# Picking marbles without replacement (alternate solution Help)

There are a set of 10 marbles in a bag :9 of which are white and the other 1 is black.

### Question

If one was to reach in the bag and pick 3 marbles all at once (i.e without replacement) what is the probability that a black marble is drawn?
The general idea seems to be to choose 2 marbles out of the 9 let the other place be filled with the black marble to give a probability of: $$\frac{\binom{9}2}{\binom{10}3}$$ But there can only be two combinations of marbles in your hand once it is out of the bag:

3 white marbles

or

2 white marbles and 1 black marble

since the marbles are indistinguishable and the sample set of answers contain only the above two possibilities should the probability not be $$\frac{1}2$$.
Note:I have assumed the number of ways of selecting 3 white marbles is not $$\binom{9}3$$ but 1 since the marbles are indistinguishable and all elements of the set $$\binom{9}3$$ are the same and their cardinality is 1.
Can someone explain why I am arriving at this alternate solution?Any sort of help is appreciated.

• The source of your mistake is in since the marbles are indistinguishable. Indeed, there are several ways to choose indistinguishable combinations. For example, there are $\binom93$ ways to choose a combination of $3$ white balls. But despite the fact that they are indistinguishable, their amount affects the probability. In order to compute probability, you need to assume that they are distinguishable. Nov 22, 2016 at 11:05

There are only two results of a scratch-it-lottery: win or loose. Since aside from one being a winning ticket and the other 299999 being losing tickets, lottery tickets are indistinguishable, then should the probability of winning not be $1/2$?

I think not.   That argument is clearly fallacious.

For the same reason, all though balls of the same colour are indistinguishable, there are still many more white than black.   The probabilities for the two results do not have equal weight.   So, indeed, considering each ball to be a distinct entities with one of two colours, and each individual ball equally likely to be selected, is the correct approach.

$$\mathsf P(W=2)=\binom 9 2\binom 11/\binom {10}3\\\mathsf P(W=3)=\binom 9 3\binom 10/\binom {10}3$$

• 'Probability is the ratio of favourable outcomes to the whole number of outcomes possible' so the only way to get $\binom{10}3$ number of outcomes is to consider the marbles to be distinct, if it was specifically given in the problem that the marbles are indistinguishable even then we would have to assume they are distinct, am I right?? Nov 23, 2016 at 5:28

The distinct number of different 3 marble collections to be chosen out of the bag is 2, but the probability of choosing 1 black marble out of the ten marbles in a 3 marble choice is $\frac{\binom{9}{2}}{\binom{10}{3}}$. This is because, you have intuitionally more chance of picking up white marbles than black. The fact you mention that probability must be $\frac{1}{2}$ is false.However, there are only two choices of 3 marbles-1 black, 2 white or all 3 white. The confusion arises by mixing up choice and chance.

• Thanks for the explanation but are you assuming that the marbles are not indistinguishable because $\binom{n}k$ is only defined for choosing k subsets out of n distinct items Nov 22, 2016 at 11:36
• @BearClaw No, $\binom{n}{k}$ is defined for choosing $k$ subsets out of $n$ item set without repacement. The distinctness does not matter. Nov 23, 2016 at 4:28