You have prepared 10 types of treats for your 5 cats. You don’t know which treat each of your cats will go for, so you have bought for each type enough treats for all your cats. Assume that each cat is equally likely to choose any type of treats, and let X be the number of pairs of cats that will choose the same type of treats. Compute E(X) and Var(X).

E(X) = 1.5

Var(X) = 2.325

Step-by-step explanation:

X - the number of pairs of cats (out of the 5 cats) to choose the same type of treat, can take values of:

X = (0, 1, 2, 3, 4, 5)

Also probability of choosing a treat, since they are all equally likely is: f(x) = 1/10

E(X) - expectation of x, is given by:

E(X) = Summation [X*f(x)]

E(X) = 0x(1/10) + 1x(1/10) + 2x(1/10) + 3x(1/10) + 4x(1/10) + 5x(1/10)

= (1/10) x (1+2+3+4+5)

E(X) = 3/2 = 1.5

Also, variance is:

Var(X) = Summation f(x)*[X - E(X)]^2

= (1/10)x(0-1.5)^2 + (1/10)x(1-1.5)^2 +(1/10)x(2-1.5)^2 +(1/10)x(3-1.5)^2 +(1/10)x(4-1.5)^2 +(1/10)x(5-1.5)^2

= (1/10)x[2.25 + 0.25 + 2.25 + 6.25 + 12.25]

Var(X) = 2.325