We start with a collection of n (distinguishable) individuals whom we want to split into three possible teams: the blue team, the red team, and the white team. Each individual is randomly assigned to exactly one of the teams according to the following procedure. An individual is assigned to the blue team with probability b, to the red team with probability r, and to the white team with probability w (where b>0, r>0, w>0, and b+r+w=1). This assignment is done independently for each individual. Let Nb, Nr, and Nw be the numbers of people on the respective teams.
1. Express your answer for this part in terms of n, b, and r only.
E[Nb+Nr]=
2. Express your answer for this part in terms of n and b only.
E[N2b]=
3. Suppose that n=5. We are told that individuals 1 and 2 belong to the blue team.
(a) Find the conditional PMF of Nb, conditioned on this information.
pNb(k∣individuals 1 and 2 on blue team)=
(b) Are Nb and Nr independent, conditioned on this information?
4. Individual i (for i=2,3,…,n−1) is happy if and only if individuals i−1,i,and i+1 belong to the same team. Find the expected number of happy individuals. (Individuals 1 and n are never happy.) Express your answer for this part in terms of n, b, r, and w.
6 answers
anyone? Im badly stuck on this :(
1) n*(b+r)
2) n*b*(1-b+n*b)
3) d, No
4) My derived answer is n*(b^3+r^3+w^3)-2*(b^2+r^2+w^2), which is not correct. I am fairy certain that the n*(b^3+r^3+w^3) component is correct. Perhaps I am not handling a corner case correctly.
The probability that (i-1) and (i+1) for a blue team is b^2; similarly r^2 and w^2 for red and white teams. Since people on both ends are not happy, the expected number of happy people on a blue team is b^2*(n*b-2). By adding the expected values of happy people from all teams we get b^2*(n*b-2) + r^2*(n*r-2) + w^2*(n*w-2). After simplification, I get n*(b^3+r^3+w^3)-2*(b^2+r^2+w^2). Any thoughts?
2) n*b*(1-b+n*b)
3) d, No
4) My derived answer is n*(b^3+r^3+w^3)-2*(b^2+r^2+w^2), which is not correct. I am fairy certain that the n*(b^3+r^3+w^3) component is correct. Perhaps I am not handling a corner case correctly.
The probability that (i-1) and (i+1) for a blue team is b^2; similarly r^2 and w^2 for red and white teams. Since people on both ends are not happy, the expected number of happy people on a blue team is b^2*(n*b-2). By adding the expected values of happy people from all teams we get b^2*(n*b-2) + r^2*(n*r-2) + w^2*(n*w-2). After simplification, I get n*(b^3+r^3+w^3)-2*(b^2+r^2+w^2). Any thoughts?
Happy people on a blue team isn't b^2*(n*b-2), it is b^2*((n-2)*b). Regards.
Thank you!
answer to 4) is (n-2)*(b^3+r^3+w^3)
How did u get b
2 answers