an urn contains m white balls ad n black balls,a ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn.a ball is again drawn at random,the probability of drawing a white ball is?
a)increases with k
b)does not depend on k
c)decreases with k
d)cannot be determined without additional information

• In the 1st turn prob. of drawing a white ball= m/(m+n)& that of a black ball= n/(m+n).
  
  If white ball is drawn in 1st turn,then prob.of drawing a white ball =(m/(m+n))*((m+k)/(m+n+k)).
  
  If black ball is drawn in 1st turn,then prob.of drawing a white ball =(n/(m+n))*(m/(m+n+k)).
  
  So prob. of drawing a white ball=((m/(m+n))*((m+k)/(m+n+k)))+((n/(m+n))*(m/(m+n+k)))= m/(m+n)i.e does not depend on k (Ans - b)
  
• 10 years agoHelpfull: Yes(39) No(3)
• ans: cannot be determined without additional information
  
  previously a ball is drawn and replaced (colour unknown- either black or white)
  
  k additional balls represents same colour as previously drawn ball - can be black or white
  
  so additional info is required
• 10 years agoHelpfull: Yes(13) No(3)
• what is the right answer
  
• 10 years agoHelpfull: Yes(3) No(0)
• ans: "decreases with k"
  
  
  as the no of ball increases, the denominator part of our soln. also increases. Hence the overall probability value decreases.
• 10 years agoHelpfull: Yes(2) No(10)
• previous one is incorrect....
  
  Case 1: firstly white ball is taken out and then agn white :
  prob. = m/(m+n)*(m+k)/(m+n+k)
  Case 2: firstly black ball is taken out and then white :
  prob. = n/(m+n)*m/(m+n+k)
  adding both cases :
  prob. obtaind = m/(m+n)
  independant of k...hence option b
• 10 years agoHelpfull: Yes(2) No(0)
• cannot be determined
  if first ball drawn is black den probability decreases with increase of k otherwise probability increases.
  hence probability depends on first draw.
• 10 years agoHelpfull: Yes(1) No(1)
• correct ans :b)
• 10 years agoHelpfull: Yes(0) No(0)
• Case 1: firstly white ball is taken out and then agn white :
  prob. = m/(m+n)*m/(m+n+k)
  Case 2: firstly black ball is taken out and then white :
  prob. = n/(m+n)*m/(m+n+k)
  adding both cases :
  prob. obtaind = m/(m+n+k)
  thus as k increases , overall probability decreases...hence option c
• 10 years agoHelpfull: Yes(0) No(2)
• d...ans is d
• 10 years agoHelpfull: Yes(0) No(1)
• sankalan gain answer is correct
  
• 9 years agoHelpfull: Yes(0) No(0)
• admin sir u plese give the right answer....
  
• 9 years agoHelpfull: Yes(0) No(0)
• Option b) does not depend on k
• 9 years agoHelpfull: Yes(0) No(0)