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Let \(\mathrm{C}\) be the collection of all integers between \(100,000\) and \(999,999\) inclusive. What is the probability that the number randomly selected from \(\mathrm{C}\) has exactly \(5\) occurrences of the digit \(3\) in the decimal representation?
Let's start by understanding what we're looking for. We have a collection C of all integers from 100,000 to 999,999 inclusive. These are all the 6-digit numbers (since 100,000 has 6 digits and 999,999 has 6 digits).
We want to find the probability that a randomly selected number from this collection has exactly 5 occurrences of the digit 3. This means that out of the 6 digit positions, exactly 5 positions must contain the digit 3, and exactly 1 position must contain some other digit (0, 1, 2, 4, 5, 6, 7, 8, or 9).
Process Skill: TRANSLATE - Converting the problem statement into clear mathematical requirements
To find the total number of 6-digit numbers in our collection, let's think about this step by step.
The smallest number is 100,000 and the largest is 999,999. So we need to count how many integers are there from 100,000 to 999,999 inclusive.
This is simply: \(999,999 - 100,000 + 1 = 900,000\)
So our total number of possible outcomes is 900,000.
Now let's think about what a favorable outcome looks like. We need exactly 5 positions to have the digit 3, and exactly 1 position to have a different digit.
Since we're dealing with 6-digit numbers, this means our number will look like: 3 3 3 3 3 X, where X is some digit that is NOT 3, and the X can be in any of the 6 positions.
For example: 333334, 333343, 333433, 343333, 433333, or 233333 are all valid patterns (though we need to be careful about the first digit since our numbers must be at least 100,000).
Let's count the favorable outcomes by considering which position gets the non-3 digit.
Case 1: The first position (leftmost) has the non-3 digit
The number looks like: X 3 3 3 3 3
Since this must be a 6-digit number (at least 100,000), the first digit X can be 1, 2, 4, 5, 6, 7, 8, or 9. It cannot be 0 (would make it a 5-digit number) or 3 (we want exactly 5 threes).
So we have 8 choices for the first position.
Case 2: Any of the other 5 positions has the non-3 digit
The number looks like: 3 X 3 3 3 3 (or any other arrangement where X is not in the first position)
For these cases, X can be any digit except 3: so X can be 0, 1, 2, 4, 5, 6, 7, 8, or 9.
That's 9 choices for each of these positions.
Since there are 5 such positions, we have \(5 \times 9 = 45\) numbers.
Total favorable outcomes = \(8 + 45 = 53\)
Process Skill: CONSIDER ALL CASES - We must separately handle the first digit due to the constraint that we need 6-digit numbers
Now we can calculate the probability using the basic probability formula:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
Probability = \(\frac{53}{900,000}\)
Looking at our answer choices, this matches option D: \(\frac{53}{900,000}\).
The probability that a randomly selected number from C has exactly 5 occurrences of the digit 3 is \(\frac{53}{900,000}\).
This corresponds to answer choice D.
Faltering Point 1: Misunderstanding the range of numbers
Students often confuse the collection C as containing all 5-digit AND 6-digit numbers, not realizing that 100,000 to 999,999 represents ONLY 6-digit numbers. This leads them to incorrectly calculate the total possible outcomes and overcomplicate the favorable outcomes counting.
Faltering Point 2: Misinterpreting "exactly 5 occurrences"
Some students might think "exactly 5 occurrences of digit 3" means the number 33333 (only 5 digits) or get confused about whether this means 5 threes with 1 other digit, or 5 threes with multiple other digits. The key insight is that we need exactly 5 positions with digit 3 and exactly 1 position with a non-3 digit in a 6-digit number.
Faltering Point 3: Overlooking the constraint for 6-digit numbers
Students may forget that the first digit cannot be 0 when counting favorable outcomes. This constraint is crucial because it affects the counting when the non-3 digit is in the first position versus other positions.
Faltering Point 1: Incorrect case analysis for the first digit
When the first position has the non-3 digit, students often incorrectly include 0 as a possibility, giving 9 choices instead of 8. They forget that including 0 in the first position would create a 5-digit number (033333), not a 6-digit number.
Faltering Point 2: Double-counting or missing cases
Students might incorrectly multiply cases or use combinations incorrectly. For example, they might think there are \(\mathrm{C}(6,1) = 6\) ways to choose which position gets the non-3 digit, and then multiply by 9 choices for each case, getting 54 total outcomes instead of properly handling the first digit constraint separately.
Faltering Point 3: Arithmetic errors in total count calculation
When calculating the total number of 6-digit numbers, students might forget to add 1 in the calculation \((999,999 - 100,000 + 1)\) or make simple arithmetic mistakes, leading to 899,999 instead of 900,000 total outcomes.
Faltering Point 1: Choosing the wrong denominator
Students who correctly find 53 favorable outcomes but incorrectly calculated the total might choose answer choice C \((\frac{53}{899,999})\) instead of the correct answer D \((\frac{53}{900,000})\). This stems from the arithmetic error in calculating total outcomes.
Faltering Point 2: Selecting unreduced or unsimplified fractions
Some students might arrive at a different numerator due to case-counting errors and then look for answer choices that seem to match their work. For example, if they calculated 54 favorable outcomes (by not properly handling the first digit constraint), they might choose C \((\frac{54}{899,999})\) thinking it matches their calculation.