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An urn contains $m$ WHITE and $n$ BLACK balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same color as that of the ball drawn. If now a ball is drawn, the probability that it is WHITE is?

  1. $(m+k)/(m+n+k)$
  2. $(n+k)/(m+n+k)$
  3. $m/(m+n+k)$
  4. $m/(m+n)$
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If the $1^{st}$ drawn ball was $white$ then probability that $2^{nd}$ drawn ball is $white$ = $\left (\dfrac{m}{m+n}*\dfrac{m+k}{m+n+k} \right ) $ $\rightarrow (i)$

If the $1^{st}$ drawn ball was $black$ then probability that $2^{nd}$ drawn ball is $white$ = $\left (\dfrac{n}{m+n}*\dfrac{m }{m+n+k}\right ) $ $\rightarrow (ii)$

$(i)+(ii) = \left (\dfrac{m}{m+n}*\dfrac{m+k}{m+n+k} \right )  +\left (\dfrac{n}{m+n}*\dfrac{m}{m+n+k}\right )=\left (\dfrac{m}{m+n}\right ) *\left (\dfrac{m+n+k}{m+n+k}\right ) =\left (\dfrac{m}{m+n}\right )$

Option D is correct.

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