A shipment of 10 items has two defective and eight nondefective items. In the inspection of the shipment, a sample of items will be selected and tested. If a defective item is found, the shipment of 10 items will be rejected.

If a sample of 3 items is selected, what is the probability that the shipment will be rejected?

I used f(x)=(nCr(2,x)*nCr(8,3-x))/nCr(10,3)
f(1)=.466667

Book answer is .5333333.

Is the book right?

Originally Posted by lisakki

A shipment of 10 items has two defective and eight nondefective items. In the inspection of the shipment, a sample of items will be selected and tested. If a defective item is found, the shipment of 10 items will be rejected.

If a sample of 3 items is selected, what is the probability that the shipment will be rejected?

I used f(x)=(nCr(2,x)*nCr(8,3-x))/nCr(10,3)
f(1)=.466667

Book answer is .5333333.

Is the book right?

I haven't done any calculations but I'll make a bet with you ....

If X is the random variable number of defectives in sample, I'll bet that you only calculated Pr(X = 1) rather than $\displaystyle \Pr(X \geq 1)$, which is equal to 1 - Pr(X = 0) by the way.

Hi:
This is a hypergeometric problem! N=10 n1=2 and n2=8 and the Probability=1-(2C0)*(8C3)/(10C3). Because if one defective component is found, the shipment will be rejected. You only need to concern about there is no defective component is found, and then use 1 minus the answer. The book answer is correct.