Hi!
I saw a exercise that I couldn't answer, but it seems so simple...
There is a chance that the book has a wrong answer (it says 5/6 is the answer), but I can't get that kind of an answer.
"There are 10 pieces, 4 of them are good. What the probability that you will pick alteast one good piece?"
So as far as I figured then here are combinations (10 3) = 10!/(3!*7!) = 120. So there are 120 ways to pick 3 pieces.
If the answer is right then there should be 100 outcomes out of 120 that you pick atleast 1 good piece (5/6), but how can you get such a number?
Hello, regdude!
Plato is absolutely correct
. . and his approach is the best one!
You can do it "the long way" . . . if you are very careful.
You are right: there are possible outcomes.There are 10 pieces, 4 of them are good.
You pick 3 of them at random.
What the probability that you will pick at least one good piece?
There are 4 good pieces (G), and 6 bad pieces (B).
How many ways can we get at least one G?
. . [1] one G, two Bs: . ways.
. . [2] two Gs, one B: . ways.
. . [3] three Gs, no B: . ways.
Hence, there are: . ways to get at least one Good.
Therefore: .
Got it?