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 balls of the same colour as that of the

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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 balls of the same colour as that of the

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 balls of the same colour as that of the ball drawn. a ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k.

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Salim Ahmed Kawsar · Answer 1 · Feb 05, 2023 15:29:48

Let U={m white, n black balls}
`E_(1)`={First ball drawn of white colour}
`E_(2)`={First ball drawn of black colour}
and `E_(3)`={Second ball of white colour}
`therefore P(E_(1))=m/(m+n)and P(E_(2))=n/(m+n)`
Also, `P(E_(3)//E_(1))=(m+k)/(m+n+k)and P(E_(3)//E_(2))=m/(m+n+k)`
`thereforeP(E_(3))=P(E_(1))cdotP(E_(3)//E_(1))+P(E_(2))cdotP(E_(3)//E_(2))`
`=m/(m+n)cdot(m+k)/(m+n+k)+n/(m+n)cdotm/(m+n+k)`
`=(m(m+k)+nm)/((m+n+k)(m+n))=(m^(2)+mk+nm)/((m+n+k)(m+n))`
`=(m(m+k+m))/((m+n+k)(m+n))=m/(m+n)`
Hence, the probality of drawing a white ball does not depend on k.

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 balls of the same colour as that of the

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