23 votes 23 votes 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? $9/10$ More than $9/10$ but less than $1$. Less than $9/10$ but more than $0$. $0$ $1$ Probability tifr2012 probability conditional-probability + – makhdoom ghaya asked Oct 30, 2015 • retagged Nov 23, 2022 by Lakshman Bhaiya makhdoom ghaya 2.9k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Chaitrasj commented Dec 6, 2018 reply Follow Share How ratio will remain same throughout the event? Since we are picking the balls the number of balls left in bag is changing, so the ratio of black: white ball will change everytime we pick a ball.. 0 votes 0 votes ayush.5 commented Jun 21, 2020 reply Follow Share Rather than picking up all the 100 balls at once, if we pick 100 balls one by one(one ball at a time), then the probability will be less than 9/10, right? Because 9/10 is based on initial 900 black and 100 red balls. But while picking one by one, the decrease in number of black balls has been more than the decrease in number of red balls. So, by time the probability 9/10 will decrease. 0 votes 0 votes Exynos commented Feb 24, 2021 reply Follow Share Someone might get confused that why answer is not in range let me tell you with respect to “Theorem of Total Probability” (just a fancy word for something very intuitive) P(101th ball is black) = P[101th ball is black/(You draw 100 white balls and 0 black balls previously)] + P[101th ball is black/(You draw 99 white balls and 1 black balls previously)] + ……….. Now this is the reason that answer wont be in a range, also instead of doing so rigorous math its better to use expectation which is used in the below answer Hope this helps 3 votes 3 votes Please log in or register to add a comment.
0 votes 0 votes Among 1000 balls 900 black balls can be choosen in 9/10 ways so, Ans will be A srestha answered Oct 30, 2015 • edited Nov 6, 2015 by srestha srestha comment Share Follow See all 9 Comments See all 9 9 Comments reply Arjun commented Nov 5, 2015 reply Follow Share that is a wrong logic :( 3 votes 3 votes srestha commented Nov 6, 2015 reply Follow Share Could u plz tell me, exactly where I have done mistake? 0 votes 0 votes Arjun commented Nov 6, 2015 reply Follow Share 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. 0 votes 0 votes srestha commented Nov 6, 2015 reply Follow Share 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 0 votes 0 votes Arjun commented Nov 6, 2015 reply Follow Share 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. 0 votes 0 votes srestha commented Nov 6, 2015 i edited by srestha Nov 6, 2015 reply Follow Share ok then ans should be 9/10 , right? But then why they are asking about "101 ball is black". then ans will be same if they asked probability of getting one black ball 0 votes 0 votes Arjun commented Nov 6, 2015 reply Follow Share yes, answer is 9/10. And it is the same for even 800th draw. Question is just for checking the correct concept. 1 votes 1 votes srestha commented Nov 6, 2015 reply Follow Share ok, I have edited it yes , there will be some more case if I check individually, corrected now :) 0 votes 0 votes Learner_jai commented Oct 12, 2017 reply Follow Share @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)' please guide this 0 votes 0 votes Please log in or register to add a comment.