A bag contains 50 tickets numbered 1, 2, 3, ..., 50. Five tickets are drawn at random and then arranged in ascending order of their numbers. What is the probability that the third ticket in this ordered list has the number 30?

Introduction / Context:
This question is a more advanced combinatorial probability problem. We select five distinct numbers from 1 to 50 and then arrange them in increasing order. We are asked for the probability that the third smallest selected number is exactly 30. This requires careful thinking about how many chosen numbers lie below and above 30.

Given Data / Assumptions:

• Tickets are numbered 1 to 50.
• Five distinct tickets are drawn at random.
• They are then arranged in ascending order: X1 < X2 < X3 < X4 < X5.
• We want P(X3 = 30).

Concept / Approach:
If the third smallest number is 30, then exactly two of the chosen numbers must be less than 30 and exactly two must be greater than 30. All choices are among distinct integers from 1 to 50. We count the number of ways to choose such a 5 element set and divide by the total number of 5 element subsets of {1, 2, ..., 50}. The ordering after selection is fixed by ascending order, so we focus on combinations.

Step-by-Step Solution:
Total ways to choose any 5 tickets from 50 = C(50, 5). Numbers less than 30: 1 to 29, so 29 numbers. Numbers greater than 30: 31 to 50, so 20 numbers. To have X3 = 30, choose 2 numbers from 1 to 29 and 2 numbers from 31 to 50, and include 30 itself. Ways to choose 2 from 29 = C(29, 2). Ways to choose 2 from 20 = C(20, 2). Favourable sets = C(29, 2) * C(20, 2). Total sets = C(50, 5). Probability = [C(29, 2) * C(20, 2)] / C(50, 5) = 551 / 15134.

Verification / Alternative check:
We can compute the numerical values: C(29, 2) = 406, C(20, 2) = 190, and C(50, 5) = 2,118,760. Then favourable sets = 406 * 190 = 77,140, and probability = 77,140 / 2,118,760. Simplifying this fraction yields 551/15134, which matches the given option.

Why Other Options Are Wrong:
1/2 and 1/9 are simple fractions that do not relate to the specific combinatorial counts here. 552/15379 is a different fraction and does not arise from the correct combination values, so it must be incorrect.

Common Pitfalls:
Common errors include ignoring the ordering condition and not enforcing exactly two numbers below and exactly two above 30. Some learners attempt to assign probabilities to individual positions without working with combinations, which leads to incorrect reasoning. Carefully translating the order condition into a statement about how many numbers lie below and above the target value is essential.

Final Answer:
Therefore, the probability that the third smallest ticket is 30 is 551/15134.