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

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

A bag contains blue and red balls. Two balls are drawn randomly without replacement. The probability of selecting a blue and then a red ball is 0.3. The probability of selecting a blue ball in the first draw is 0.6. What is the probability of drawing a red ball, given that the first ball drawn was blue?

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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 $\frac{1}{2}=0.5$.

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

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