A bag has \(r\) red balls and \(b\) black balls. All balls are identical…
2021
A bag has \(r\) red balls and \(b\) black balls. All balls are identical except for their colours. In a trial, a ball is randomly drawn from the bag, its colour is noted and the ball is placed back into the bag along with another ball of the same colour. Note that the number of balls in the bag will increase by one, after the trial. 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?
Answer: A. \(\dfrac{r}{r+b} \) — ConceptIn a Pólya urn, the drawn ball is replaced and one additional ball of the same colour is added. The colour proportion for the next draw is therefore…
- A.
\(\dfrac{r}{r+b} \) - B.
\(\dfrac{r}{r+b+3}\) - C.
\(\dfrac{r+3}{r+b+3} \) - D.
\(\left( \dfrac{r}{r+b} \right) \left ( \dfrac{r+1}{r+b+1} \right) \left( \dfrac{r+2}{r+b+2} \right) \left( \dfrac{r+3}{r+b+3} \right)\)
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Show answer & explanation
Correct answer: A
Concept
In a Pólya urn, the drawn ball is replaced and one additional ball of the same colour is added. The colour proportion for the next draw is therefore random after each trial.
However, its expected value is invariant: the unconditional probability of a colour on every draw equals that colour’s initial proportion.
Application
Let p0 = r/(r+b), the initial red proportion, and let Xn be the number of red balls after n trials.
Conditioned on the first n outcomes, the chance of red on trial n+1 is Xn/(r+b+n). Hence the unconditional chance is E[Xn]/(r+b+n).
If every earlier draw has marginal red probability p0, then E[Xn] = r + n·p0, because each red draw adds exactly one red ball.
Therefore P(red on trial n+1) = (r+n·p0)/(r+b+n). Substituting p0 = r/(r+b) simplifies this expression to p0, completing the induction.
For the fourth trial, n=3, so P(red on trial 4) = (r+3r/(r+b))/(r+b+3) = r/(r+b).
Cross-check and contrast
Condition on the first three draws. Each possible history gives a fourth-draw fraction, and averaging those fractions over their own probabilities returns the same initial red proportion.
r/(r+b+3) is conditional on the first three draws being black.
(r+3)/(r+b+3) is conditional on the first three draws being red.
The product of four successive fractions is the joint probability that all four draws are red, not a one-draw marginal probability.
Result
The probability of drawing a red ball in the fourth trial is r/(r+b).