To solve the problem, we will analyze the two quantities step by step.
### Given:
- Set of integers: {7, 14, 25, 27, 33, 29, 30}
- Quantity I: Probability that the average of 12, 9, and M is at least 17.
- Quantity II: \( \frac{1}{3} \)
### Step 1: Calculate the condition for the average
The average of the numbers 12, 9, and M is given by:
\[
\text{Average} = \frac{12 + 9 + M}{3}
\]
We want this average to be at least 17:
\[
\frac{12 + 9 + M}{3} \geq 17
\]
### Step 2: Multiply both sides by 3
To eliminate the fraction, multiply both sides by 3:
\[
12 + 9 + M \geq 51
\]
### Step 3: Simplify the equation
Combine the constants on the left side:
\[
21 + M \geq 51
\]
### Step 4: Isolate M
To find the minimum value of M, subtract 21 from both sides:
\[
M \geq 51 - 21
\]
\[
M \geq 30
\]
### Step 5: Identify valid values of M
Now we need to check which values of M from the set {7, 14, 25, 27, 33, 29, 30} satisfy this condition:
- The values that satisfy \( M \geq 30 \) are: 30 and 33.
### Step 6: Count the favorable outcomes
The favorable outcomes for M are:
- 30
- 33
Thus, there are 2 favorable outcomes.
### Step 7: Calculate total outcomes
The total number of outcomes in the set is 7 (since there are 7 numbers in the set).
### Step 8: Calculate the probability
The probability that M meets the condition is given by:
\[
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{7}
\]
### Step 9: Compare with Quantity II
Now we compare Quantity I with Quantity II:
- Quantity I: \( \frac{2}{7} \)
- Quantity II: \( \frac{1}{3} \)
To compare \( \frac{2}{7} \) and \( \frac{1}{3} \), we can find a common denominator or cross-multiply:
- \( 2 \times 3 = 6 \)
- \( 1 \times 7 = 7 \)
Since \( 6 < 7 \), we have:
\[
\frac{2}{7} < \frac{1}{3}
\]
### Conclusion
Thus, we conclude that:
- Quantity I is less than Quantity II.
### Final Answer
The answer is: Quantity I is less than Quantity II.
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