Thank for the help. 2. Firing pins A engineer samples 5 from a large lot of manu

Question

Thank for the help.

2. Firing pins A engineer samples 5 from a large lot of manufactured firing pine and checks for defects. Unknown to the inspector, three of five pins in the samples are defective. The engineer will test the pins until a defective firing pin is observed - in which case no more pins will be tested and the entire lot will be rejected. Let Y be the number of firing pins the QC engineer must test before the lot is rejected a. List the 10 possible arrangements of the 2 good and 3 bad firing pins. Assume the pins are placed in a clip/holder, to keep them in order. b. Compute the probability of rejecting the lot on the first test, starting at the top of the holder? c. Compute the probability of rejecting the lot on the second test, starting at the top of the holder? d. Compute the probability of rejecting the lot on the third test, starting at the top of the holder? e. Compute the same probabilities if the engineer starts testing from bottom of the holder.

Explanation / Answer

(a)

Let G shows the good Pin and B shows the bad pin. Following is the list of all possible arrangements:

S = {BBBGG, BBGBG, BGBBG, GBBBG, GBBGB, GBGBB, GGBBB, BBGGB, BGGBB, BGBGB}

(B)

Let right most position shows the top of the holder. Out of 10 possible arrangements, 6 has first pin bad. So the required probabaility is 6 /10 =0.6.

(c)

Let right most position shows the top of the holder. Out of 10 possible arrangements, 3 has first pin good and second pin bad. So the required probabaility is 3 /10 =0.3.

(d)

Out of 10 possible arrangements, 1 has two first pins good and third pin bad. So the required probability is 1 /10 =0.1.

(e)

Let Y shows the number of pins fired. Here Y can take values 1, 2, or 3. Let X shows the cost of pin fired.

When Y=1 then X =$200 so

P(X=$200) = 0.6 from part (b)

When Y=2 then X =2*$200 =$400 so

P(X=$400) = 0.3 from part (c)

When Y=3 then X =3*$200 =$600 so

P(X=$600) = 0.1 from part (d)

So the expected cost for testing is

E(X) = $200*0.6 + $400*0.3 + $600*0.1 = $300

Hence, required expected coat is $300.

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