Question

A small college has only 4 students and offers only 3 majors: Art, English, and Physics.

Suppose that the 4 students are assigned to these 3 majors at random with probabilities P[Art] = 1/4, P[English] = 1/2, P[Math] = 1/4.

Let A, E, and M denote the numbers of students assigned to Art, English, and Math, respectively

(i) What is the probability distribution of A? E? M? State the expected values and variances of each.

(ii) What is the joint probability distribution of (A, E, M)? State the joint probability mass function f(a, e, m) = P[A = a, E = e, M = m].

(iii) What are Cov(A, E), Cov(A, M), Cov(E, M)? Are A, E, and M independent?

(iv) Find the probability that each major has at least 1 student.

Answer 1

i) Let A be the number of students assigned to Art .Then each of the four students are either independently assigned to Art with probability p=0.25 or not assigned to Art with probability 1−p=0.75.
It gives idea that A is the sum of 4 i.i.d Bernoulli trial and hence
A ~ Binomial(4,p=0.25)
Mean = n*p = 4*0.25 = 1
Variance = n*p*q = 0.75

Similarly for E
Let E be the number of students assigned to English. & P(English) = 1/2 = p, q= 0.5
E ~binomial (4,0.5)
Mean = n*p = 2
Variance = n*p*q = 1

For M
Let M be the number of students assigned to Math. & P(Math) = 1/4 = p ,q= 3/4
M ~binomial (4,0.25)
Mean = n*p = 1
Variance = n*p*q = 0.75

ii) joint probability distribution of (A,E,M) is
Multinomial distribution with parameters n =4 and P(art) =p1 = 0.25, P(English)= p2 = 0.5, P(Math) = p3 = 0.25.

The joint probability mass function
f(a, e, m) = P[A = a, E = e, M = m]

iii) Cov(A, E) = -np1p2 = -4*0.25*0.5 = -0.5
Cov(A, M)= -np1p3 = -4*0.25*0.25 = -0.25
Cov(E, M) = -np2p3 = -4*0.5*0.25 = -0.5

Checked A, E, and M are not independent
Because cov is not zero.

iv)

= 0.7383

= 0.7383

= 0.3125