Thread: Probability Problem

Dear Forum , I am having trouble figuring out the following , any help would be appreciated...
A shipment of 60 inexpensive watches, including 9 that are defective, is sent to a department store. The receiving department selects 10 at random (from the 60 received) for testing, and rejects the entire shipment if 1 or more in the sample is found defective. What is the probability that the shipment will be rejected?

thanks - AC-

Originally Posted by AlgebraicallyChallenged

Dear Forum , I am having trouble figuring out the following , any help would be appreciated...
A shipment of 60 inexpensive watches, including 9 that are defective, is sent to a department store. The receiving department selects 10 at random (from the 60 received) for testing, and rejects the entire shipment if 1 or more in the sample is found defective. What is the probability that the shipment will be rejected?

thanks - AC-

Let X be the number of defective units in the sample of 10.

P(X= or > 1)= P(X=1) + P(X=2) +...+ P(X=9) since 9 is the maximum number of defective units in the shipment.

P(X=1) is the probability of getting EXACTLY one defective unit, then all the other nine work.

= (9/60) (51/59)(50/58) (49/57) continue until you have 10 terms.
Doing this for all nine cases is a lot of work so we write:

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

P(X < 1) = P(X=0) which is the probability of getting 0 defective units. This means all 10 units chosen at random will work:

51/60 is the probability of the first one working
50/59 is the probability of the second one working

... write them all out, and multiply them since we are assuming independence.

1 - the product above is your answer.

Good luck!

Originally Posted by AlgebraicallyChallenged

Dear Forum , I am having trouble figuring out the following , any help would be appreciated...
A shipment of 60 inexpensive watches, including 9 that are defective, is sent to a department store. The receiving department selects 10 at random (from the 60 received) for testing, and rejects the entire shipment if 1 or more in the sample is found defective. What is the probability that the shipment will be rejected?

thanks - AC-

I have a long solution if the receiving find the first defective they will reject so the probability is
(d)+(n,d)+(n,n,d)+(n,n,n,d)+(n,n,n,n,d)+(n,n,n,n,n ,d)+........(n,n,n,n,n,n,n,n,n,d) that is it

so the probability that they will find the defective from the first one selected then they will reject so the probability of this is



or they will find the defective one from the second selection so the probability will be




or they will find the defective one from the third selection so the probability will be



or from the fourth selection



or from the fifth selection


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find them all then find the sum of them all or equal add

Originally Posted by apcalculus

Let X be the number of defective units in the sample of 10.

P(X= or > 1)= P(X=1) + P(X=2) +...+ P(X=9) since 9 is the maximum number of defective units in the shipment.

P(X=1) is the probability of getting EXACTLY one defective unit, then all the other nine work.

= (9/60) (51/59)(50/58) (49/57) continue until you have 10 terms.
Doing this for all nine cases is a lot of work so we write:

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

P(X < 1) = P(X=0) which is the probability of getting 0 defective units. This means all 10 units chosen at random will work:

51/60 is the probability of the first one working
50/59 is the probability of the second one working

... write them all out, and multiply them since we are assuming independence.

1 - the product above is your answer.

Good luck!

your solution is correct if the question said what is the probability they will get one defective but the question said what is the probability they will reject if they find the defective (any time ) they will reject so for them they will not continue in selecting if they find one defective just one they will reject as I understand from the question that's it

Is that not easy?