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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?

  1. $9/20$
  2. $9/19$
  3. $1/2$
  4. $10/19$
  5. None of the above
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15 votes
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, $B$.
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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) = $\frac{\binom{18}{10}}{\binom{19}{10}}$ = $\frac{9}{19}$

0 votes
0 votes
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
Answer:

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