in a room der r 6 married couples. 4ppl r selected at random. find probability dat xact 1 couple is selected.

• suppose in four people selected there is one couple so other two people can be selected in following way-
  1-when other two people are ladies -5c2
  2-when other two people are gents -5c2
  3-when other two people are 1 lady and 1 gents-in this case for one lady there can be four gents because we can not have two couple in the group.so for five ladies no. of ways one gent and one lady can be selected is 5*4=20.
  thus for one couple no. of ways=5c2+5c2+20.
  for six couples no. of ways=6*(5c2+5c2+20).
  And total no. of ways to select four people=12c4.
  solution-6*(5c2+5c2+20)/12c4=16/33.
• 12 years agoHelpfull: Yes(39) No(1)
• As 6 couples are there so we can select 1 couple(means 2 persons) by 6C1
  Now to select remaining 2 persons we will have 3 cases:-
  Select 2 males 6C1*5C2
  Select 2 females6C1*5C2
  3. Select 1 male and 1 female(Leaving already selected pertner)6C1*5C1*4C1
  So,
  Slection of 4 persons will be:-
  Now, Selecting 4 persons without any constraint will be 12C4
  So, total probability will be:-
  
  [6C1*5C2 + 6C1*5C2 + 6C1*5C1*4C1]/12C4
  P = 16/33
• 12 years agoHelpfull: Yes(19) No(0)
• 6c1/12c4=6/495
• 12 years agoHelpfull: Yes(5) No(16)
• There are 6*2=12 person in the room
  among which 1 couple is selected in 6c1 ways..
  that means
  12-1*2=10 people are left
  among them other 2 are selected=10c2
  xact 1 couple=6c1*10C2
  that means probability should be(6C1*10c2)/(12c4)
• 12 years agoHelpfull: Yes(5) No(4)
• there are 3 cases
  1) 1couple + 2 women=6C1*5C2
  2) 1couple + 2men=6C1*5C2
  3) 1couple + 1woman +1man=6C1*5C1*5C1
  sample space=12c4
  
• 12 years agoHelpfull: Yes(5) No(5)
• As 6 couples are there so we can select 1 couple(means 2 persons) by 6C1
  Now to select remaining 2 persons we will have 3 cases:-
  1. Select 2 males
  2. Select 2 females
  3. Select 1 male and 1 female(Leaving already selected pertner)
  So,
  Slection of 4 persons will be:-
  for Case 1: 6C1*5C2
  for Case 2: 6C1*5C2
  for Case 3: 6C1*5C1*4C1 (4C1 is taken bcz we have removed the pweson of same couple)
  Now, Selecting 4 persons without any constraint will be 12C4
  So, total probability will be:-
  
  [6C1*5C2 + 6C1*5C2 + 6C1*5C1*4C1]/12C4
  i.e P = 16/33
• 12 years agoHelpfull: Yes(5) No(0)
• ans is 72/11
  there are 6 couples
  so 6 males and 6 females
  now , we have to select exact 1 couple, so there should be 1 male and 1 female
  so selecting 1 male and 1 female partner is 6c1*6c1. NOW remaining 2 people can be selected in 10c2 ways... (as out of 12 , 2 belong to couples which are selected, ) so total favourable cases- 6c1*6c1*10c2
  and total cases- 12c2
  so probablity is = (6c1*6c1*10c2)/12c2= 72/11
  
• 12 years agoHelpfull: Yes(2) No(10)
• in the book answer to this question is 16/33. but am not getting it.
• 12 years agoHelpfull: Yes(1) No(1)
• suppose in four people selected there is one couple so other two people can be selected in following way-
  1-when other two people are ladies -5c2
  2-when other two people are gents -5c2
  3-when other two people are 1 lady and 1 gents-in this case for one lady there can be four gents because we can not have two couple in the group.so for five ladies no. of ways one gent and one lady can be selected is 5*4=20.
  thus for one couple no. of ways=5c2+5c2+20.
  for six couples no. of ways=6*(5c2+5c2+20).
  And total no. of ways to select four people=12c4.
  solution-6*(5c2+5c2+20)/12c4=16/33
• 12 years agoHelpfull: Yes(1) No(1)
• i think answer is 2/33..........because out of 6 male or female 3 are selected... then in the 3 persons selected corresponding other males or females are to be selected......hence (6c3 * 3c1)/(12c4)
• 12 years agoHelpfull: Yes(0) No(1)
• @ Ravindra Joshi
  
  Your answer is 72/11 > 1. Hence wrong. Probability cannot be greater than 1.
• 12 years agoHelpfull: Yes(0) No(0)
• suppose in four people selected there is one couple so other two people can be selected in following way-
  1-when other two people are ladies -5c2
  2-when other two people are gents -5c2
  3-when other two people are 1 lady and 1 gents-in this case for one lady there can be four gents because we can not have two couple in the group.so for five ladies no. of ways one gent and one lady can be selected is 5*4=20.
  thus for one couple no. of ways=5c2+5c2+20.
  for six couples no. of ways=6*(5c2+5c2+20).
  And total no. of ways to select four people=12c4.
  solution-6*(5c2+5c2+20)/12c4=16/33.
• 10 years agoHelpfull: Yes(0) No(0)