1. ## Probability with combinations

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?

2. Originally Posted by regdude
Hi!
(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?"
The answer in the text is correct.
There are $_{6}\mathcal{C}_3=\binom{6}{3}$ ways to select all 'bad' items.
$1-\frac{_{6}\mathcal{C}_3}{_{10}\mathcal{C}_3}=\frac {5}{6}$.

3. Can you please explain why there are (thats 20) ways to chose 'bad' pieces?

4. Originally Posted by regdude
Can you please explain why there are (thats 20) ways to chose 'bad' pieces?
There are ten items in all. Four are 'good'. So six are 'bad'.
If we select three items and all are 'bad' that is choosing three from six.
The probability of that happening is $\frac{1}{6}$, all ‘bad’.
At least one ‘good’ is the complement of all ‘bad’: $1-\frac{1}{6}$.

5. Oh, right.
Thanks.

6. 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.

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?
You are right: there are ${10\choose3} = 120$ possible outcomes.

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: . ${4\choose1}{6\choose2} \:=\:4\cdot15 \:=\:60$ ways.

. . [2] two Gs, one B: . ${4\choose2}{6\choose1} \:=\:6\cdot6 \:=\:36$ ways.

. . [3] three Gs, no B: . ${4\choose3}{6\choose0} \:=\:4\cdot1 \:=\:4$ ways.

Hence, there are: . $60 + 36 + 4 \:=\:100$ ways to get at least one Good.

Therefore: . $P(\text{at least one G}) \;=\;\frac{100}{120} \;=\;\frac{5}{6}$

Got it?

7. Thanks, your way seems to be more clearer to me, because this one I know (since I understand combinations more then the actual probability theory).