Step 1: Understanding the Concept:
We have a set of tickets partitioned into odd and even numbers. Drawing "one at a time" conventionally implies drawing without replacement. We need to find the probability of two mutually exclusive specific sequences occurring: either Odd-Even-Odd or Even-Odd-Even. The total probability will be the sum of the probabilities of these two individual sequences.
Step 2: Key Formula or Approach:
1. Identify counts: Total tickets $= 9$. Odd tickets $= 5$ (1, 3, 5, 7, 9). Even tickets $= 4$ (2, 4, 6, 8).
2. Probability of a sequence (without replacement): $P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B)$.
3. Total Probability $= P(\text{OEO}) + P(\text{EOE})$.
Step 3: Detailed Explanation:
First, let's calculate the probability of the sequence {Odd, Even, Odd}, denoted as $P(\text{OEO})$.
- Probability 1st ticket is Odd: $P(\text{O}_1) = \frac{5}{9}$
- Probability 2nd ticket is Even (given 1st was Odd): $P(\text{E}_2|\text{O}_1) = \frac{4}{8}$ (since 4 even tickets remain out of 8 total remaining)
- Probability 3rd ticket is Odd (given 1st Odd, 2nd Even): $P(\text{O}_3|\text{O}_1 \cap \text{E}_2) = \frac{4}{7}$ (since 4 odd tickets remain out of 7 total remaining)
\[ P(\text{OEO}) = \frac{5}{9} \times \frac{4}{8} \times \frac{4}{7} \]
\[ P(\text{OEO}) = \frac{5}{9} \times \frac{1}{2} \times \frac{4}{7} = \frac{20}{126} \]
Next, let's calculate the probability of the sequence {Even, Odd, Even}, denoted as $P(\text{EOE})$.
- Probability 1st ticket is Even: $P(\text{E}_1) = \frac{4}{9}$
- Probability 2nd ticket is Odd (given 1st was Even): $P(\text{O}_2|\text{E}_1) = \frac{5}{8}$ (5 odd remain out of 8)
- Probability 3rd ticket is Even (given 1st Even, 2nd Odd): $P(\text{E}_3|\text{E}_1 \cap \text{O}_2) = \frac{3}{7}$ (3 even remain out of 7)
\[ P(\text{EOE}) = \frac{4}{9} \times \frac{5}{8} \times \frac{3}{7} = \frac{60}{504} \]
Let's keep the denominator the same to make addition easier:
\[ P(\text{EOE}) = \frac{4}{9} \times \frac{5}{8} \times \frac{3}{7} = \frac{1}{9} \times \frac{5}{2} \times \frac{3}{7} = \frac{15}{126} \]
Since these two sequences are mutually exclusive events, we add their probabilities:
\[ \text{Total Probability} = P(\text{OEO}) + P(\text{EOE}) \]
\[ \text{Total Probability} = \frac{20}{126} + \frac{15}{126} = \frac{35}{126} \]
Now, simplify the fraction. Both the numerator and denominator are divisible by 7:
\[ \frac{35 \div 7}{126 \div 7} = \frac{5}{18} \]
Step 4: Final Answer:
The required probability is $\frac{5}{18}$.