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

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