Question

Harry Potter approaches with a strange bag full of balls, numbered 1 to k. As you reach in to pick one, he notes that they are not all equally likely because of magic: ball 1 is least likely to be chosen, with probability c, where c is some constant. Ball 2 has probability 2c, Ball 3 has probability 3c, . . . , Ball k − 1 has probability (k − 1)c, and Ball k has probability kc.

1. What is the expected value of the ball number you pick? Your answer can’t use the constant c, but will use k.

Answer 1

[k*(k+1)*(2*k+1)] / 6

Explanation:

We have balls numbered as: 1, 2, 3, ... , k with probabilities as: c, 2*c, 3*c, ... , k*c

Let Y be the discrete random variable defined as: Y = ball number

We know that Expected value of discrete Random Variable is:

E[X] = Σ₁ⁿ xₐ*f(xₐ) ,where f(xₐ) is probability of xₐ

then,

E[Y] = 1*c + 2*2*c + 3*3*c + ... + k*k*c

E[Y] = c*(1 + 2*2 + 3*3 + ... + k*k)

E[Y] = c*(1^2 + 2^2 + 3^2 + ... + k^2)

consider c = 1 (because it's constant so you can suppose any you wish)

E[Y] = 1^2 + 2^2 + 3^2 + ... + k^2

using formula of first n squares natural numbers (as attached picture)

E[Y] = [k*(k+1)*(2*k+1)] / 6