Four machines are known to contain exactly two faulty ones. Machines are tested one by one in random order until both faulty machines are identified. What is the probability that only two tests are needed?

Introduction / Context:
We randomly order four machines containing exactly two faulty units. We test until both faulty ones are found. “Only two tests” means the first two tested are precisely the two faulty machines (in any order).

Given Data / Assumptions:

• Total machines: 4, faulty: 2, good: 2.
• Random order of testing; each permutation equally likely.
• Success event: the first two positions are the two faulty machines.

Concept / Approach:
Compute the probability that position 1 is faulty and position 2 is faulty given the first. Multiply sequential conditional probabilities.

Step-by-Step Solution:

Verification / Alternative check:
Permutation counting: number of orders with the two faulty first equals 2! * 2! (arrangements of faulty then good) out of 4! total? More directly, probability that a specific 2-set occupies the first two slots is C(2,2)C(2,0)/C(4,2) = 1/6; same result.

Why Other Options Are Wrong:
1/3, 1/2, 1/4 mis-handle sequential conditioning or overcount permutations.

Common Pitfalls:
Assuming independence between first and second picks; the second probability is conditional because sampling is without replacement.

Final Answer:
1/6