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

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.


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


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.

Tags:

Probability

Probability Theory

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