[HELP] Probability question (1 Viewer)

[trapizi]

I'm confusing with this question

5 girls 3 boys sitting at a circular table. Find the probability of at least 2 boys sitting next to each other.

Answer: 5/7

 

[faisalabdul16]

Okay so you take two cases.

One where 2 boys sit next to each other and the other when all 3 sit next to each other.

Case 1)

From the three boys you choose 2. So 3C2. Then you fix one of the boys positions anywhere. The second boy has 2 spots to choose and sit next to the other guy. Either his left hand side or right. So far, 3C2 x 2. Now two boys are together, the third must not sit beside any of the boys and therefore he only has 4 places to choose from. Then, because there's no restrictions on girls, they can be arranged in 5! ways.

So for two boys next to each other, number of ways = 3C2 x 2 x 4 x 5! = 2880

Case 2) three boys together

Let boys be ABC AND GIRLS be dashes

If you place seats in a line it'll be ABC-----
Bold parts = the 8 seats. The three boys sitting together can be arranged in 3! ways. Then the rest of the girls can be arranged in 5! ways because it's a circle. So total ways of 3 boys together = 3! X 5! = 720

Total number of ways = 3600

Probability = 3600/7! (7! Because without restrictions that's total arrangements)

= 3600/5040
=5/7

 

[braintic]

Another way - don't treat the boys (or the girls) as distinct, so there is no need to choose 2 from 3.

We are forming a "circular word" from the letters GGGGG BBB.

Case I - all the boys together - only 1 possibility

Case 2 - exactly two boys together - place GBBG - then there are 4 places to place the 4th boy, so 4 arrangements.

Case 3 - exactly one girl between two of the boys - place GBGBG - then there are 2 possibilities for the remaining 2 seats

It is impossible to arrange them with at least 2 girls between every pair of boys, so no more cases.

Answer: (1+4)/(1+4+2) = 5/7

 

[trapizi]

Thanks guys, really helpful.

 

[braintic]

Another way - don't treat the boys (or the girls) as distinct, so there is no need to choose 2 from 3.

We are forming a "circular word" from the letters GGGGG BBB.

Case I - all the boys together - only 1 possibility

Case 2 - exactly two boys together - place GBBG - then there are 4 places to place the 4th boy, so 4 arrangements.

Case 3 - exactly one girl between two of the boys - place GBGBG - then there are 2 possibilities for the remaining 2 seats

It is impossible to arrange them with at least 2 girls between every pair of boys, so no more cases.

Answer: (1+4)/(1+4+2) = 5/7

Just realised this is most likely BS. The different 'words' are not equally likely.

 

[Carrotsticks]

The 5!3! is for all the boys together.

Now we split the boys into 2 groups, one with 2 and another with just 1. These 2 groups may not be next to each other to ensure that we are considering purely the case when it's 2 boys next to each other.

The girls can be arranged in a circular fashion in 4! ways.

There are 5 positions where we may place the 2 'groups' of boys such that exactly 2 are next to each other, so 5C2.

Permute the boys = 3!

Swap the groups of boys = 2!

 

[dunjaaa]

Yep, that's what I did using the insertion method