# Drawing without replacement: why is the order of draw irrelevant?

I am trying to wrap my head around this problem:

Daniel randomly chooses balls from the group of $$6$$ red and $$4$$ green. What is the probability that he picks $$2$$ red and $$3$$ green if balls are drawn without replacement.

What I remember from my college days that the probability is found by this formula:

$$P(A)=\frac{\binom{6}{2}\binom{4}{3}}{\binom{10}{5}}=\frac{5}{21}$$

Is this correct? I am trying to understand why this works. Wouldn't probability depend on the order of balls drawn as the number of balls is changing after each draw? I get how we obtain numerator and denominator, I just feel that the probability should be dependent on the order. For example, the probability to pick red first is $$\frac{6}{10}$$ so the probability for the second draw becomes $$\frac{5}{9}$$ for red and $$\frac{4}{9}$$ for green. But if the first picked ball is green, the probability for the second draw becomes $$\frac{6}{9}$$ for red and $$\frac{3}{9}$$ for green. What am I missing?

• Usually the "nominator" is called numerator. – callculus Apr 18 at 12:12
• @callculus: yes, of, course, I need coffee :) – Vasya Apr 18 at 12:20
• The order doesn't matter because $P(A)$ is ultimately the sum of various conditional probabilities, and addition is both commutative and associative. – chepner Apr 18 at 14:04

## 4 Answers

If you took into consideration the order, you would get the same result but the calculation would be a bit more difficult:

• all possible sequences of $$5$$ balls respecting order (as if they were distinguishable): $$10\cdot 9\cdot 8\cdot 7 \cdot 6$$
• all possible selections of $$\color{red}{2}$$ out of $$\color{red}{6}$$ red balls: $$\color{red}{\binom{6}{2}}$$
• all possible selections of $$\color{green}{3}$$ out of $$\color{green}{4}$$ green balls: $$\color{green}{\binom{4}{3}}$$
• all possible arrangements of the selected $$\color{red}{2}+\color{green}{3}$$ balls: $$5!$$

All together $$\frac{\color{red}{\binom{6}{2}}\cdot \color{green}{\binom{4}{3}} \cdot 5!}{10\cdot 9\cdot 8\cdot 7 \cdot 6} = \frac{\color{red}{\binom{6}{2}}\cdot \color{green}{\binom{4}{3}}}{\frac{10!}{5!\cdot 5!}}= \frac{5}{21}$$

The probability of picking a red ball first and then a green ball is $$\frac{6}{10} \cdot \frac{4}{9}$$ The probability of picking a green ball first and then a red ball is $$\frac{4}{10} \cdot \frac{6}{9}$$ Notice that the numbers in the denominator are the same, while the numbers in the numerator are the same but in reverse order? Multiplication is commutative.

Another way of looking at this: we don't care about the process you go through in picking the balls, as long as it is fair: each possible outcome (i.e. each possible subset of $$5$$ of the $$10$$ balls, where we consider the balls as in principle distinguishable) has the same probability. If this is the case, you just need to count the number of outcomes that belong to the event you're considering, and divide by the total number of outcomes.

• Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective! – Vasya Apr 18 at 12:25

You can comprehend the calculation in a simpler way with smaller numbers.

Daniel randomly chooses balls from the group of $$3$$ red and $$2$$ green. What is the probability that he picks $$2$$ red and $$2$$ green if balls are drawn without replacement.

Indeed we have to regard the order. There are $$\frac{4!}{2!\cdot 2!}=6$$ ways to draw 2 red and 2 green balls:

$$\color{green}g\color{green}g\color{red}r\color{red}r, \color{green}g\color{red}r\color{green}g\color{red}r, \color{green}g\color{red}r\color{red}r\color{green}g, \color{red}r\color{green}g\color{green}g\color{red}r, \color{red}r\color{green}g\color{red}r\color{green}g, \color{red}r\color{red}r\color{green}g\color{green}g$$

Each way has the same probability: $$\frac{3}{5}\cdot \frac{2}{4}\cdot \frac{2}{3}\cdot \frac{1}{2} \quad (ggrr)$$

Multiplying with 6 (ways) we get $$6\cdot \frac{3}{5}\cdot \frac{2}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}=\frac{3}5=0.6$$

Using binomial coefficients we get $$\frac{\binom{3}{2}\cdot \binom{2}{2}}{\binom{5}{4}}=\frac{3\cdot 1}{5}=\frac35=0.6$$

And we get the same result.

There is a principle called "conservation of expected evidence" that says that if you have events A and B, then when you calculate the probability of A without knowing whether B happens, the result should be the same as the expected value of the probability over the possible results of B.

In this case, let A be the probability that the second ball is red, and B be the probability the first one is green. The principle says that P(A) = P(A|B)P(B)+P(A|~B)P(~B). That is, if you split A into two cases of A and B versus A and not B, the total probability should just be the probability of A. If you roll a die and flip a coin, the probability of getting a 1 one the die should change if you split it into P(die=1,coin=heads) plus P(die=1,coin=tails).

We have the following values for those probabilities:

P(A) = $$\frac6{10}$$
P(A|B) = $$\frac5 9$$
P(B) = $$\frac6{10}$$
P(A|~B) = $$\frac 6 9$$
P(~B) = $$\frac 4 {10}$$

So the equation is $$\frac6{10} = \frac5 9 \frac6{10}+\frac 6 9 \frac 4 {10}=\frac{30+24}{9*10} = \frac{54}{9*10}=\frac{9*6}{9*10}=\frac 6 {10}$$

If you have ten cards, 6 red and 4 green, and you shuffle them, would the probability of the first one being red be any different from the probability of the second one being red?