oxon wrote:

Interesting problem so though I'd share it with you guys.

Three items are removed from a box of 12 items. All three items are inspected. The box is rejected if more than one item is found to be faulty.

If there are three faulty items in the box, what is the probability that the box is accepted?

Let A be a bad ball and B be a good one.

Box is rejected if the 2 or 3 balls that are drawn are bad.

Possible combinations for the box to be rejected=

ABA, BAA, AAA.

Hence, the probability for the box to be rejected

(3*9*2)/(12*11*10)+ (9*3*2)/(12*11*10)+(3*2*1)/(12*11*10)=19/220

hence the probability that it will be accepted= 1-19/220=201/220.

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Let's get it right!!!!