A box contains 12 oranges, of which one-third are bad (i.e., 4 bad and 8 good). If three oranges are drawn at random without replacement, what is the probability that at least one of the three is good?

Introduction / Context:
“At least one good” is most efficiently computed via the complement event: “all drawn are bad.” The composition is small, making the computation straightforward with combinations.

Given Data / Assumptions:

• Total oranges = 12.
• Bad = 4; Good = 8.
• Three are drawn uniformly at random without replacement.

Concept / Approach:
P(≥1 good) = 1 − P(all 3 are bad). The “all bad” event exists only if we draw all three from the 4 bad oranges.

Step-by-Step Solution:
Total ways to choose 3 = C(12, 3) = 220.Ways to choose 3 all bad = C(4, 3) = 4.P(all bad) = 4 / 220 = 1 / 55.Therefore P(≥1 good) = 1 − 1/55 = 54/55.

Verification / Alternative check:
Direct enumeration of mixed cases is longer and error-prone compared to the clean complement method.

Why Other Options Are Wrong:
1/55 is the probability that all three are bad, not “at least one good”; 45/55 and 3/55 are arithmetic mismatches; 53/55 still undercounts.

Common Pitfalls:
Confusing “one-third bad” as 1/3 probability per draw without replacement; combinatorics must reflect finite counts.

Final Answer:
54/55