1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ How do you count $P(A)$? Is the last 3 from the orders WBB, BWB and BBW? $\endgroup$
    – Henno Brandsma
    Jul 11, 2017 at 16:52

3 Answers 3

Reset to default
0
$\begingroup$

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.

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 $1/2$?)

$\endgroup$
1
  • $\begingroup$ No, the probability depends on how the lottery is played. Thanks, i understand better now. $\endgroup$
    – Little Rookie
    Jul 12, 2017 at 5:05
0
$\begingroup$

It is okay to label and list all possible outcomes, but it becomes extremely tedious and even near impossible when you have large sets. When you label the balls you actually change the nature of the variable and thus requires a different solution route. Yet, you seem to understand this as both of your approaches were sufficient for this problem.

So to directly answer your question, there is no need to label the balls in these kinds of problems. In fact, I recommend using the approach you did for $P(B)$ as this will be the most useful in difficult counting/probability problems and in this particular case order does not matter.

$\endgroup$
5
  • $\begingroup$ As i have stated in the example of selecting 2 black balls from the 5 black balls, the number of ways would be $\dbinom{5}{2}$. By saying that the number of ways is $\dbinom{5}{2}$, isn't that we already accept that the black balls are distinguishable from one another? Otherwise, there would only be 1 way to select 2 black balls since any combination of 2 black balls is the same. $\endgroup$
    – Little Rookie
    Jul 11, 2017 at 16:37
  • $\begingroup$ @LittleRookie In your way of counting it would not matter if we had a 100 or 5 black balls? $\endgroup$
    – Henno Brandsma
    Jul 11, 2017 at 16:39
  • $\begingroup$ Combinations are for grouping things. We only care about the color of the ball, not the number you assign. When you use the combination $\dbinom{x}{y}$ you are acknowledging there are $x$ distinct number of balls. Whether you can distinguish one ball from another is irrelevant to the problem. Now if the problem asked you what is the probability that the black balls selected are labeled $1$ and $3$, then it would be different. Here we only care about color. $\endgroup$
    – MasterYoda
    Jul 11, 2017 at 16:46
  • $\begingroup$ @HennoBrandsma if i think all the black balls are identical, then there would only be one way to choose 2 black balls irregardless of the number of black balls provided in the first place (given that it is at least 2). $\endgroup$
    – Little Rookie
    Jul 11, 2017 at 16:54
  • $\begingroup$ Why should i think that the black balls are actually distinct and not identical? What should i think of the "black balls" ? $\endgroup$
    – Little Rookie
    Jul 11, 2017 at 16:55
0
$\begingroup$

It's normal to label the items and maybe assume an order. Otherwise you couldn't easily count things (by product rules etc.). In the end you just discard the labels (and the order, if need be). It doesn't matter to the probability whether we throw 2 marked dice or two unmarked ones, right? But the marked ones allow us to see that there really are 2 different ways to throw $(1,2)$, and so the probability to throw sum $3$ is twice that of sum $2$. Labelling shows us the real underlying truth.

Using labelling we show that there are $5 \choose 2$ ways to pick 2 objects out of 5, without order and no laying back. We can us this as a lemma in other more complicated problems.

So for the ball problem we have to pick one white ball and two blacks and the order per colour does not matter. So we have ${6 \choose 1}$ and ${5 \choose 2}$ respectively, and the total number of options without order is their product. The lemma provides us with ${11 \choose 3}$ total options. Hence the quotient.

$\endgroup$
7
  • $\begingroup$ By saying $\binom{5}{2}$, we deemed the black balls as distinct? So which means that i should not assume the black balls as identical unless it was stated? Since by the only information given is the colour (black) of the balls, it could be that they are of different sizes and shapes? $\endgroup$
    – Little Rookie
    Jul 11, 2017 at 17:28
  • $\begingroup$ You do assume them to be different first. Then you ignore the differences again. That last step justifies the indistinguishable balls part. If we didn't do that we'd way overcount $\endgroup$
    – Henno Brandsma
    Jul 11, 2017 at 18:06
  • $\begingroup$ Why do i ignore the differences again? $\endgroup$
    – Little Rookie
    Jul 12, 2017 at 5:06
  • $\begingroup$ @LittleRookie because in the end you only look at the colours not what balls exactly. $\endgroup$
    – Henno Brandsma
    Jul 12, 2017 at 5:41
  • $\begingroup$ Okay, so what you meant is that we deemed the balls as distinguishable with the known similarity of colour. Any differences or similarities they may have in common besides the colour does not matter to us. $\endgroup$
    – Little Rookie
    Jul 12, 2017 at 6:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .