Question

An urn contains four balls numbered 2, 2, 5 and 6. If a person selects a set of two balls at random, what is the expected value of the sum of the number of the balls?

Answer 1

Suppose we select two distinct balls

then we have the following possibilities of selecting two balls

(2,2), (2,2) ............, i.e. 2 ways to get a sum of 4...............(2+2 = 4 and 2+2 = 4)

(2,5), (2,5), (5,2), (5,2), i.e. 4 ways to get a sum of 7...............(5+2 = 7 and 2+5 = 7)

(2,6), (2,6), (2,6), (2,6), i.e. 4 ways to get a sum of 8...............(2+6 = 8 and 6+2 = 8)

(6,5), (5,6) , i.e. 2 ways to get a sum of 11...............(5+6 = 11 and 6+5 = 11)

total number of possible outcomes = 2(for a sum of 4) + 4(for a sum of 7) + 4(for a sum of 8) + 2 (for a sum of 11)

= 2 + 4 + 4+2

= 12

So, P(getting a sum of 4) = (number of ways to get a sum of 4)/total number = 2/12

similarly,

P(getting a sum of 7) = (number of ways to get a sum of 7)/total number = 4/12

P(getting a sum of 8) = (number of ways to get a sum of 8)/total number = 4/12

P(getting a sum of 11) = (number of ways to get a sum of 11)/total number = 2/12

We know that expected value E[x] = sum of all individual sum values by their respective probability

= 4*(2/12) +7*(4/12) +8*(4/12) +11*(2/12)

= 0.6667 + 2.3333 + 2.6667 + 1.8333

= 7.500 (rounded to 4 decimals)