Solution: We are selecting 5 distinct integers independently and uniformly at random from $ \0, 1, 2, \ldots, 50\ $, and let $ z $ be the sum of the two largest. We want $ \mathbbP(z \geq 100) $. - RTA

Understanding the Context

The Combinatorial Setting

We draw 5 distinct integers uniformly at random from $ S = \{0, 1, 2, \ldots, 50\} $. Let the sampled set be $ \{a_1, a_2, a_3, a_4, a_5\} $, sorted so that:
$$
0 \leq a_{(1)} < a_{(2)} < a_{(3)} < a_{(4)} < a_{(5)}
$$
Then $ z = a_{(4)} + a_{(5)} $, the sum of the two largest values. We seek
$$
\mathbb{P}(z \geq 100) = \frac{\ ext{Number of 5-subsets where } a_{(4)} + a_{(5)} \geq 100}{\binom{51}{5}}
$$

Key Observations

Key Insights

1. Maximum possible sum: The largest two distinct values are 49 and 50, so $ z_{\max} = 99 $.
   Therefore, $ z \leq 99 $.
   This means $ \mathbb{P}(z \geq 100) = 0 $.

Why? Even in the most favorable case—sampling $ 49, 50, $ and three lower values, the two largest sum to $ 49 + 50 = 99 $, which is strictly less than 100.

Thus, no outcome satisfies $ z \geq 100 $. The event is impossible.

Technical Clarification

Final Thoughts

While the problem involves sophisticated combinatorics—counting 5-element subsets with $ a_{(4)} + a_{(5)} \geq 100 $—the constraint $ a_{(5)} \leq 50 $, $ a_{(4)} \leq 49 $, and $ a_{(4)} + a_{(5)} \leq 99 $ rules out 100 or more entirely.

Some might infer that rounding or sampling error could allow higher sums, but since only integers from 0 to 50 are allowed, and we take distinct values, achieving $ z \geq 100 $ is mathematically impossible.

Why This Matters

Understanding boundary behavior in discrete probability is crucial in fields like:
- Risk analysis (e.g., maximum losses from top 5 samples)
- Algorithm design (maximal element selection)
- Monte Carlo simulations involving order statistics

Even when draws are uniform and independent, domain bounds constrain outcomes. This example illustrates the importance of rigorous boundary checking before engaging in complex probability computations.