Probability - two possible points of view?

[Lotto]

TL;DR

We have 10 ball, 6 are blue, 4 are red. What is the probability that if we take two balls, the ball will have the same colour?

I know it is a simple problem, but I am confused by the fact that we can look at the problem from two points of view.

In general, we differentiate between the balls of the same colour, I mean we could mark them B1, B2, ... and B6, the same way R1, R2, R3 and R4. Then we could say that the probability that we choose two blue balls can be $\frac{6 \cdot 5}{10 \cdot 9}$, we just have possibilities such as (B1,B2) or (B2, B1) or (B2, B3) etc. So the position of the balls we take matters. In the same way we can calculate the probability for the red balls.

But we can also calculate the probability for the blue balls as $\frac{{6}\choose{2}}{{10}\choose{2}}$, so here the positions don't matter - B1, B2 and B2, B1 is the same possibility.

So, why can we look at the problem from two points of view? In one case, positions matter, in the second case they don't. Why? Let's say this is a case when we take the balls one by one. But if we arranged it so that we chose the two balls simultaneously, the probability would be still the same, because we would use the binomial coefficient again.

 

[pbuk]

In general, we differentiate between the balls of the same colour, I mean we could mark them B1, B2, ... and B6, the same way R1, R2, R3 and R4. Then we could say that the probability that we choose two blue balls can be $\frac{6 \cdot 5}{10 \cdot 9}$, we just have possibilities such as (B1,B2) or (B2, B1) or (B2, B3) etc.

Edit: [Yes, there are of course $\cancel{{{6}\choose{2}} = 6 \cdot 5 = 30}$ such possibilities.]

Yes, there are of course $6 \cdot 5 = 30$ such possibilities.

So the position of the balls we take matters.

Why do you say that? The position of the balls we take does NOT matter because for every $(B_i,B_j)$ possibility we have counted we have also counted $(B_j,B_i)$ within the total of 30.

 

[PeroK]

So, why can we look at the problem from two points of view? In one case, positions matter, in the second case they don't. Why? Let's say this is a case when we take the balls one by one. But if we arranged it so that we chose the two balls simultaneously, the probability would be still the same, because we would use the binomial coefficient again.

The probability of getting two blue balls is the same, regardless of whether we consider B1-B2 different from B2-B1 or not.

 

[Ibix]

So, why can we look at the problem from two points of view? In one case, positions matter, in the second case they don't. Why?

Your answer is the probability that they're both blue, by the way, not the probability that they're the same colour.

The reason you can look at it both ways is hinted at by the fact that the 2! cancels out in the second expression for probability. Caring about ordering doubles the number of ways of drawing two blue balls, yes, but also doubles the number of ways of drawing two balls overall.

 

[Lotto]

Your answer is the probability that they're both blue, by the way, not the probability that they're the same colour.

The reason you can look at it both ways is hinted at by the fact that the 2! cancels out in the second expression for probability. Caring about ordering doubles the number of ways of drawing two blue balls, yes, but also doubles the number of ways of drawing two balls overall.

Or course I have to also add the probability for the red balls, I didn't say that $\frac 13$ is the final result.

 

[Lotto]

Yes, there are of course ${{6}\choose{2}} = 6 \cdot 5 = 30$ such possibilities.


Why do you say that? The position of the balls we take does NOT matter because for every $(B_i,B_j)$ possibility we have counted we have also counted $(B_j,B_i)$ within the total of 30.

Isn't ${{6}\choose{2}}$ 15?

 

[pbuk]

Isn't ${{6}\choose{2}}$ 15?

Good catch - I have edited my post.

 

[sophiecentaur]

TL;DR Summary: We have 10 ball, 6 are blue, 4 are red. What is the probability that if we take two balls, the ball will have the same colour?

So, why can we look at the problem from two points of view?

You can!
When I first learned GCE Statistics we were introduced to the concepts of Permutations and Combinations. It seems to me that your question revolves about this. The probability of a result is the ratio of the possible wanted result to all possible results.
There are many hits from a search but I found this link which could be what you need. Sorry if it's too basic.