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Xavier Aptitude Test 2011 Paper

Question : 28 of 101

Marks: +1, -0

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 ?

Validate

Solution:

(d) As the mechanic has decided to check two machines,

therefore, if he identifies either both defective machines or both non-defective then the probability that he is able to catch the last bus is sum of the two.

Following table gives the possible outcome and their corresponding probabilities.

	Outcome	Probability
1	def $\;\;\;\;$ def	$\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\frac{1}{6}$
2	def $\;\;\;\;$ non def	$\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)=\frac{1}{3}$
3	non def $\;\;\;\;$ def	$\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)=\frac{1}{3}$
4	non - def $\;\;\;\;$ non - def	$\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\frac{1}{6}$

In the first and last cases, the mechanic would have identified the defective machines in time to catch the bus.

∴ Probability that he is able to catch the last bus is

\frac{1}{6}+\frac{1}{6}=\frac{1}{3}

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