Latest TCS Aptitude Question SOLUTION: in a room der r 6 married couples. 4ppl r selected at random. find probability dat xact 1 couple is selected.

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

Asked In TCS aanchal kushwaha (12 years ago)

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Sanjay Singh  (12 years ago)

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.

Is this solution Helpfull? Yes (39) | No (1)  

Ankit Gupta  (12 years ago)

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

Is this solution Helpfull? Yes (19) | No   

taruna  (12 years ago)

6c1/12c4=6/495

Is this solution Helpfull? Yes (5) | No (16)  

Sourav Bandyopadhyay  (12 years ago)

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)

Is this solution Helpfull? Yes (5) | No (4)  

abhijeet patil  (12 years ago)

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

Is this solution Helpfull? Yes (5) | No (5)  

raj  (12 years ago)

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

Is this solution Helpfull? Yes (5) | No   

ravindra joshi  (12 years ago)

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

Is this solution Helpfull? Yes (2) | No (10)  

aanchal kushwaha  (12 years ago)

in the book answer to this question is 16/33. but am not getting it.

Is this solution Helpfull? Yes (1) | No (1)  

sagar  (12 years ago)

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

Is this solution Helpfull? Yes (1) | No (1)  

srikhakollu bharghav  (12 years ago)

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)

Is this solution Helpfull? Yes | No (1)  

Narendra Kulkarni  (12 years ago)

@ Ravindra Joshi

Your answer is 72/11 > 1. Hence wrong. Probability cannot be greater than 1.

Is this solution Helpfull? Yes | No   

sowmya  (10 years ago)

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.

Is this solution Helpfull? Yes | No   

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