TIFR CSE 2012 | Part A | Question: 20

Question

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?

• A. $9/10$
• B. More than $9/10$ but less than $1$.
• C. Less than $9/10$ but more than $0$.
• D. $0$
• E. $1$

Comments

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..

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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.

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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

Answer 1

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 

Comments

Can we think of it in this way: We have drawn 100 balls, but we don't know what was the color of those balls. So for us, the probability is still the constant calculating before the experiment of drawing of 100 balls was conducted(9/10). Hence, now the probability is 9/10.

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Even if we randomly draw 999 balls from the bag, then also the probability of next ball(last one) to be balck is 9/10.

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but since we are removing the  balls then prob will vary everytime. And in the worst case when we draw all 100 white balls p(101 being black )=1 and when all 100 are blacks then p(black)=800C1

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What sorcery is this! Not in my best dreams would have I been able to hink in this way. I am really stunned and depressed and interested at the same time.

Answer 2

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

Answer 3

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 = $\dfrac{^{999} C_{100}}{^{1000} C_{100}}$

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

                                                 = $\dfrac{9}{10}$

Comments

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Answer 4

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, $^{1000}C_{900}$. 

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

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