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cptholt
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There are four machines in a factory. At exactly 8 pm, when the mechan Updated on: 24 Jul 2017, 18:58
Originally posted by cptholt on 24 Jul 2017, 18:30.
Last edited by cptholt on 24 Jul 2017, 18:58, edited 1 time in total.
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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



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bowlalife
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There are four machines in a factory. At exactly 8 pm, when the mechan 24 Jul 2017, 18:36
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For the mechanic to leave at 832 and if it takes him 20 minutes to reach the bus stop. He needs to finish his work by 8:12.

Finishing work by 8:12 requires him to fix both the faulty machines. Thus probability of catching the bus at 8:32 is simply the probability of fixing both the machines.

Probability of fixing both machines is given by -

P(first) x P(second) = 2/4x1/3 which is 1/6.

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There are four machines in a factory. At exactly 8 pm, when the mechan 24 Jul 2017, 19:18
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bowlalife
For the mechanic to leave at 832 and if it takes him 20 minutes to reach the bus stop. He needs to finish his work by 8:12.

Finishing work by 8:12 requires him to fix both the faulty machines. Thus probability of catching the bus at 8:32 is simply the probability of fixing both the machines.

Probability of fixing both machines is given by -

P(first) x P(second) = 2/4x1/3 which is 1/6.

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There is a complementary case. If both the checked machines are not defective, the last two are defective.

P(Case 2) = 1/6

P(Finding defective machine) = 1/6+1/6 = 1/3
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There are four machines in a factory. At exactly 8 pm, when the mechan 24 Jul 2017, 19:21
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theperfectgentleman
bowlalife
For the mechanic to leave at 832 and if it takes him 20 minutes to reach the bus stop. He needs to finish his work by 8:12.

Finishing work by 8:12 requires him to fix both the faulty machines. Thus probability of catching the bus at 8:32 is simply the probability of fixing both the machines.

Probability of fixing both machines is given by -

P(first) x P(second) = 2/4x1/3 which is 1/6.

Sent from my ONE A2003 using GMAT Club Forum mobile app

There is a complementary case. If both the checked machines are not defective, the last two are defective.

P(Case 2) = 1/6

P(Finding defective machine) = 1/6+1/6 = 1/3
Didn't even think of that. Assumption is that he spends 0 minutes on any machine that isn't required to be fixed? Great question. Any source?

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There are four machines in a factory. At exactly 8 pm, when the mechan 24 Jul 2017, 19:46
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theperfectgentleman
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

In order for the mechanic to catch the last bus at 8:32 PM he must be successful in either identifying the defective machines or the non-defective machines in his first and second attempt, either way he will be done in 12 mins to catch his last bus.

Probability that the first machine is defective = \(\frac{2}{4}\).

Probability that the second machine is defective = \(\frac{1}{3}\).

Probability that the first and second machines are defective \(= \frac{2}{4} * \frac{1}{3} = \frac{1}{6}\).

On the other hand, if the first two machines are not defective, then the other 2 machines are defective with 100% probability. Hence we have,

Probability that the first machine is not defective = \(\frac{2}{4}\).

Probability that the second machine is not defective = \(\frac{1}{3}\).

Probability that the first and second machines are not defective = \(\frac{2}{4} * \frac{1}{3} = \frac{1}{6}\).

Required probability = \(2*\frac{1}{6} = \frac{1}{3}\). Ans - D.



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