A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is
A. 4/19 B. 7/19
C. 12/19 D. 21/95

• 
  P( None is defective)
  = 16C2 / 20C2
  = (16x15/2x1 ×2x1/20x19)
  = 12/19.
  P( at least one is defective)
  = (1- 12/19)
  = 7/19.
• 12 years agoHelpfull: Yes(49) No(0)
• this will be..
  {(4c1*16c1)/20c2}+{4c2/20c2}=7/19 ans.
  
  at least 1 defective should be chosen from the bulb set. that is either one or two defective bulbs can be chosen.
• 12 years agoHelpfull: Yes(27) No(1)
• 20c2-16c2=70
  p=70/20c2=70/190=7/19
  'B'
• 12 years agoHelpfull: Yes(7) No(1)
• two balls are choosen at randomly from the 20 electric bulbs then the probablity is =20c2
  in that 4 are defective,two balls are choosen from that=4c2
  then the total probability is =4c2/20c2=2/95
• 12 years agoHelpfull: Yes(1) No(22)
• 7/19 kjhbjhbjhknknkjbjbklnkn
  
• 7 years agoHelpfull: Yes(0) No(3)
• B
  c(4,1)*c(16,1)+c(4,2)=70
  c(20,2)=190
  p=70/190
  =7/19
• 7 years agoHelpfull: Yes(0) No(0)