Conditional Probability Math Example 4

Solution

1. 1
   Use $P(\text{both blue} \mid \text{at least one blue}) = \frac{P(\text{both blue})}{P(\text{at least one blue})}$.
2. 2
   Count unordered pairs: total pairs $= \binom{10}{2} = 45$.
3. 3
   Both blue pairs: $\binom{4}{2} = 6$.
4. 4
   At least one blue pair = total pairs $-$ both red pairs $= 45 - \binom{6}{2} = 45 - 15 = 30$.
5. 5
   So the conditional probability is $\frac{6}{30} = \frac{1}{5}$.