From a collection of 12 bulbs of which 6 are defective 3 bulbs are chosen at random for three sockets in a room.Find the probability that the room is lighted if one bulb is sufficient to light the room.

• 1-(6C3/12C3) = 10/11
• 9 years agoHelpfull: Yes(11) No(0)
• Prob= [6c3*6c0+6c2*6c1+6c2*6c1]/[12c3]=10/11
• 9 years agoHelpfull: Yes(9) No(0)
• total ways of choosing three bulbs=12C3
  if 0 defective bulb is choosen= 6C3
  if 1 ...........................................= 6C1*6C2
  if 2 .......................................... = 6C2*6C1
  total ways in which room room is lighted = 6C3+(6C1*6C2)+(6C2*6C1)
  reqd prob= [6C3+(6C1*6C2)+(6C2*6C1)]/12C3 = 10/11
  => reqd prob= 10/11
  
• 9 years agoHelpfull: Yes(5) No(0)
• To lighten the room at least one non defective ball is required:
  ways of selecting at least one non defective ball,lets say n= (6C1 * 6C2)+(6C2*6C1)+6C3
  total ways of selecting 3 ball ,lets say s=12C3
  probability=n/s= 10/11
• 9 years agoHelpfull: Yes(1) No(0)
• prob= 10/11
• 9 years agoHelpfull: Yes(1) No(0)
• 2*(6C1*6C2/12C3)+(6C3/12C3)
  =7/11
• 9 years agoHelpfull: Yes(0) No(3)
• room can be light in 3 ways
  if(i good bulb select) or (2 good bulb selected) or (all good bulb selected)
  (1/2) + (1/2*5/11) + (1/2*5/11*2/5)=9/11 ans
• 9 years agoHelpfull: Yes(0) No(4)
• (6C1+6C2+6C3)/12C3 ... simply take the probability of each case of taking 1 bulb, 2 and 3 out separately and add them ..
• 9 years agoHelpfull: Yes(0) No(1)
• Prob =10/11
• 9 years agoHelpfull: Yes(0) No(0)
• total no of combination=12c3=220
  case1, 1bulb is on=11c2=55
  case2,2 bulbs on=10c1=10
  case3, 3bulbs on=6c3=20
  total favourable case=85
  p=85/220=17/44
• 9 years agoHelpfull: Yes(0) No(0)
• total number of chances,n(S)=12;
  chances of lighting room n(A)=6;
  therefore probability of lighting the room= 6/12=1/2
  
  
• 8 years agoHelpfull: Yes(0) No(0)