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There are $1000$ balls in a bag, of which $900$ are black and $100$ are white. I randomly draw $100$ balls from the bag. What is the probability that the $101$st ball will be black?

1. $9/10$
2. More than $9/10$ but less than $1$.
3. Less than $9/10$ but more than $0$.
4. $0$
5. $1$
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Can this question be modeled using any standard probability distrbution?

Here we are having a total of $1000$ Balls, out of which we firstly draw $100$ balls , and then $101^{st}$ ball..

Firstly we have to find expected number of white and black balls in drawn $100$ balls , as both can occur in 100 balls..

We have a situation like this:

Expected number of white balls =$n\times \dfrac{W}{N} = 100\times \left(\dfrac{100}{1000}\right) = 10$

Expected number of black balls =$n\times \dfrac{B}{N} = 100\times \left(\dfrac{900}{1000}\right) = 90$

So,  we have drawn $100$ balls($90$ black, $10$ white)

Left balls = ($810$ Black , $90$ White) = $900$ total

Now,

probability for ${101}^{st}$ ball to be black = $\dfrac{810}{900} =\dfrac{9}{10}.$

So, option (A) is Correct

edited

This is Probability with prediction type question.In this probability tells us how likely something is to happen in the long run.

We can calculate prob by looking at the outcomes of an experiment or by reasoning about the possible outcomes.

Now in this Qn given that the ratio of Black and White balls is 9:1.

It means we can predict that if I draw 1 ball then number of black balls =1 * 9/10

if I draw 2 balls then the number of black balls =2 * 9/10

...........................................................................

if I draw k balls then the number of black balls =k * 9/10

if I draw 100 balls then the number of black balls =100 * 9/10 = 90 and number of white balls =100 *1/10 =10

So after 100 balls drawn we have 900 - 90 =810 black balls and 90 white balls.

the probability that the 101st ball will be black = 810 /900 = 9/10.

## The correct answer is (A) 9 /10

Among 1000 balls 900 black balls can be choosen in 9/10 ways

so, Ans will be A
edited by
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that is a wrong logic :(
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Could u plz tell me, exactly where I have done mistake?
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You assumed the case that all 100 balls could have been white. But all 100 balls could have been black also. So, this approach is not correct- we have to consider the probability of a ball being white/back for the first 100 balls also.
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yes first 100 ball is white that is the worst case

In other case where from 1000 balls we are picking randomly that is with probability 9/10

I have taken both case as consideration
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yes, but your answer is w.r.t. special weightage to worst case. The same argument you used to say >9/10 can be used to say <9/10. Actually it just remains the same for all the balls.
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ok then ans should be 9/10 , right?

then ans will be same if they asked probability of getting one black ball
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yes, answer is 9/10. And it is the same for even 800th draw. Question is just for checking the correct concept.
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ok, I have edited it

yes , there will be some more case if I check individually, corrected now :)
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@srestha ma'am,  you have mentioned that you have considered the worst case scenario  of getting first 100 ball is white, but if we take this then we get probability of black ball in next draw will be 1(bz we only left with black ball)'

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