3 red, 2 white and 3 green balls are available in a bag. 2 balls are drawn one by one from the bag. Probability of one ball having white & other green will be.

• The probability of selecting 1 white ball from 8 balls is 2/8.
  The balls are drawn one by one. so now 7 balls are left. The probability of selecting 1 green ball from rest 7 is 3/7.
  so probability of one ball being green & other white is 2/8* 3/7= 3/28
  
• 9 years agoHelpfull: Yes(2) No(1)
• Probability of first ball having white colour = 3/8
  and that of second ball having green colour = 3/7
  so probability = 2/8 * 3/7 = 6/56
  and no of methods to make arrangement = 2*1 = 2 (WG and GW)
  so net probability = 5/56 * 2 = 12/56 = 3/14
• 9 years agoHelpfull: Yes(1) No(1)
• n(s)=8c2
  n(e)=2c1*3c1
  p=2c1*3c1/8c2
• 9 years agoHelpfull: Yes(0) No(1)
• The answer is 1
  Total probability is 8c2is 28
  because we have to select 2 balls out of eight balls
  And selecting white balls is 2c1
  and selecting green balls is 3c1
  so (2c1*3c1)/8c2
• 9 years agoHelpfull: Yes(0) No(0)