An urn contains $6$ white and $4$ black balls. A fair die is rolled and that number of balls are chosen from the urn. Find the probability that the balls selected are white.

I know the basic way to go about solving the problem.

Let $W$ be the event of finally drawing all white balls. Let $P(n)$ denote the probability of appearance of $n$ on the die.

We want: $$P(W) = P(1)P(W\mid 1)+ P(2)P(W\mid 2)+\dots \implies P(W) = \dfrac{1}{6}\left(\sum_{i=1}^6P(W\mid i)\right)$$

Now, I am actually facing trouble in computing $P(W/i)$. I saw author's method and in it he has used $P(W\mid i) = \dfrac{^6C_i}{^{10}C_i}$

but I fail to understand how that can be true when all white balls are identical and all black balls are identical.

Here, $^6C_i$ denotes the combination of $i$ different things from 6 different objects, doesn't it? How can that be used here?

  • Doesn't it look pretty much as a Hypergeometric distribution were the number of draws is a rv? $K=6$, $N=10$, $n=k$ is a random variable. Does it make any sense to you? – Ramiro Scorolli Nov 14 at 8:13
  • I don't know what's a hypergeometric distribution @RamiroScorolli – Abcd Nov 14 at 8:13
  • Basically you have $N$ balls(10 balls in total), $K$ represent success (6 white balls), you draw $n$ balls (where n is the number obtained with the dice) and you are expecting $k$ successes. (basically n cause you want all to be white). en.m.wikipedia.org/wiki/Hypergeometric_distribution – Ramiro Scorolli Nov 14 at 8:20
  • I'm not completely sure but I think that this could be the $P(W/I)$ you are looking for – Ramiro Scorolli Nov 14 at 8:20
  • How about something similar with a smaller amount of balls? If you have two identical black balls and an otherwise similar but white ball in a bag, and you pull out one in random, what's the probability of getting a black ball? Is it 1/2 because there are only two colors (and you couldn't tell the black ones apart)? – ilkkachu Nov 14 at 14:43
  • The probability of drawing $i$ white balls is the probability of drawing a white ball and then (without replacement) drawing a second white ball, and so on up to $i$

  • The first ball drawn has a probability of $\frac{6}{10}$ of being white

  • Given that the first ball drawn is white, the second ball drawn has a probability of $\frac{5}{9}$ of being white

  • An so on up to the $i$th ball having a probability of $\frac{6-i+1}{10-i+1}$ of being white

  • So the probability all $i$ balls drawn are white is $$\frac{6 \times 5 \times \cdots \times(6-i+1)}{10 \times 9 \times \cdots \times(10-i+1)} = \dfrac{\frac{6!}{(6-i)!}}{\frac{10!}{(10-i)!} }= \dfrac{\frac{6!}{(6-i)!i!}}{\frac{10!}{(10-i)!i!}} = \dfrac{^6C_i}{^{10}C_i}$$

  • 1
    Wow, this is amazing. – Abcd Nov 14 at 8:18
  • So does this also imply that considering them as distinct is a sort of "trick" that is always going to work? – Abcd Nov 14 at 8:22
  • @Abcd If you are going to use a counting calculation, then you want each event to be of equal probability, and treating the balls as physically distinct even when they look the same does this. Otherwise you risk saying that the probability of all white when $i=6$ might be higher than when $i=4$, since when $i=6$ you can get $4$, $5$ or $6$ white balls while with $i=4$ you can get $0$, $1$, $2$, $3$ or $4$ white balls; you need to be able say what the differing probabilities of the various possible outcomes are – Henry Nov 14 at 8:55

If you draw the $i$ balls sequentially $$P(W\mid i)=\frac{6}{10}\cdot\frac{5}{9}\cdots\frac{6-i+1}{10-i+1}$$ If you draw the $i$ balls simultaneously $$P(W\mid i)=\frac{\dbinom{6}{i}\cdot\dbinom{4}{0}}{\dbinom{10}{i}}$$ Now, can you convince yourself that these two are equivalent?

Since I don't see it anywhere, let me address this part of your question, rather than specific case:

how that can be true when all blue balls are identical and all black balls are identical.

Here, $^6C_i$ denotes the combination of $i$ different things from 6 different objects, doesn't it? How can that be used here?

Let's make a simple thought experiment. Let's say that rather than having indistinguishable white and black balls you have 10 balls numbered 1 to 10. Balls numbered 1 to 6 are white and those numbered 7 to 10 are black.

Now you have 6 white balls and 4 black balls that are distinguishable. How does that change your probability?

The numbers on balls change nothing in probability as long as the only thing you're concerned is ball colour. The resulting probability has to be the same as if there were no numbers. We may simply ignore the numbers on the balls and follow the original problem. We still have 6 white balls and 4 black balls so the old approach holds. Whatever other way you count the probability it has to provide the same result.

But now your balls are distinguishable so you can apply methods specific to distinguishable balls, specifically use combination. The results, as shown earlier will be the same.

This is why to indistinguishable balls you can always apply approach "Let's assume the balls are distinguishable..."

Let me answer from a different direction.

The reason you must use the equations which treat the balls as distinguishable is that thousands of repeated experiments have shown that the statistical behavior of all macroscopic objects is that of distinguishable items. There is no "proof" of this, since it's a physical reality. (by comparison, Bosons often do act as indistinguishable, leading to cool stuff in the super-cold regime).

Now, in all statistics, the wording of the question is critical (see arguments about the Monty Hall problem!). If your teacher says "Assume all balls of a given color are truly indistinguishable," then it's a theoretical problem and you basically divide by the number of permutations of distinguishable objects. But unless indistinguishability is specifically given as a premise, objects are distinguishable.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.