ISI2017-MMA-21

Question

There are four machines and it is known that exactly two of them are faulty. They are tested one by one in a random order till both the faulty machines are identified. The probability that only two tests are required is

• A. $\left(\dfrac{1}{2}\right)$
• B. $\left(\dfrac{1}{3}\right)$
• C. $\left(\dfrac{1}{4}\right)$
• D. $\left(\dfrac{1}{6}\right)$

Comments

@sukanya :given answer is wrong.

correct answer will be $\frac{1}{3}$

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$\frac{2C1*1C1}{4C2} = \frac{2}{6} = \frac{1}{3}$

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Out of 4 , it's given that exactly two are faulty therefore two are non faulty
now, ways to permute them will be 4!/(2!*2!) = 6

so total sample space contain 6 items now probab. that in two test we detect the faulty machines = (probab. of detecting two faulty m/c in 1st two tests) OR (probab. of detecting two unfaulty m/c in 1st two tests)
                            
                      = (1/6) + (1/6)
                      = (2/6)
                      =(1/3)

Answer 1

Without loss of generality. There are $C(4,2)$ = $6$ ways to identify two of the machines.  Only two tests are required. Thus,  $p$= $2/6$=$1/3$