8 couples (husband and wives) attend a dance show nach baliye in a popular tv channel.A lucky draw in which 4 persons picked up for a prize is held,then probablity that atleast one couple will be selected is:

a)8/39
b)15/39
c)12/13
d)none of these

• P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)
  = 1-(16c1*14c1*12c1*10c1/16c4) = 15/39
• 9 years agoHelpfull: Yes(14) No(2)
• there are 8 couples...i.e. 16 persons...
  so total outcomes for choosing 4 persons is 16C4 = 1820
  p(atleast 1 couple) = p(1 couple or 2couples)
  for 1 couple -> we have 8 choices to choose a couple,
  for remaining 2 non couple persons, we can select any 1 person from 14 people in 14 ways
  now the 4th person can't be spouse of 3rd person, so we have only 12 persons to choose from
  so total choices for 3rd and 4th persons are 14*12/2 (divided by 2 as we are counting a pair twice) and total choices including 1st and 2nd persons are 8*14*12/2
  which is 672
  Now for 2 couples we have 8c2 = 28 choices
  so total favuorable outcomes = 672 + 28 = 700
  700/1820 = 5/13, or 15/39
  
• 9 years agoHelpfull: Yes(12) No(0)
• Req. probability= 1 - Probability in which none of the couples are selected
  So, prob = 1- (8c4 + 8c4)/(16c4) = 12/13
• 9 years agoHelpfull: Yes(2) No(3)
• @ p.ajay..... calculate your answer ...itz coming in negative , so itz wrong
• 9 years agoHelpfull: Yes(0) No(2)
• 1-((2*8C4)/16C4)=0.92=12/13
• 9 years agoHelpfull: Yes(0) No(4)
• none of these my ans is 3/8
• 9 years agoHelpfull: Yes(0) No(4)