Urn Ball Probability problem from Sheldon Ross' Probability Textbook Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that
(b) at least 2 red balls are withdrawn;
I understand that one way to do the problem is to subtract the probability that 0 or 1 red ball is drawn from 1 to get the probability of at least 2 red balls being drawn.
However, I don't understand why this method fails:
Pick 2 red balls, then pick any 5 balls of the remaining 44. For the position of the red balls to not matter, take $\binom{7}{2}$.
The total number of possibilities is $\binom{46}{7}$.
So P(at least 2 red) = $\frac{\binom{12}{2} \binom{44}{5} \binom{7}{2}}{\binom{46}{7}}$
This is greater than 1, so it's definitely wrong.
 A: There are two reasons why your approach is wrong, as explained below:

*

*Confusion about specifying that position does not matter.


*Overcounting.
The difficulty is that you haven't intuitively grasped the pertinent concepts.
Label the red balls as $R_1, R_2, \cdots, R_{12}$.
Label the blue balls as $B_1, B_2, \cdots, B_{16}$.
Label the green balls as $G_1, G_2, \cdots, G_{18}.$
You are trying to compute the probability that of the $7$ balls taken, at least $2$ are red.
So, this probability will be expressed as
$$\frac{N}{D} ~: ~D = \binom{46}{7}. \tag1 $$
In (1) above, the denominator represents the total number of distinct ways of selecting $7$ balls out of the $46$ balls.
Your attempted enumeration was
$$N = \binom{12}{2} \times \binom{44}{5} \times \binom{7}{2}.$$
I will discuss the two mistakes separately:

$\underline{\text{Confusion about specifying that position does not matter}}$
Your computation of $D$ represents that you line up the $46$ balls in a row, and then choose any $7$ of them, regardless of the order that they are selected.
Your computation of $N$ includes the factor of $~\displaystyle \binom{7}{2}$.  Your reasoning is that this is supposed to adjust the numerator, $N$, so that order of selection in $N$ also does not matter.
The idea is right, but the remedy is wrong.  You are right that since order of selection is deemed irrelevant in the denominator, it must also be deemed irrelevant in the numerator.
However, after asserting that the number of satisfying ways of selecting the balls is $~\displaystyle \binom{12}{2} \times \binom{44}{5},~$ you then assert that the factor of $~\displaystyle \binom{7}{2}~$ is needed.
Why? 
Assuming that you were right (which you are not, as discussed in the next section) about the computation of $~\displaystyle \binom{12}{2} \times \binom{44}{5},~$ such a computation would already be consistent with the denominator's computation of $~\displaystyle \binom{46}{7}.$
The factor of $~\displaystyle \binom{7}{2}~$ relates to determining which of the $7$ ball-positions (i.e. first ball selected, second ball selected, ..., seventh ball selected) are red.  This relates to order of selection being regarded as relevant.  This is inconsistent with the $~\displaystyle \binom{46}{7}~$ computation.
So, the $~\displaystyle \binom{7}{2}~$ factor should be eliminated.

$\underline{\text{Overcounting}}$
Based on the discussion in the previous section, suppose that you alter your computation of $N$, so that it is now
$$N = \binom{12}{2} \times \binom{44}{5}. \tag2 $$
This is still wrong.
In order to compute $N$ correctly, you must identify each situation where at least $2$ of the $7$ balls are red.  Further, for each such situation, you must count that situation exactly once.
That is, you must not count the same situation more than once.
Consider the following satisfying distribution of the balls:
$$\text{Situation-1:} ~~R_1, R_2, R_3, B_1, B_2, B_3, B_4.$$
The $~\displaystyle \binom{12}{2}~$ factor represents that you are selecting $2$ of the $12$ red balls.  In Situation-1, which two of the $3$ red balls are the ones that are supposed to be represented by the $~\displaystyle \binom{12}{2}~$ factor?
That is, by your computation of
$$\binom{12}{2} \times \binom{44}{5}$$
you have that :

*

*Situation-1 is counted once when the two red balls are $R_1, R_2.$


*Situation-1 is counted once when the two red balls are $R_1, R_3.$


*Situation-1 is counted once when the two red balls are $R_2, R_3.$
So, Situation-1 is counted three times, instead of only being counted once.  So, you have over-counted Situation-1.  Similar considerations apply to any of the other satisfying situations that involve more than $2$ red balls.
So, the entire approach is flawed, and results in over-counting.

$\underline{\text{Correct Enumeration}}$
Although not specifically requested by your posted question, it is relevant to show how the probability should be counted.
Let $f(n)$ denote the number of ways that $\color{red}{\text{exactly}}$ $n$ red balls might be selected, from the $7$ balls taken.  Here, $n$ is any element in $\{0,1,2,3,4,5,6,7\}$.
Then,
$$f(n) = \binom{12}{n} \times \binom{34}{7 - n}.$$
That is, there are exactly $~\displaystyle \binom{12}{n}~$ ways of selecting $n$ red balls from the $12$, and then there are exactly $~\displaystyle \binom{34}{7-n}~$ ways of selecting $(7-n)$ non-red balls from the $34$ non-red balls.
As you indicated in your posting, one approach to computing the probability would be
$$1 - \frac{f(0) + f(1)}{\binom{46}{7}}.$$
The alternative approach to computing the probability is
$$\frac{f(2) + f(3) + \cdots + f(7)}{\binom{46}{7}}.$$
A: Your method overcounts all cases where there are more than two red balls drawn (before accounting for order). For example, restricted to the case where five red balls are drawn, you overcount the number of cases by $\binom53=10$.
When accounting for order you should also be dividing by the number of equivalence classes rather than multiplying, which explains why your obtained result is larger than $1$. But a factor of $\binom72$ to account for order is irrelevant here since the draws are ultimately unordered; the size of the equivalence classes here are not equal across different numbers of red balls drawn – you cannot use a single divisor and have to consider each number of red balls drawn separately.
A: You don't want to handle all the possible red cases that are ${\ge2}$. Instead, we should look at ${1-P(N_{red}<2)=1-P(N_{red}=0)-P(N_{red}=1)}$. So, all that's left to do is find the probability that we don't draw a red or that we draw exactly ${1}$ red.
As you've stated, the number of ways to choose any 'ole 7 balls without replacement is ${46 \choose 7}$.
Now, there are ${34}$ non-red balls. So the number of ways you could choose ${N_{red}=0}$ is ${34 \choose7}$, the probability of which is ${P(N_{red} = 0)=\frac{34 \choose 7}{46 \choose 7}}$.
There are 12 red balls, in order to choose exactly 1 then you must choose exactly 6 non-red balls. The number of ways to do this is ${{12 \choose 1}{34 \choose 6}}$, the probability of which is ${P(N_{red}=1)=\frac{{12 \choose 1}{34 \choose 6}}{46 \choose 7}}$.
Putting it all together, we have:
$${P(N_{red} \ge 2)=1-P(N_{red}<2)=1-P(N_{red}=0)-P(N_{red}=1)=1-\frac{34 \choose 7}{46 \choose 7}-\frac{{12 \choose 1}{34 \choose 6}}{46 \choose 7}}$$
