An urn contains 5 red and 7 green balls. A ball is drawn at random and its color is noted. The ball is placed back into the urn along with another ball of same color. The probability of getting a red ball in the next draw is _____.
My approach:
I think both of these events are independent and hence probability of drawing a red ball random remains same which is equal to $\frac{5}{12}$.
is it correct?
Best Answer
\begin{align} & \Bbb P(\text{2nd draw red}) \\ = & \Bbb P(\text{2nd draw red} \; | \; \text{1st draw red}) \cdot \Bbb P(\text{1st draw red}) + \\ & \Bbb P(\text{2nd draw red} \; | \; \text{1st draw green}) \cdot \Bbb P(\text{1st draw green}) \\ = & \frac{6}{13} \cdot \frac{5}{12} + \frac{5}{13} \cdot \frac{7}{12} \end{align}