A group of four components is known to contain two defectives. An inspector tests the componentsone at a time until the two defectives are located. Once she locates the two defectives, shestops testing, but the second defective is tested to ensure accuracy. Let Y denote the number ofthe test on which the second defective is found.a) Find the probability distribution for Y .b) If the cost of testing a component is $2 and the cost of repairing adefective is $4, find the expected total cost for testing and repairing the lot.c)Find the variance for the total cost for testing and repairing of the lot.

Answer:a) See belowb) $13.67c) 0.5555Step-by-step explanation:Let us denote with C a correct component and with D a defective one.We have then the 6 possibilities of arrangement:[tex] \bf \left[\begin{array}{ccc}Arrangement&Y&Cost\;test+repair\\CCDD&4&2+2+6+4=14\\CDCD&4&2+6+2+4=14\\DCCD&4&6+2+2+4=14\\DCDC&3&6+2+6=14\\DDCC&2&6+6=12\\CDDC&3&2+6+6=14\end{array}\right][/tex]a) Find the probability distribution for Y .P(Y=4) = 3/6 = ½P(Y=3) = 2/6 = 1/3P(Y=2) = 1/6b) If the cost of testing a component is $2 and the cost of repairing a defective is $4, find the expected total cost for testing and repairing the lot.The expected total cost for testing and repairing the lot is the average of testing and repairing(5*14 +12)/6 = $13.6666≅ $13.67c)Find the variance for the total cost for testing and repairing of the lot.[tex] \bf Variance=\frac{5(14-13.6666)^2+(12-13.6666)^2}{6}=0.5555[/tex]