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?
- A.
\(\dfrac{r}{r+b} \\\) - B.
\(\dfrac{r}{r+b+3}\\\) - C.
\(\dfrac{r+3}{r+b+3} \\\) - D.
\(\left( \dfrac{r}{r+b} \right) \left ( \dfrac{r+1}{r+b+1} \right) \left( \dfrac{r+2}{r+b+2} \right) \left( \dfrac{r+3}{r+b+3} \right)\)
Attempted by 3 students.
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Correct answer: A
Answer: The probability of drawing a red ball in the fourth trial is r/(r+b).
Reason (induction using expected counts):
Base case: On the first trial the probability of red is r/(r+b).
Inductive hypothesis: Assume that for each trial up to n the marginal probability of drawing red is r/(r+b).
Let R_n be the number of red balls after n trials. Each trial that results in red increases R_n by 1, so by linearity of expectation,
E[R_n] = r + sum_{i=1}^n P(red on trial i) = r + n * r/(r+b).
The probability of red on trial n+1 equals the expected fraction of red balls at that time:
P(red on trial n+1) = E[ R_n / (r+b+n) ] = E[R_n] / (r+b+n) = [r + n * r/(r+b)]/(r+b+n) = r/(r+b).
By induction this holds for every trial number; in particular for the fourth trial the probability is r/(r+b).
Remark: Some other expressions in the choices are conditional or sequence probabilities. For example, (r+3)/(r+b+3) is the probability the fourth draw is red given the first three draws were all red, and the product of factors equals the probability that all four draws are red. The unconditional (marginal) probability for the fourth draw is r/(r+b).