ket_gmat wrote:hi
Its probability problem.
Please find attached problem.
Can any one suggest me good study material for probability

its D 2/5

Total arrangements = 5!= 120

both couples are together = 3!*2!*2! = 24......................................1
Only one couple is together = 2*(2!*2!*3!) = 48...............................2

Thus prob.f atleast one couple together is 24+48/ 120 = 72/120= 3/5

Therefore

probability of no couple together is =1- 3/5 = 2/5

Hope its clear.. do let me know if u have any doubts,

both couples are together = 3!*2!*2! = 24......................................1
Only one couple is together = 2*(2!*2!*3!) = 48...............................2

hi sudhir,,, can u pls elaborate these two steps,,,,

thanks,,

raunekk wrote:

both couples are together = 3!*2!*2! = 24......................................1
Only one couple is together = 2*(2!*2!*3!) = 48...............................2

hi sudhir,,, can u pls elaborate these two steps,,,,

thanks,,

here it goes raunekk....

assume the 2 couples are AB ,CD the single person is X

let AB and CD sit together . Assume AB and CD both as one unit
thus its (AB) and (CD) and E
number of ways is 2!*2!*3! = 24............................1

Now assume only one couple is sitting together ... for instance only AB

thus its

( AB) as one unit and CDE

4!* 2 = 48 ....( u can also write 2*(2!*2!*3!) = 48 ).....................2

hope its makes more sense now ... do let me know if u feel u need more clarification...

sudhir3127 wrote:

raunekk wrote:

both couples are together = 3!*2!*2! = 24......................................1
Only one couple is together = 2*(2!*2!*3!) = 48...............................2

hi sudhir,,, can u pls elaborate these two steps,,,,

thanks,,

here it goes raunekk....

assume the 2 couples are AB ,CD the single person is X

let AB and CD sit together . Assume AB and CD both as one unit
thus its (AB) and (CD) and E
number of ways is 2!*2!*3! = 24............................1

Now assume only one couple is sitting together ... for instance only AB

thus its

( AB) as one unit and CDE

4!* 2 = 48 ....( u can also write 2*(2!*2!*3!) = 48 ).....................2

I think this will also include some cases which has already been counted in case 1
what about those?

hope its makes more sense now ... do let me know if u feel u need more clarification...

@ stop@800 agree

total - 5!

Combinations both coulples are together (AB)(E)(CD) = 3!*2*2 = 24
1st couple is together
(ABE)(CD) = 3!*2*2 = 24

2nd couple is together
(AB)(ECD) = 2*3!*2 = 24
But we alredy counted arrangement (AB)(E)(CD) in 1st statement 3!*2*2 = 24

I have some doubts as stop@800 does

Total arrangement for five people: 5! = 120

let the five people be ABCDE
AB are couple: (AB)CDE
all instances when AB sit together: 4!*2 = 48
taking AB as one unit - ie total 4 units. Also BA can arrange in 2 ways

other couple is CD: total arrangements - AB(CD)E = 4!*2 = 48

when both couples sit together = (AB)(CD)E = 3!*2*2 = 24

total arangements whenatleat one couple sits together = 48+48-24=72

because the arrangement of both sitting together is already included

ways when none of the couple sits together = 120 - 72 = 48

probability = 48/120 = 2/5