A factory produces fuses which are packaged in boxes of 10. Three fuses from each box are chosen at random for inspection. The box is rejected if at least one of these three fuses is defective. What is the probability that a box containing 5 defective fuses will be rejected?

Question

anonymous · Answer

First fuse not defective = 5/10 (10 total, 5 not defective).
Second fuse not defective = 4/9 (9 total, 4 not defective).
Third fuse not defective = 3/8 (8 total, 3 not defective).

p = 5/10 * 4/9 * 5/8 = 1/12

given 1-p = p', p' = 1 - 1/12 = 11/12

anonymous · Answer

this is probably a dumb question but i can't understand the actual situation in the question. what's with the 5 fuses? isn't it talking about inspecting only 3 fuses per box?:

anonymous · Answer

The 5 defective fuses is a made up scenario for the problem.  It's saying: out of the 10 fuses per box, 5 are defective and 5 are not defective.  Then, it's asking you to find the probability of a box of 5 defectives being rejected if you're picking 3 at a time and rejection of a box happens if 1 of those 3 is defective.

I hope that makes sense.  It is a little difficult to wrap the head around.

anonymous · Answer

ooh okay, that makes sense. thank you so much :)

anonymous · Answer

Very welcome!