in Probability retagged ago by
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20 votes
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?

  1. $\dfrac{r}{r+b} \\$
  2. $\dfrac{r}{r+b+3}\\$
  3. $\dfrac{r+3}{r+b+3} \\$
  4. $\left( \dfrac{r}{r+b} \right) \left ( \dfrac{r+1}{r+b+1} \right) \left( \dfrac{r+2}{r+b+2} \right) \left( \dfrac{r+3}{r+b+3} \right)$
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4 Comments

answer may be D or none of them but your key is showing answer is A. how?
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@Arpit Somani, bro did you get any information regarding whether that D is also correct, as per the mentioned intuition in the below comments, D fares as well, dk whether r8 or wrong :(
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today i find the solution on GO and it is well expalined. till yesterday i thought this was an ambiguous question because of no proper solution available anywhere. option a is correct.
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4 Answers

30 votes
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

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4 Comments

Can we apply the principle of symmetry here?
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@ 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?

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@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

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In question, All red balls are identical. So to choose 1 ball out of r balls, no. of ways should 1 only.

Thus P(getting 1 red ball) should be 0.5 and should not depend on the r and b. As All balls are identical except for their colour.

Please correct me if I am wrong.
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11 votes
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 $\frac{r}{r+b}$

4 Comments

plz, can someone clarify me out what's the answer to this Question, coz it seems D is also a viable choice.
 

@Arjun Sir,

@ACCURACY, @zxy123
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only A is correct, rest all will get -0.66
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and for future refrence to all of you guys, if instead of just one extra ball we add n number of extra balls of same color, and asked for any $i^{th}$ ball being red or black, ans remains the same as of ball being first red or black respectively.
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7 votes
7 votes

Guys if anything is wrong let me know

5 votes
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 $(a)$.

Answer:

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