A box holds 6 red balls and 6 blue balls. Six balls are drawn at once. What is the probability that at least 2 of them are red?

Step-by-step solution

1. 1

   Count the sample space. The balls are drawn all at once, so order does not matter and every set of $6$ balls out of the $12$ is equally likely:

   $\binom{12}{6}=924.$

2. 2

   Use the complement instead of five separate cases. "At least $2$ red" covers $2,3,4,5,6$ reds — five counts to compute. Its complement covers only two: $0$ reds or $1$ red. So compute

   $P(\text{at least }2\text{ red})=1-P(0\text{ red})-P(1\text{ red}).$

   Whenever "at least $k$" leaves fewer cases on the other side, the complement is the shorter road.

3. 3

   Count the draws with no red ball. All six drawn balls must be blue, and there are exactly $6$ blue balls, so there is only one such set:

   $\binom{6}{0}\binom{6}{6}=1\cdot 1=1.$

4. 4

   Count the draws with exactly one red ball. Choose $1$ of the $6$ reds and $5$ of the $6$ blues; the two choices are independent, so multiply:

   $\binom{6}{1}\binom{6}{5}=6\cdot 6=36.$

   Together the complement contains $1+36=37$ of the $924$ possible draws.

5. 5

   Subtract to get the answer.

   $P=1-\frac{37}{924}=\frac{924-37}{924}=\frac{887}{924}\approx 0.9600.$

   The fraction is already in lowest terms ($924=2^{2}\cdot 3\cdot 7\cdot 11$ shares no factor with $887$, which is prime). A brute-force enumeration of all $924$ subsets counts $887$ with two or more reds, confirming the result — and a probability near $96\%$ makes sense, since drawing half the box while avoiding almost all of one colour is very unlikely.