Question

A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box one at a time, the probability that they are alternatively either {odd, even}or {even, odd} is

• A. \( \frac{5}{17} \)
• B. \( \frac{4}{17} \)
• C. \( \frac{3}{16} \)
• D. \( \frac{4}{18} \)

Hint:

When calculating the probability of drawing tickets in a specific order, consider the total number of possible outcomes and the desired outcomes.

Answer 1

The Correct Option is D

Step 1: Probability of drawing odd/even tickets.
There are 5 odd and 4 even numbered tickets. We calculate the probability of drawing three tickets alternatively as odd and even.

Step 2: Conclusion.
Thus, the correct probability is \( \frac{4}{18} \), which simplifies to \( \frac{2}{9} \). Hence, the correct answer is option (D).

Final Answer: \[ \boxed{\text{(D) } \frac{4}{18}} \]