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Suppose that $X$ is chosen uniformly from $\{1,2,\ldots,100\}$ and given $X =x$, $Y$ is chosen uniformly from $\{1,2,\ldots,x\}. $Then $P(Y = 30)=$

  1. $\dfrac{1}{100}$
  2. $\dfrac{1}{100} \times \left(\dfrac{1}{30} + \ldots+\dfrac{1}{100}\right)$
  3. $\dfrac{1}{30}$
  4. $\dfrac{1}{100} \times \left(\dfrac{1}{1} + \ldots +\dfrac{1}{30}\right)$
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At first $X$ is chosen uniformly from $1-100$, $P($choosing $X)=\frac {1}{100}$

Now, $Y$ is chosen from the set$\{1,2....x\}$, so this set should definitely contain atleast $30$ elements from $1-30$, then only $Y=30$ can be chosen, $P(Y=30)=\frac{1}{30}$

We can expand the set to contain $31$ elements, then $P(Y=30)=\frac {1}{31}$

......

We can expand the set till it contains all the $100$ elements, then $P(Y=30)=\frac {1}{100}$

So, $P(Y=30)=\frac{1}{100}*({\frac{1}{30}}+{\frac{1}{31}}+{\frac{1}{32}}+......+{\frac{1}{100}})$

Correct Answer: $B$
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