Key Concepts and Formulas

1. Probability Definition: The probability of an event $E$ is the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space. $P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$
2. Fundamental Principle of Counting (Multiplication Rule): If an event can occur in $n_1$ ways, and a subsequent event can occur in $n_2$ ways, and so on, then the total number of ways for all events to occur in sequence is $n_1 \times n_2 \times \dots \times n_k$. This principle is essential for calculating the size of our sample space and the number of favorable outcomes.
3. "At Least" Concept: The phrase "at least two digits are odd" means we need to consider all scenarios where the number of odd digits is two or more. For a 3-digit number, this translates to two distinct, mutually exclusive cases:
   • Exactly two digits are odd.
   • Exactly three digits are odd. The total number of favorable outcomes will be the sum of the outcomes from these individual cases.

---

Step-by-Step Solution

We are tasked with finding the probability that a randomly selected 3-digit number has at least two odd digits. Let's define the characteristics of the digits first:

• 3-digit number: A number ranging from 100 to 999. This implies the hundreds digit (first digit) cannot be 0.
• Odd digits: {1, 3, 5, 7, 9}. There are 5 choices for an odd digit.
• Even digits: {0, 2, 4, 6, 8}. There are 5 choices for an even digit.

Let the 3-digit number be represented as $d_1 d_2 d_3$, where $d_1$ is the hundreds digit, $d_2$ is the tens digit, and $d_3$ is the units digit.

Step 1: Determine the Total Number of Possible Outcomes (Sample Space)

We first need to count all possible 3-digit numbers.

• For $d_1$ (hundreds digit): It cannot be 0. So, $d_1$ can be any digit from 1 to 9.
  • Number of choices for $d_1 = 9$ (i.e., {1, 2, 3, 4, 5, 6, 7, 8, 9}).
• For $d_2$ (tens digit): It can be any digit from 0 to 9.
  • Number of choices for $d_2 = 10$ (i.e., {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}).
• For $d_3$ (units digit): It can also be any digit from 0 to 9.
  • Number of choices for $d_3 = 10$ (i.e., {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}).

Using the Multiplication Rule, the total number of distinct 3-digit numbers is: $\text{Total Outcomes} = (\text{Choices for } d_1) \times (\text{Choices for } d_2) \times (\text{Choices for } d_3)$ $\text{Total Outcomes} = 9 \times 10 \times 10 = 900$

Step 2: Determine the Number of Favorable Outcomes

We need to count 3-digit numbers with "at least two odd digits." This breaks down into two mutually exclusive cases:

Case A: Exactly three digits are odd (Odd, Odd, Odd) In this case, all three digits ($d_1, d_2, d_3$) must be odd.

• For $d_1$ (hundreds digit): Must be odd. Since $d_1$ cannot be 0, and all odd digits (1, 3, 5, 7, 9) are non-zero, all 5 odd choices are valid.
  • Number of choices for $d_1 = 5$ (i.e., {1, 3, 5, 7, 9}).
• For $d_2$ (tens digit): Must be odd.
  • Number of choices for $d_2 = 5$ (i.e., {1, 3, 5, 7, 9}).
• For $d_3$ (units digit): Must be odd.
  • Number of choices for $d_3 = 5$ (i.e., {1, 3, 5, 7, 9}).

Number of outcomes for Case A = $5 \times 5 \times 5 = 125$.

Case B: Exactly two digits are odd (and one digit is even) This case involves one even digit and two odd digits. We must consider the position of the single even digit, keeping in mind the rule that $d_1$ cannot be 0. Let 'O' represent an odd digit and 'E' represent an even digit.

• Subcase B1: The even digit is in the hundreds place (E O O)

  • For $d_1$ (hundreds digit): Must be even AND non-zero. The even digits are {0, 2, 4, 6, 8}. Excluding 0, we have 4 choices.
    • Number of choices for $d_1 = 4$ (i.e., {2, 4, 6, 8}).
  • For $d_2$ (tens digit): Must be odd.
    • Number of choices for $d_2 = 5$ (i.e., {1, 3, 5, 7, 9}).
  • For $d_3$ (units digit): Must be odd.
    • Number of choices for $d_3 = 5$ (i.e., {1, 3, 5, 7, 9}). Number of outcomes for Subcase B1 = $4 \times 5 \times 5 = 100$.

• Subcase B2: The even digit is in the tens place (O E O)

  • For $d_1$ (hundreds digit): Must be odd. All odd digits are non-zero.
    • Number of choices for $d_1 = 5$ (i.e., {1, 3, 5, 7, 9}).
  • For $d_2$ (tens digit): Must be even. The digit 0 is allowed in the tens place.
    • Number of choices for $d_2 = 5$ (i.e., {0, 2, 4, 6, 8}).
  • For $d_3$ (units digit): Must be odd.
    • Number of choices for $d_3 = 5$ (i.e., {1, 3, 5, 7, 9}). Number of outcomes for Subcase B2 = $5 \times 5 \times 5 = 125$.

• Subcase B3: The even digit is in the units place (O O E)

  • For $d_1$ (hundreds digit): Must be odd.
    • Number of choices for $d_1 = 5$ (i.e., {1, 3, 5, 7, 9}).
  • For $d_2$ (tens digit): Must be odd.
    • Number of choices for $d_2 = 5$ (i.e., {1, 3, 5, 7, 9}).
  • For $d_3$ (units digit): Must be even. The digit 0 is allowed in the units place.
    • Number of choices for $d_3 = 5$ (i.e., {0, 2, 4, 6, 8}). Number of outcomes for Subcase B3 = $5 \times 5 \times 5 = 125$.

Total number of outcomes for Case B (exactly two odd digits) is the sum of outcomes from Subcases B1, B2, and B3: $\text{Total for Case B} = 100 + 125 + 125 = 350$

Now, we sum the outcomes from Case A and Case B to get the total number of favorable outcomes: $\text{Total Favorable Outcomes} = \text{Outcomes (Case A)} + \text{Outcomes (Case B)}$ $\text{Total Favorable Outcomes} = 125 + 350 = 475$

Step 3: Calculate the Probability

Using the probability definition from Key Concepts and Formulas: $P(\text{at least two odd digits}) = \frac{\text{Total Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$ $P(\text{at least two odd digits}) = \frac{475}{900}$

To simplify the fraction, we divide both the numerator and the denominator by their greatest common divisor, which is 25: $475 \div 25 = 19$ $900 \div 25 = 36$ So, the probability is: $P(\text{at least two odd digits}) = \frac{19}{36}$

---

Common Mistakes & Tips

1. First Digit Constraint: Always remember that the hundreds digit ($d_1$) of a 3-digit number cannot be 0. This is a common pitfall, especially when dealing with even digits in the first position.
2. Zero as an Even Digit: The digit '0' is an even digit. It can be used in the tens ($d_2$) or units ($d_3$) place, but not in the hundreds ($d_1$) place. Failing to account for this correctly can lead to errors in counting.
3. "At Least" Interpretation: Break down "at least N" problems into a sum of mutually exclusive "exactly N", "exactly N+1", etc., cases. This systematic approach helps avoid missing scenarios.

---

Summary

To determine the probability that a randomly selected 3-digit number has at least two odd digits, we first established the total number of possible 3-digit numbers as 900. Next, we identified the favorable outcomes by considering two distinct cases: numbers with exactly three odd digits (125 numbers) and numbers with exactly two odd digits (350 numbers). Summing these cases yielded 475 favorable outcomes. Finally, by dividing the total favorable outcomes by the total possible outcomes, we arrived at the probability of $\frac{475}{900}$, which simplifies to $\frac{19}{36}$.

The final answer is $\boxed{{{19} \over {36}}}$, which corresponds to option (A).