A box holds 5 red balls and 7 blue balls. Five balls are drawn at random without replacement. What is the probability that at least one blue ball is drawn?

Step-by-step solution

1. 1

   Count the sample space. There are $5+7=12$ balls and the draw is unordered and without replacement, so the number of equally likely selections is $\binom{12}{5}=\frac{12\cdot11\cdot10\cdot9\cdot8}{5!}=\frac{95040}{120}=792.$

2. 2

   Switch to the complement instead of summing cases. "At least one blue" would need the blue counts $1,2,3,4,5$ added up - five separate terms. Its complement is the single event "zero blue", i.e. all five drawn balls are red, so the complement rule $P(\text{at least one blue})=1-P(\text{no blue})$ turns five computations into one.

3. 3

   Count the all-red selections. There are exactly $5$ red balls, so drawing $5$ reds means taking every red ball: $\binom{5}{5}\binom{7}{0}=1\cdot 1=1.$ That is why the complement is so small here - there is only one way to miss blue entirely.

4. 4

   Assemble the probability. $P(\text{no blue})=\frac{1}{792},\qquad P(\text{at least one blue})=1-\frac{1}{792}=\frac{792-1}{792}=\frac{791}{792}.$

5. 5

   Interpret the number. $\tfrac{791}{792}=0.998737\ldots\approx 99.87\%$. The intuition matches: blues outnumber reds and you are taking nearly half the box, so avoiding blue altogether is close to impossible.