praetorian123 wrote:

AkamaiBrah wrote:

praetorian123 wrote:

There are 4 couples. You arrange these people in a line. What is

the probability that no husband is with his wife.

While i got to the answer, it doesnt match...please help

my solution and answer to follow

thanks

praetorian

Where did you get this problem. It is quite difficult, IMO. In any case I ended up with 11/21

This was posted in another group...he told us that its from a Probability book and as we found out, its way too hard for a GMAT problem.

But i guess its better to solve these hard conceptual problems.

Well, the answer given is 12/35

I also didnt get the answer right...Please tell me what concept i got wrong...

Calculate the possibilities where couples can be together...and

then subtract the probability from 1.

Heres my attempt :

# of ways in which 1 couple are together = 4P1=4

# of ways in which 2 couples are together = 4P2=12

# of ways in which 3 couples are together = 4P3=24

# of ways in which 4 couples are together = 4P4=24

Total # of ways in which the couple can be together = 4*12*24*24

Total # of ways of arranging 8 people = 8!

4*12*24*24 / 8! = 24/35

So the prob that no husband is with his wife = 1- 24/35 = 11/35

My Answer : 11/35 ..please comment

thanks

praetorian

Oops. I made a bad math error. However, I know how to solve this.

Your way is not correct even though you ended up close. There is no way to justify your multiplying all of the possibilities together.

The correct solution is as follows:

Let call the couples A, B, C, and D.

We want to know the probability that NO couple are together. This is the same as calculating the probability of NOT (A or B or C or D).

However, since these events are not mutually exclusive, we need to subtract out the parts that overlap. Without going into a lot of math, the probability will look like this:

P(no couples) = 1 - (4C1*p(one particular couple) - (4C2*p(two particular couples) - (4C3*p(three particular couples) - 4C4*p(4 couples))))

(repeated application of the OR rule with events that are not mutually exclusive -- look it up. Or imagine a venn diagram and how you would count the total number inside (you cannot draw a 4 section venn-diagram on paper, so imagine it for 3 sections))

= 1 - (4*(7!*2) - (6*(6!*4) - (4*(5!*8) - 4!*16))) / 8!

= 1 - (40320 - 17280 + 3840 - 384) / 40320

= 1 - 26496/40320

= 1 - 23/35

= 12/35

_________________

Best,

AkamaiBrah

Former Senior Instructor, Manhattan GMAT and VeritasPrep

Vice President, Midtown NYC Investment Bank, Structured Finance IT

MFE, Haas School of Business, UC Berkeley, Class of 2005

MBA, Anderson School of Management, UCLA, Class of 1993