A good part after a defective one

Question

A bin holds 10 good parts and 3 defective parts, 13 in total. The first part pulled turned out to be defective and was set aside. You now pull one more part.

Given the defective part already removed, what is the probability the next part is good?

Accepted Answer

The condition changes the bin. After removing 1 defective, what remains is 10	ext{ good}+2	ext{ defective}=12	ext{ parts}. The chance the next pull is good is just the good fraction of the current bin: P(	ext{next good})=\frac{10}{12}=\frac{5}{6}\approx 0.833. Takeaway. A single conditional draw is easy once you recount: it is just the target fraction of what is left. Pulling a defective out first raised the good fraction from 	frac{10}{13} to 	frac{10}{12}.

A good part after a defective one

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