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?
Best Answer
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.