Q). There are two bags. One bag contains 4 white and 2 black balls. Second bad contains 5 white and 4 black balls. 2 balls are transferred from first bag to second bag. Then one ball is taken from the second bag. The probability that the ball is white is ?

1) 42⁄165
2) 5⁄165
3) 48⁄165
4) 19⁄33

• Ans = d
  prof:
  (prob of trans 2 white ball and prob of taking 1 white ball from 2nd bag) 0r
  (prob of trans 2 black ball and prob of taking 1 white ball from 2nd bag) or
  (prob of trans 1 white ball & 1 black ball and prob of taking 1 white ball from 2nd bag)
  ==> ((4c2 * 7c1) + (2c2 * 5c1) + (4C1 * 2c1 * 6c1)) / (6c2 * 11c1)
  = 19/33
  
• 10 years agoHelpfull: Yes(53) No(2)
• (4c2*7c1+2c2*5c1+4c1*2c1*6c1)/6c2*11c1
  =95/165=19/33
• 10 years agoHelpfull: Yes(8) No(6)
• answer is d......(5c1*2c1 + 6c1*2c1*4c1 + 7c1*4c2)/(6c2*11c1)=95/165 =19/33...
• 10 years agoHelpfull: Yes(5) No(8)
• please elaborate your solutions.
• 10 years agoHelpfull: Yes(5) No(1)
• 6c2*11c1 came like this->
  From 6 balls in first bag we select 2 and as two balls transferred to other bag,number became 11 and from that select 1
• 10 years agoHelpfull: Yes(5) No(1)
• Ans = d
  prof:
  (prob of trans 2 white ball and prob of taking 1 white ball from 2nd bag) 0r
  (prob of trans 2 black ball and prob of taking 1 white ball from 2nd bag) or
  (prob of trans 1 white ball & 1 black ball and prob of taking 1 white ball from 2nd bag)
  ==> ((4c2 * 7c1) + (2c2 * 5c1) + (4C1 * 2c1 * 6c1)) / (6c2 * 11c1)
  = 19/33
  just check the above statments as we are doing "or" operation we should add all the "or" statments
  if the statments are such that if you encounter "and" then do mul of indivisual statments
• 10 years agoHelpfull: Yes(3) No(2)
• how did 6c2*11c1 came ?? kindly explain
• 10 years agoHelpfull: Yes(2) No(1)
• @RAMKUMAR
  
  We need to select(move) 2 balls out of (4+2) from bag1 == 6c2 ----->(1)
  
  and after adding those 2 balls to bag2 total no of balls in bag2= 5+4+2 = 11
  
  then we need to pick 1 ball out of 11 balls = 11c1 ----->(1)
  
  sooooo.... (6c2)*(11c1)
• 10 years agoHelpfull: Yes(1) No(2)
• Ans = d
  
  (prob of trans 2 white ball and prob of taking 1 white ball from 2nd bag)= 0r
  (prob of trans 2 black ball and prob of taking 1 white ball from 2nd bag) or
  (prob of trans 1 white ball & 1 black ball and prob of taking 1 white ball from 2nd bag)
  
  ==> ((4c2 * 7c1) + (2c2 * 5c1) + (4C1 * 2c1 * 6c1)) / (6c2 * 11c1)
  = 19/33
  
  Here 6c2*11c1 is taken for all favourable cases i.e from 1st bag 2 balls are drawn in total from 6 and finally from 2nd bag 1 out of 9+2=11 is drawn .
  
• 9 years agoHelpfull: Yes(1) No(1)
• plz elavorated....i dont getted
• 10 years agoHelpfull: Yes(0) No(1)
• explain the answers plzzz...
• 10 years agoHelpfull: Yes(0) No(1)
• explain the answers plzzz...
• 10 years agoHelpfull: Yes(0) No(1)