# Probability of selecting a ball from a collection of indistinguishable balls

Suppose a box contains $$5$$ indistinguishable white balls and $$2$$ indistinguishable black balls. What is the probability that a selected ball from the box is white?

This question certainly appears trivial. If the balls were all distinguishable, the answer would have been $$5/7$$. However, what would the answer be if balls of the same colour are indistinguishable?

If the answer remains $$5/7$$, why is the answer independent of whether balls of the same colour are distinguishable or not?

• The answer remains 5/7 because we are supposing that the balls are indistinguishable after the selection. Distinguishability does not affect our selection. May 20, 2023 at 6:40
• Consider a box with 5 indistinguishable white balls and 2 indistinguishable black balls. Get seven labels numbered from 1 to 7, and place them on the balls such that each ball has a unique label. Ie "ball nr. 1, , ball nr. 4, etc." . Now all balls are distinguishable. But having a label placed on a ball does not affect the probability of it being black or white. Ie. the labels are ignored/irrelevant for our current problem. May 20, 2023 at 6:44
• @user3257842 If the balls were not labelled, there would be only one possible outcome in which the ball is white, since we have only one way of obtaining a white ball. Also the total number of possible outcomes would be two i.e either black or white. Thus, the indistinguishable and distinguishable cases appear inequivalent.
– Toba
May 20, 2023 at 6:55
• How do we know for sure that labelling the balls doesn't affect the probability? For instance labelling the Es in the word BEEN as E1 and E2 changes the number of possible arrangements of the letters in the word.
– Toba
May 20, 2023 at 7:06
• Labeling the balls doesn't affect the probability because we're only interested in the color attribute of the balls. Ie. in the final result we check the color, nothing else. The ball is either black or white. We ignore any other attribute that would allow us to distinguish a ball from another, such as a label. This is not true in your word example: $E_{1}E_{2}NB$ is counted as different from $E_{2}E_{1}NB$. The difference comes from the labels, therefore they affect the result. What we don't ignore are the relative quantities of balls possessing the relevant color attribute: 5W/2B. May 20, 2023 at 7:34

so P(white ball selected) (which was what the question asked) $$= \frac27$$ (and because you then worked out for black balls instead )
P(black ball selected) $$= \frac57$$
so you have$$B_1...B_5,\;and\; W_1,W_2$$ and you want to compute P(black ball selected)
P($$k{th}\;$$ black ball selected) $$= \frac17$$
P(some black ball selected) $$= P(B_1) +P(B_2) +...P(B_5)=\frac57$$