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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?

  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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5 Answers

Best answer
34 votes
34 votes

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:

HG

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 by
8 votes
8 votes

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

6 votes
6 votes
Just thinking in another way to solve this

Since there are 1000 balls and out which 100 are white balls. Consider that these 1000 balls are equivalent to 1000 array indexes starting from 1 upto 1000.

We need to fill all the array locations with white balls except $101^{st}$ location. That means there 999 locations which can be filled in $^{999} C_{100}$ ways.

Hence required probability = $\dfrac{^{999} C_{100}}{^{1000} C_{100}}$

                                                 = $\dfrac{999!}{100! * 899!}  X \dfrac{100! * 900!}{1000!}$

                                                 = $\dfrac{9}{10}$
3 votes
3 votes

To find the probability of $101^{th}$ ball be black can be given as the ratio of number of favourable cases where $101^{th}$ position is black ball to the total cases where $101^{th}$ can be anything (black or white).

$\underline{\textbf{Number of Favourable Cases}}$

We set the $101^{th}$ position ball as black, now we are left with $899$ black balls and $100$ white balls. We can choose $0$ of $899$ black balls and place it on the left of $101^{th}$ position, so to choose $0$ position(s) of $100$ we can do $^{100}C_0$, and for the leftover positions on left we can fill white balls, same thing we can do with leftover black balls (here, $899-0=899$) on right of $101^{th}$ position. Choose $899$ of $899$ positions and set black balls there, and for the rest of the positions we can fill white balls. 

One thing to note is, we are not concerned with arrangement of balls as they are identical within their class, hence just selecting the positions to place them will get the job done. 

So, when we choosed $0$ black balls, number of favourable cases are $^{100}C_0 \ ^{899}C_{899}$. Just like that if we choose $1$ black ball then we will have $^{100}C_1 \ ^{899}C_{898}$, for $2$ black balls $^{100}C_2 \ ^{899}C_{897}$ and so on.

$$\begin{aligned}\text{Total Favourable cases} &= ^{100}C_0 \ ^{899}C_{899} + ^{100}C_1 \ ^{899}C_{898} + \dots + ^{100}C_{100} \ ^{899}C_{799}\\
&= \sum_{x=0}^{100} \ ^{100}C_x \ ^{899}C_{899-x} \\
&= \sum_{x=0}^{100} \ ^{100}C_{100-x} \ ^{899}C_{x}  \\
&= ^{100+899}C_{100} \\
&= ^{999}C_{100}\\\end{aligned}$$

$\sum_{x=0}^{100} \ ^{100}C_x \ ^{899}C_{899-x} = ^{999}C_{100}$ using Vandermonde's identity

$\underline{\textbf{Total Cases}}$

Total cases we can find by selecting $900$ positions to place blacks balls of $1000$ positions. Hence, $^{1000}C_{900}$. 

$$Probability = \dfrac{^{999}C_{100}}{^{1000}C_{100}} = \dfrac{9}{10}$$

$\textbf{Option (A) is correct}$

Answer:

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