There are 2 outcomes: {1, 2}.
2. **Count of odd numbers:**
- There are 3 outcomes: {1, 3, 5}.
3. **Count of outcomes that are both less than 3 and odd:**
- The only outcome that fits both criteria is 1, so there is 1 outcome: {1}.
Now we apply the inclusion-exclusion principle:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Where:
- \(P(A)\) is the probability of rolling a number less than 3,
- \(P(B)\) is the probability of rolling an odd number, and
- \(P(A \cap B)\) is the probability of rolling a number that is both less than 3 and odd.
Calculating the probabilities:
\[
P(A) = \frac{2}{6} = \frac{1}{3}
\]
\[
P(B) = \frac{3}{6} = \frac{1}{2}
\]
\[
P(A \cap B) = \frac{1}{6}
\]
Now substituting back into our formula:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{6}
\]
To add these fractions, we need a common denominator. The least common multiple of 3, 2, and 6 is 6.
Converting each fraction:
\[
P(A) = \frac{1}{3} = \frac{2}{6}
\]
\[
P(B) = \frac{1}{2} = \frac{3}{6}
\]
\[
P(A \cap B) = \frac{1}{6}
\]
Now, substituting:
\[
P(A \cup B) = \frac{2}{6} + \frac{3}{6} - \frac{1}{6} = \frac{2 + 3 - 1}{6} = \frac{4}{6} = \frac{2}{3}
\]
Thus, the probability of rolling a number less than 3 or an odd number is
\[
\frac{2}{3}
\]
So the final answer is:
**Response: \(\frac{2}{3}\)**