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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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2 Answers

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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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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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