TIFR CSE 2018 | Part A | Question: 15

Question

Suppose a box contains $20$ balls: each ball has a distinct number in $\left\{1,\ldots,20\right\}$ written on it. We pick $10$ balls (without replacement) uniformly at random and throw them out of the box. Then we check if the ball with number $\text{“1"}$ on it is present in the box. If it is present, then we throw it out of the box; else we pick a ball from the box uniformly at random and throw it out of the box.

What is the probability that the ball with number $\text{“2"}$ on it is present in the box?

• A. $9/20$
• B. $9/19$
• C. $1/2$
• D. $10/19$
• E. None of the above

Comments

$\frac{\binom{18}{10}}{\binom{20}{10}} + \frac{\binom{18}{9}}{\binom{20}{10}}*\frac{9}{10}$ = $\frac{9}{19}$

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nitish explain how you calculate

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A = {1,2,3,4.............20}

case 1:probability of slecting 10 balls excluding 1 and 2 = $\binom{18}{10}$ /$\binom{20}{10}$, now there wil be definately ball '1' with probability =1, remove it, and now definately there is ball '2' with probabilty=1,

so overall probailty for case 1= ($\binom{18}{10}$ /$\binom{20}{10}$)*1*1 = $\binom{18}{10}$ /$\binom{2}{10}$

case 2:probability of selecting 10 balls including ball '1' and excluding ball 2 = $\binom{18}{9}$ /$\binom{20}{10}$, now there are 10 balls left, and ball '1' is not present in urn, so we will randomly select one ball and throw out, therefore we can choose 1 ball out of 10(excluding ball '2') with prob = $\frac{9}{10}$

so overall probability for case 2 = ($\binom{18}{9}$ /$\binom{20}{10}$)*$\frac{9}{10}$

prob(case 1) + prob(case 2) = $\binom{18}{10}$ /$\binom{2}{10}$ + ($\binom{18}{9}$ /$\binom{20}{10}$)*$\frac{9}{10}$ =$\frac{9}{19}$

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cool.

Answer 1

Required probability has $2$ parts.

a) When both $1$ and $2$ remain after $10$ are chosen. Later $1$ will be chosen with probability $1$ and hence, $2$ will remain.

b) When $1$ is thrown along with some other $9$, and $2$ remains. Later we have to multiply with probability that $2$ will not be thrown.

$P$(2 remains after all the events) = $P$(1 and 2 are not thrown out) + $P$(1 is thrown out , 2 remains) * $P$(from remaining 10 2 is not chosen)

$\rightarrow$ $\frac{_{10}^{18}\textrm{C} + _{9}^{18}\textrm{C} * \frac{9}{10} }{ _{10}^{20}\textrm{C}} $

$\rightarrow \frac{9}{19}$

Hence, $B$.

Comments

@Shivansh

$\frac{\binom{18}{9}}{\binom{19}{9}}*\frac{9}{10}$, this is not correct,

if you are choosing ball '1' with some other '9' balls then also total ways of selecting 10 balls will remain $\binom{20}{10}$,

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then what is the meaning of making sure that '1' is defintely thrown out.

It is this condition that is limiting the total no. of ways.

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@Joshi_nitish, you are right.

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ok. tell me one thing..

you have 10 numbers{1,2,3...10}. you have to draw 2 numbers randomly, what is probability that one number is definately 1?

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it will be $$\frac{\binom{9}{1}}{\binom{10}{2}}$$ rt?

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yes, exactly this was what i was trying to tell you.

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Can someone easily explain this question? I am not getting anyone of the answer

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I think this answer is wrong. The denominator in the probability cannot be $\binom{20}{10}$

We have two cases in the event here.

Case1) If in the first 10 draws if we have thrown out 1 then we go for randomly picking again (this will be the 11th time) from the remaining 10 balls and throw out the randomly selected ball.

For this the number of ways = $\binom{19}{9}$ $\times$ $\binom{10}{1}$

Case2) If in the first 10 draws if we did not get 1 then the remaining 10 balls in the box will definitely have a '1', so we need not go for random picking again. We just have to pick the 1 left in the box and throw it out.

For this the number of ways = $\binom{19}{10}$ $\times$ $\binom{1}{1}$

Now, we will calculate the number of ways of randomly choosing so that 2 remains in the box.

Even for this we have 2 cases.

Case1) If both '1' and '2' are not thrown out after the first 10 random pickings. In this case we need not go for 11th random picking.

For this the number of ways = $\binom{18}{10}$ $\times$ $\binom{1}{1}$

Case2) If '1' is thrown out and '2' is not thrown out in the first 10 random pickings. This case requires 11th random picking, even here we will not be choosing '2' to be thrown out.

For this the number of ways = $\binom{18}{9}$ $\times$ $\binom{9}{1}$

 

So, finally the required probability = [$\binom{18}{10}$ $\times$ $\binom{1}{1}$ + $\binom{18}{9}$ $\times$ $\binom{9}{1}$] $\div$ [$\binom{19}{10}$ $\times$ $\binom{1}{1}$ + $\binom{19}{9}$ $\times$ $\binom{10}{1}$]

= (9/19) .....(I have skipped numerical calculations here :P)

 

Please let me know if anything is incorrect. :)

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@joshi_nitish

I think the selected answer is incorrect (or maybe I'm making some mistake, but I'm not able to recognize it)

The solution can be written as

Case 1: [C(18,10)] / [C(20,10)] -----------> We don't just pick 10 times, we even go for picking the 11th time when we cannot find '1' remaining in the box. But this is not considered here. The denominator is considering only 10 picks no matter what.

Case 2: [C(18,9) * C(9,1)] / [C(20,10) * C(10,1)] -----------> We don't pick for the 11th time if in the remaining balls we have '1'. But here in the denominator it is considering 11th random pick no matter what.

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But the interesting thing is that both methods are giving same answer 9/19. But how? :O

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What I have used is this:

Probability = Total number of favorable outcomes $\div$ Total number of outcomes of the event

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https://math.stackexchange.com/a/3449208/545704

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Well, if MSE says it's correct, then it has to be correct, but @joshi_nitish 's approach is also correct I think.(Hiding my above comments.)

Answer 2

If we observe this question carefully we see that in both cases, ball-1 has to be taken out. So out of 11 balls we need to take out, ball-1 is fixed.

So our sample space becomes $\binom{19}{10}$.

Now ball-2 always needs to be present, therefore $\binom{18}{10}$ ways

P(ball-2 always present) = $\frac{\binom{18}{10}}{\binom{19}{10}}$ = $\frac{9}{19}$

Answer 3

In both the cases, “ball with no.1” is thrown out of the box.

the box remain with 19 balls, out of which 10 balls more are also thrown out.


we assume, “ball with no. 2” is among those 10 which is thrown out.

probability that the “ball with no. 2” is not present = 10/19


Required probability= 1-10/19   (ball no. 2 is present)

                    =9/19