A box holds 6 red balls and 6 blue balls. Six balls are drawn at random without replacement. What is the probability that at least one is blue?

Step-by-step solution

1. 1

   Switch to the complement immediately. "At least one blue" would need the cases of exactly $1,2,3,4,5,6$ blue — six separate counts. Its complement is a single case: zero blue. So use

   $P(\text{at least one blue})=1-P(\text{no blue})$

2. 2

   Count the sample space. There are $6+6=12$ balls and order does not matter, so the number of possible six-ball selections is

   $\binom{12}{6}=\frac{12!}{6!\,6!}=924$

3. 3

   Count the favourable-to-the-complement outcomes. "No blue" means all six drawn balls are red — and there are exactly $6$ red balls, so the only way is to take every red ball:

   $\binom{6}{6}=1$

   This is why the answer comes out so close to $1$: there is a single bad selection out of $924$.

4. 4

   Assemble the probability.

   $P(\text{no blue})=\frac{1}{924},\qquad P(\text{at least one blue})=1-\frac{1}{924}=\frac{923}{924}$

5. 5

   Convert to a decimal and sanity-check. $\dfrac{923}{924}=0.998918\ldots\approx99.89\%$. That is believable: drawing half the box and somehow avoiding all six blue balls requires the one perfect red-only draw, so near-certainty is exactly what to expect ✓.

6. 6

   Cross-check by direct summation. Adding the hypergeometric probabilities for $1$ through $6$ blue gives $\frac{36+225+400+225+36+1}{924}=\frac{923}{924}$, matching the complement result ✓.