This is an example from a book on probability.
If 3 balls are “randomly drawn” from a bowl containing 6 white and 5 black balls,
what is the probability that one of the balls is white and the other two black?
Let $A$ be the event where the order of drawing balls mathers, $P(A) =\dfrac{6\cdot 5\cdot 4 \cdot 3}{11\cdot 10\cdot 9}$.
Let $B$ be the event where the order of drawing balls does not mather, $P(B)=\dfrac{\binom{6}{1} \binom{5}{2}}{\binom{11}{3}}$.
Both leads to the same probability, as $P(A) = P(B) \cdot \dfrac{3!}{3!}$.
My question is on the counting part of my above solution.
I have deemed the white balls as distinguishable among themselves, and so do the black balls.
For example, there are 6 white balls, each labelled from 1 to 6, and the black balls labelled from 1 to 5.
For example, there are $\dbinom{5}{2}$ ways to choose 2 black balls from the 5 black balls where order does not mather.
If i would to list all the possible outcomes, it would be $\bigg\{\{b_1,b_2\} , \{b_1,b_3\}, \{b_1,b_4\}, \{b_1,b_5\}, \{b_2,b_3\}, \{b_2,b_4\}, \{b_2,b_5\}, \{b_3,b_4\}, \{b_3, b_5\}, \{b_4,b_5\}\bigg\}$
So i have been counting in this manner since my first course in probability, which i have "label" or distinguish the balls of the same colour among themselves.
But what is the reason that this is the coorect way to count in the first place?
If i were to replace the white balls as men, and the black balls as women, then it is intuitive that the each man is different, and each woman is different. But this is a different scenario.
Best Answer
No, it is basically the same situtation. You are selecting from physically distinct objects which may share an identifying property.
Being unable to visually distinguish balls of the same colour does not mean that they are somehow the same ball.
When measuring the probability for an event, it is not sufficient to merely count the distinguishable outcomes in the sample space; you have to weigh the probability for each of those outcomes.
( When buying a ticket in a lottery there are two distinct outcomes: either you win xor you lose. Is the probability that you win usually equal to $1/2$?)