Question

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A quality-control plan for the concrete in a nuclear reactor containment vessel calls for casting 6 cylinders for each batch of 10 yd poured and testing them as follows: 1 at 7 days, 1 at 14 days, 2 at 28 days, 2 more at 28 days if any of the first four cylinders is "inadequate". The required strength is a function of age. If the cylinder to be tested is chosen at random from those remaining (i.e. with equal likelihoods) a)- What is the probability that all six will be tested if in fact one inadequate cylinder exists in the six? b)- If the batch will be "rejected" if two or more inadequate cylinders are found, what is the likelihood that it will not be rejected given that exactly two are in fact inadequate? (Rejection will lead to more expensive coring and testing of concrete in place) c)-A "satisfactory" concrete batch gives rise to an inadequate cylinder with probability p-0.1 there will be one or more inadequate cylinders in the six when the batch is "satisfactory"? (Assume independence of the quality of the individual cylinders) rejected? What is the probability that an unsatisfactory batch (in particular, say, p = 0.3) will not be rejected? Clearly a quality control plan wants to keep both these probabilities low, while also keeping the cost of testing small.

Answer 1

A.

P(All six will be tested) = P(Inadequate cylinder gets tested in the first 4 selected for testing)

Total ways of selecting 4 cylinders from 6 = $\fn_phv \binom{6}{4}$

Total ways of always selecting Inadequate cylinder among the 4 from the 6 = $\fn_phv \binom{5}{3}$

P(All six will be tested) = $\fn_phv \binom{5}{3} \div \binom{6}{4}$

= 2/3

B.

P(Batch will not be rejected) = P(No inadequate cylinder gets tested in the first 4 testing given that 2 are inadequate)

= $\fn_phv \binom{4}{4} \div \binom{6}{4}$

= 1/15

C.

P(Inadeqaure cylinder) = p = 0.1

P(adequate cylinder) = q = 0.9

Let X = occurence of inadequate distibution

It becomes a binomial distribution.

n = 6

P(X>=1) = 1-P(X=0)

= 1- $\fn_phv \binom{6}{0}*0.1^{0}*0.9^{6}$

=1- 0.53

= 0.47

D.

P(Inadeqaure cylinder) = p = 0.1

P(adequate cylinder) = q = 0.9

X = Inadequte cylinders

P(batch will be rejected) = P(Two or more inadequate cylinders are found)

= P(rejection|X = 2) + P(X>=3)

Since occurance of 2 inadequate cylinders does reject the batch only if all the six cylinders ar tested thus, P(X=2) has to be multiplied with probability that the particular batch is rejected.

P(X>=3) = 0.01585

P(X=2) = 0.098415

P(rejection|X = 2) = (1-1/15)*0.098415 = 0.091

P(batch will be rejected) = 0.1073

P(Inadeqaure cylinder) = p = 0.3

P(adequate cylinder) = q = 0.7

X = Inadequte cylinders

P(Unsatisfactory batch will not be rejected) = P(X<=2)

= P(X<=1) + P(Not rejected|X=2)

P(X<=1) = 0.420175

P(X=2) = 0.324135

P(Not rejected|X=2) = 1/15*0.324135 = 0.022

P(Unsatisfactory batch will not be rejected) = 0.44