Agroup of four components is known to contain two defectives. an inspector tests the components
one at a time until the two defectives are located. once she locates the two defectives, she
stops testing, but the second defective is tested to ensure accuracy. let y denote the number of
the 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 a
defective 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.

a) See below

b) $13.67

c) 0.5555

Step-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:

a) Find the probability distribution for Y .

P(Y=4) = 3/6 = ½

P(Y=3) = 2/6 = 1/3

P(Y=2) = 1/6

b) 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≅ $13.67

c)Find the variance for the total cost for testing and repairing of the lot.

false

step-by-step explanation:

c

step-by-step explanation:

i think because what else can be a sample of every member