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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

  1. $\left(\dfrac{1}{2}\right)$
  2. $\left(\dfrac{1}{3}\right)$
  3. $\left(\dfrac{1}{4}\right)$
  4. $\left(\dfrac{1}{6}\right)$
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4 Comments

@sukanya :given answer is wrong.

correct answer will be $\frac{1}{3}$
0
0
$\frac{2C1*1C1}{4C2} = \frac{2}{6} = \frac{1}{3}$
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0
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)
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5 Answers

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