# Probability without replacement question

I have a bag full of 10 marbles: 2 black, 1 blue, 1 yellow, 3 green, 1 brown, and 2 purple. I draw 5 marbles one at a time without replacement. What is the probability of a black marble to be in the five drawn?

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You have a bag full of 10 marbles: 2 black, 8 non-black. ... – Frenzy Li Aug 14 '12 at 2:30

Think of the marbles as having, in addition to colour, an ID number that makes them distinct.

There are two interpretations of "one black:" A: at least one black, and B: exactly one black. The probabilities are of course different. My preferred interpretation of the wording is A. Edit: With the change of wording to "a black" it is clearly A that is meant, but for your interest I will keep the analysis of B.

A: At least one black: It is easier to find first the probability of no black.

There are $\binom{10}{5}$ ways to choose $5$ marbles, all equally likely. Note that there are $\binom{8}{5}$ ways to choose $5$ marbles from the $8$ non-black. So the probability that all the balls are non-black is $$\frac{\binom{8}{5}}{\binom{10}{5}},$$ and therefore the probability of at least one black is $$1-\frac{\binom{8}{5}}{\binom{10}{5}}.$$

B: Exactly one black: There are $\binom{2}{1}$ ways of choosing one black from the two available. For each such way, there are $\binom{8}{4}$ ways to choose the non-blacks to go with it. So the total number of ways to pick exactly one black, and the rest non-black, is $\binom{2}{1}\binom{8}{4}$. Thus our probability is $$\frac{\binom{2}{1}\binom{8}{4}}{\binom{10}{5}}.$$

Remark: We used general techniques. For A, there is a simpler way. The probability that the first marble chosen is non-black is $\frac{8}{10}$. Given the first was non-black, the probability that the second is non-black is $\frac{7}{9}$, since there are $7$ non-blacks left in a total of $9$. So the probability the first two are non-black is $\frac{8}{10}\cdot\frac{7}{9}$. Continue in this way. The probability all five are non-black is $$\frac{8}{10}\cdot\frac{7}{9}\cdot\frac{6}{8}\cdot\frac{5}{7}\cdot\frac{4}{6}.$$ As in the earlier discussion, subtract the above from $1$.

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Great answer, thanks for the time it took to write all this out! A quick question though, how would I calculate the probability of getting both black marbles from the bag in the 5 drawn? – ericbowden Aug 14 '12 at 5:45
The black marbles can be chosen in $1$ way, or if you prefer, $\binom{2}{2}$ ways (which is $1$). The remaining $3$ marbles can be chosen in $\binom{8}{3}$ ways, for a total of $\binom{2}{2}\binom{8}{3}$. Divide this by $\binom{10}{5}$. – André Nicolas Aug 14 '12 at 5:56
Exactly what I needed. One last question it seems that by using the Binomial Coefficient to calculate combinations that the numbers can get factorially large. Is there a feasible way to calculate exactly 2 black marbles from a bag of say a 1000 marbles with 5 picks? – ericbowden Aug 15 '12 at 22:32
Well, they do get large. You know what the answer would look. Take the ugliest term, $\binom{1000}{5}$. You may be thinking of the formula $\frac{1000!}{5!995!}$ where a couple of the factorials are ridiculously huge. But $1000!=(1000)(999)(998)(997)(996)(995!)$ and the $995!$s cancel. Moral is that by laying out calculation properly, it can be very doable by simple calculator. But sometimes it is hard. Not many years ago, people used good approximate but easy to calculate formulas. Still do a lot, but mid-range calculations are easy to do exactly with the right software. – André Nicolas Aug 15 '12 at 22:43

Hint: You have a $\frac 8{10}$ chance that the first marble drawn is not black. Assuming it is not black, you have a $\frac ??$ chance that the next one is not black. Then you have a chance $\frac 8{10}\times \frac ??$ that they are both non-black. This will give you the chance of getting no black marbles, so $1-$this will be the chance of at least one black marble.

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