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Home  >>  CBSE XII  >>  Math  >>  Probability
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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.

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  • 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'

 

answered Mar 1, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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