Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
An urn contains m white and n black balls. A ball is drawn at random and is put back into urn along with k additional balls of the same color as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white?
An urn contains m white and n black balls. A ball is drawn at random and is put back into urn along with k additional balls of the same color as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white?
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.
Let W1 (B1) be the event that a white (a black) ball is drawn in the first draw and let W be the event that a white ball is drawn in the second draw. Then
P(W) = P(B1). P(W| B1) + P(W1). P (W| W1)
= n/m + n. m/m + n + k + m/m + n. m + k/m + n + k
= m(n + m + k)/(m + n) (m + n + k) = m/m + n