Probability of drawing a red ball, given that the first ball drawn was blue?

Let $n$ be the total number of balls in the bag, $b$ be the number of blue balls and $r$ be the number of red balls. Note that $b+r=n$. The probability of selecting a blue and then a red ball is
\[(1)         \frac{b}{n} \times \frac{r}{n-1}=0.3.\]
The probability of selecting a blue ball in the first draw is 
\[ (2)     \frac{b}{n}=0.6.\]
Combining (1) and (2) we get 
\[\frac{r}{n-1}=\frac{1}{2}.\]
Note that the left hand side ($\frac{r}{n-1}$) is the probability of drawing a red ball, given that the first ball drawn was blue. Thus the answer is $\frac{1}{2}=0.5$.