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.
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?
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
1. To express E[Nb+Nr] in terms of n, b, and r only, we consider that the number of people on the blue team, Nb, follows a binomial distribution with parameters n and b. Similarly, the number of people on the red team, Nr, follows a binomial distribution with parameters n and r.
The expected value of Nb is given by E[Nb] = n * b.
The expected value of Nr is given by E[Nr] = n * r.
Therefore, the expected value of Nb+Nr is E[Nb+Nr] = E[Nb] + E[Nr] = n * b + n * r = n * (b + r).
2. To express E[N2b] in terms of n and b only, we consider that the number of individuals assigned to the blue team squared, N2b, follows a binomial distribution with parameters n and b. This is because N2b represents the number of individuals assigned to the blue team in the first selection, and then again in the second independent selection.
The expected value of N2b is given by E[N2b] = n * b * b = n * b^2.
3. (a) To find the conditional PMF of Nb, conditioned on individuals 1 and 2 being on the blue team, we first need to determine the new probability distribution.
Since individuals 1 and 2 are on the blue team, we know that these two spots are already filled. Therefore, we have n-2 individuals left to distribute among the remaining two teams (red and white) with probabilities r and w.
The remaining n-2 individuals follow a binomial distribution with parameters n-2, r/(r+w). Let's denote this random variable as N'.
The conditional PMF of Nb, given individuals 1 and 2 on the blue team, can be written as:
pNb(k | individuals 1 and 2 on blue team) = P(Nb = k | individuals 1 and 2 on blue team)
= P(N' = k-2)
= (n-2)C(k-2) * (r/(r+w))^(k-2) * (w/(r+w))^(n-2-(k-2))
(b) To determine if Nb and Nr are independent, conditioned on individuals 1 and 2 being on the blue team, we need to check if the joint PMF of Nb and Nr can be factored into the product of their individual conditional PMFs.
If Nb and Nr are independent, the joint PMF would simplify to:
p(Nb=k, Nr=m | individuals 1 and 2 on blue team) = pNb(k | individuals 1 and 2 on blue team) * pNr(m | individuals 1 and 2 on blue team)
If this equation holds, then Nb and Nr are independent. However, we would need to calculate the joint PMF to verify independence.
4. To find the expected number of happy individuals, we need to sum the probabilities of each individual being happy.
For an individual i (2 ≤ i ≤ n-1) to be happy, individuals i-1, i, and i+1 must belong to the same team.
Let's consider three cases based on the team assignment probabilities:
Case 1: Individuals i-1, i, and i+1 are on the blue team.
Case 2: Individuals i-1, i, and i+1 are on the red team.
Case 3: Individuals i-1, i, and i+1 are on the white team.
The probability of each case occurring can be calculated as follows:
Case 1: b^3 (since each individual is assigned to the blue team with probability b)
Case 2: r^3 (since each individual is assigned to the red team with probability r)
Case 3: w^3 (since each individual is assigned to the white team with probability w)
The expected number of happy individuals can be calculated as the sum of the probabilities multiplied by the number of individuals in each case:
E[Number of Happy Individuals] = (n-2) * (b^3 + r^3 + w^3)