A bag has \(r\) red balls and \(b\) black balls. In a trial, a ball is randomly drawn, its colour is noted, and the ball is placed back into the bag along with another ball of the same colour. A sequence of four such trials is conducted. Which one of the following choices gives the probability of drawing a red ball in the fourth trial?
Show Hint
- \( \dfrac{r}{r+b} \)
- \( \dfrac{r}{r+b+3} \)
- \( \dfrac{r+3}{r+b+3} \)
- \( \left(\dfrac{r}{r+b}\right)\left(\dfrac{r+1}{r+b+1}\right)\left(\dfrac{r+2}{r+b+2}\right)\left(\dfrac{r+3}{r+b+3}\right) \)
The Correct Option is A
Solution and Explanation
Step 1: Identify the process.
The experiment follows a Polya's urn model, where after each draw, a ball of the same colour is added back to the bag.
Step 2: Use symmetry of Polya's urn.
In Polya's urn scheme, the probability of drawing a red ball at any trial remains equal to the initial fraction of red balls in the bag.
Step 3: Apply to the fourth trial.
Thus, the probability of drawing a red ball in the fourth trial is
\[
\frac{r}{r+b}.
\]
% Final Answer
Final Answer: \[ \boxed{\dfrac{r}{r+b}} \]
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