Two numbers are chosen without replacement from S = {1, 2, 3, 4, 5, 6}. What is the probability that the minimum of the two numbers is less than 4?

Introduction / Context:
This is a complementary counting problem. Instead of directly counting outcomes where the minimum is less than 4, it is often easier to count the opposite event and subtract from 1.

Given Data / Assumptions:

• Set S = {1,2,3,4,5,6}.
• Two distinct numbers are drawn without replacement.
• Total unordered pairs = C(6,2) = 15 (order does not matter for “minimum”).

Concept / Approach:
Let E be the event “min < 4”. Its complement E^c is “both numbers ≥ 4”. The subset {4,5,6} contains 3 numbers; choosing both from it gives pairs with minimum ≥ 4. Count pairs in E^c, then compute 1 − P(E^c).

Step-by-Step Solution:

Verification / Alternative check:
Direct count for min < 4: any pair that includes at least one of {1,2,3}. Counting shows 12 out of 15 pairs, i.e., 4/5, consistent with the complement approach.

Why Other Options Are Wrong:
14/15 would require only one “bad” pair, but there are 3; 1/15 or 1/5 reflect counting the complement or partial cases only.

Common Pitfalls:
Confusing ordered vs unordered counting; the event depends only on the set of two numbers, not their order, so combinations are appropriate.

Final Answer:
4/5