Question

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?

• A. \( \dfrac{r}{r+b} \)
• B. \( \dfrac{r}{r+b+3} \)
• C. \( \dfrac{r+3}{r+b+3} \)
• D. \( \left(\dfrac{r}{r+b}\right)\left(\dfrac{r+1}{r+b+1}\right)\left(\dfrac{r+2}{r+b+2}\right)\left(\dfrac{r+3}{r+b+3}\right) \)

Hint:

In Polya's urn model, the probability of drawing a colour remains invariant across trials.

Answer 1

The Correct Option is A

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: \[ \boxed{\dfrac{r}{r+b}} \]