GATE CSE 2021 Set 2 | Question: 33

Question

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

Comments

https://projects.iq.harvard.edu/files/stat110/files/selected_solutions_blitzstein_hwang_probability.pdf

31. Page 5 (page 7 of Pdf)

 

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Please watch this video for better understanding of the questions https://youtu.be/ZEFXWtQ5jjo 

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I still don’t understand

Answer 1

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

Comments

@ani0135

choosing 1 ball from r red identical wall is $\binom{r}{1}$ and p(getting 1 red ball from r + b balls) is $\binom{r}{1}$/r+b 

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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.

@ani0135 ;

Choosing $1$ red ball from $r$ identical red ball can be in 1 way only. That’s correct, But it doesn’t have anything to do with the given question.

You draw a ball from a bag containing 10 identical balls(except their color). In which of the following case the probability of that ball being red is more??

1. Bag contains 9 Red balls, 1 Blue balls

2​​​​​. Bag contains 1 Red ball, 9 Blue balls

??

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@Deepak Poonia Sir,

I got this question already. But seeing that comment I just replied.

 

Answer 2

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

On solving you get $\frac{r}{r+b}$

Comments

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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@zxy123 any resource to study this tree method in probability with which u have drawn this tree and further solved ??

Answer 3

Guys if anything is wrong let me know

Answer 4

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 $(a)$.

Comments

yeah that can be but if two option matches with same value then you have take another values :)