Question

There are four machines in a factory. At exactly 8 pm, when the mechanic is about to leave the factory, he is informed that two of the four machines are not working properly. The mechanic is in a hurry, and decides that he will identify the two faulty machines before going home, and repair them next morning. It takes him twenty minutes to walk to the bus stop. The last bus leaves at 8:32 pm. If it takes six minutes to identify whether a machine is defective or not, and if he decides to check the machines at random, what is the probability that the mechanic will be able to catch the last bus?

• A. 0
• B. 1/6
• C. 1/4
• D. 1/3
• E. 1

Hint:

When success can occur by either identifying all defective or all non-defective, always consider both cases in probability.

Answer 1

The Correct Option is D

Step 1: Time constraints.
- Last bus leaves at 8:32 pm. - Walking time = 20 min. - Therefore, mechanic must finish identification by 8:12 pm. - He has 12 minutes (from 8:00 to 8:12) to identify machines. Since it takes 6 minutes per machine, he can check exactly 2 machines.

Step 2: Probability setup.
There are 4 machines in total, out of which 2 are defective. The mechanic checks 2 machines at random. He can catch the bus if: - Both machines he checks are defective (so he identifies the 2 defective ones). - Both machines he checks are non-defective (so he indirectly identifies the defective ones by elimination).

Step 3: Case probabilities.
- Case (i): Both defective. \[ P = \frac{\binom{2}{2}}{\binom{4}{2}} = \frac{1}{6} \] - Case (ii): Both non-defective. \[ P = \frac{\binom{2}{2}}{\binom{4}{2}} = \frac{1}{6} \] - Case (iii): One defective, one non-defective. \[ P = \frac{\binom{2}{1}\binom{2}{1}}{\binom{4}{2}} = \frac{4}{6} = \frac{2}{3} \]

Step 4: Successful cases.
Only cases (i) and (ii) are successful. \[ P(\text{success}) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \]

Final Answer:
\[ \boxed{\tfrac{1}{3}} \]