Dark Mode

5,560 views

20 votes

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?

- $\dfrac{r}{r+b} \\$
- $\dfrac{r}{r+b+3}\\$
- $\dfrac{r+3}{r+b+3} \\$
- $\left( \dfrac{r}{r+b} \right) \left ( \dfrac{r+1}{r+b+1} \right) \left( \dfrac{r+2}{r+b+2} \right) \left( \dfrac{r+3}{r+b+3} \right)$

30 votes

Best answer

Ans : $\frac{r}{r+b}$

Exp :

**Assume** that **after** $i_{th}$ iteration, we have $r$ red balls and $b$ black balls.

So, probability of choosing red ball in $(i+1)_{th}$ iteration will be :

$=\frac{r}{r+b}$

So, Probability of choosing red ball in $(i+2)_{th}$ iteration will be :

(note that, in $(i+1)_{th}$ iteration, we could either pick black or red ball. )

$\frac{r}{r+b} \times \frac{r+1}{r+b+1} + \frac{b}{r+b} \times \frac{r}{r+b+1} $

$=\frac{r}{r+b}$

So, in the beginning we have $r$ red balls, and $b$ green balls. So, in every iteration, the probability of choosing a red ball will be $\frac{r}{r+b}$ .

A beautiful question!! Whatever iteration we take, even in $100_{th}$ iteration, probability of picking a red ball will be same.

Refer following link. This GATE question can be found here as it is, along with many possible variations.

http://www.stat.yale.edu/~pollard/Courses/600.spring08/Handouts/Symmetry%5BPolyaUrn%5D.pdf

@Deepakk Poonia (Dee) Sir, the way i interpreted from the wordings of the question as mentioned in above comments with regard to Option D being correct along with opt A..is it completely incorrect?

1

@arjun Sir,

Yes, we can.

Refer following link. This GATE question can be found here as it is, along with many possible variations.

http://www.stat.yale.edu/~pollard/Courses/600.spring08/Handouts/Symmetry%5BPolyaUrn%5D.pdf

2

11 votes

$\text{P(red ball after 3 insertions) = } \\\frac{r}{r+b}\times[\frac{r+1}{r+b+1}\times[\frac{r+2}{r+b+2}\times\frac{r+3}{r+b+3} + \frac{b}{r+b+2}\times\frac{r+2}{r+b+3}] + \\\frac{b}{r+b+1}\times[\frac{r+1}{r+b+2}\times\frac{r+2}{r+b+3} + \frac{b+1}{r+b+2}\times\frac{r+1}{r+b+3}]] + \\\frac{b}{r+b}\times[\frac{r}{r+b+1}\times[\frac{r+1}{r+b+2}\times\frac{r+2}{r+b+3} + \frac{b+1}{r+b+2}\times\frac{r+1}{r+b+3}] + \\\frac{b+1}{r+b+1}\times[\frac{r}{r+b+2}\times\frac{r+1}{r+b+3} + \frac{b+2}{r+b+2}\times\frac{r}{r+b+3}]]$

On solving you get $\frac{r}{r+b}$

0

5 votes

For questions like these, it's good to take some values of variables and then solve them. So, try out one or two different values of $r$ and $b$ for which all the four options are giving different values. I tried out with $r=b=1$, but then $(b)$ and $(d)$ are giving the same values. So, I tried $b=2$ and $r=1$, and got different values for all the four options. Used the tree method shown below to get the answer $\frac{1}{3}$ which is given by option $(a)$.