Question

An apparently easy probabilities problem.

So here it is :

5 people sit in a circular table, two of them are john and carl, ¿What is the probability for jhon and carl don´t be sit together.

A. 60% B. 70% C.80% D. 90%

My problem here is that the answer i get by permutations don't agree with the answer if i pick a particular case where they are the two first to get seated. And the question do not specify any order..

Any idea thanks.

Answer 1

john has to sit somewhere, so the question is, what is the probability carl sits next to him. there are two ways he can sit next to john, on the left or on the right
of the 4 seats there are \(4!=24\) ways for the four other people to sit
of those 2 have carl next to john

Answer 2

ok maybe that is wrong. of those 24 ways, \(2\times 3!=12\) have carl next to john

Answer 3

hmm i am getting 50% but that is not one of the choices, so i am doing something wrong

Answer 4

actually i think i am going to stick with my answer

Answer 5

an even simpler way to think about it is this
put john in a seat
then carl has choice of 4 seats
two are next to john
two aren't

Answer 6

Well , that makes sense for me.

But, considering that the probabilitiy for one of them tho sit first is 2/5, and the probability for the other to be the next is 1/4, so the probability for the two sit together first is 1/10, that gives you an answer of 90%, wich is completly incoherent with our firt answer.

And, if the posible combinations of two people on 5 chairs is
5
C = 5!/3!
2

Then we can let the posible combinations of two people to do not sit together as
4
C = 4!/2!
2

And that gives you another answer, 12/20 = 3/5 = 60%.

So the answers contradict each other.
Where i am going wrong?

Answer 7

a third way
compute the probability that john and carl sit together
the number of ways 5 people can be seated in a circular table is \(4!=24\) if john and carl sit together, we can treat them as one unit making 3 people instead of 4, and that makes \(3!\) ways they can be seated next to each other, except that you have to multiply by 2 be because they can be right left of left right
i still get \(\frac{2\times 3!}{4!}=\frac{12}{25}\)

Answer 8

the line should read
"we can treat them as one unit making 4 people instead of 5"

Answer 9

i am going to stick with that answer. if it is wrong, i would love to know why

Answer 10

probability that they don't sit together = 1-P(do sit together)

Answer 11

C(5,2) = 5!/(5-2)!*(2!) = 5!/(3!*2!) = (5*4)/2 = 10

Answer 12

@satellite73 I am also getting 50%. this is how i did it

I don't think that we can use combinations since we are given two people's names and seating is different.

so we have



P(not sitting together) = 1- P(sitting together)

# of ways people can sit together where order matters = P(5,2) = 5!/(5-2)1 = 5 * 4 = 20



so we have P(sitting together) = P((A AND B) OR (B AND C) OR (C AND D)...)

these events are mutually exclusive so the following thm holds. P(A OR B) = P(A) + P(B)

so

P(sitting together) = P(A AND B) + P(B AND C) + P(C AND D) + ...

P(A AND B) = 2/P(5,2) = 2/20

P(sitting together) = (2/20) + (2/20) + (2/20) + (2/20) + (2/20) = (5*2)/20 = 10/20 = .5


so P(not sitting together) = 1- P(sitting together) = 1-.5 = .5 = 50%