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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,A ball is again drawn at random.Show that the probability of drawing a white ball now does not depend on k.

Toolbox:
• n urn contain m white (W) and n Black (B) balls
• Let $E_1$= balls drawn is white in 1$^{st}$ draw
• $E_2$= balls drawn is Black in 1$^{st}$ draw
• A drawing a White ball in 2$^{nd}$ draw
• P(A)=P($A/E_1$)+P($E_2$)P($A/E_2$)
P($E_1)=\large\frac{m}{m+n}$
P($E_2)=\large\frac{n}{m+n}$
Now if first ball is drawn is White'k' white ball are added Total balls are m+n+k
P($A/E-1$)=P(getting white in 2$^{nd}$ / 1$^{st}$ ball drawn is'w')
=$\large\frac{m+k}{m+n+k}$
P($A/E-2$)=P(getting white in 2$^{nd}$ / 1$^{st}$ ball drawn is'b')
=$\large\frac{m}{m+n+k}$
P(A)=$\Large\frac{m}{m+n}$$\times$$\frac{m+k}{m+n+k}$$\times$$\frac{n}{m+n}$$\times$$\frac{m}{m+n+k}$
=$\Large\frac{m(m+n+k)}{(m+n)(m+n+k)}$
=$\large\frac{m}{m+n}$
this probability is independent of 'k'

edited Jun 4, 2013