Solution:
When two dice are thrown, the possible outcomes can be represented by a sample space of 36 outcomes.

Let A be the event that the sum of the numbers on the upper side of the dice is k, where k = 2, 3, ..., 12.

Then, the probability of A is given by:

P(A) = number of outcomes in A / total number of outcomes

The number of outcomes in A can be determined by counting the number of ways that k can be obtained as the sum of two numbers on the dice. For example, if k = 7, there are six ways to obtain this sum:

1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1

Therefore, the number of outcomes in A is:

number of outcomes in A = 6 if k = 7
number of outcomes in A = 5 if k = 6 or 8
number of outcomes in A = 4 if k = 5 or 9
number of outcomes in A = 3 if k = 4 or 10
number of outcomes in A = 2 if k = 3 or 11
number of outcomes in A = 1 if k = 2 or 12

The total number of outcomes is 36, so the probabilities of the events A are:

P(A=2) = 1/36
P(A=3) = 2/36
P(A=4) = 3/36
P(A=5) = 4/36
P(A=6) = 5/36
P(A=7) = 6/36
P(A=8) = 5/36
P(A=9) = 4/36
P(A=10) = 3/36
P(A=11) = 2/36
P(A=12) = 1/36

The expected value of the sum of the numbers on the upper side of the dice is given by:

E(X) = Σk P(X=k)

where k is the sum of the numbers on the upper side of the dice.

Using the probabilities calculated above, we can find that:

E(X) = 2×1/36 + 3×2/36 + 4×3/36 + 5×4/36 + 6×5/36 + 7×6/36 + 8×5/36 + 9×4/36 + 10×3/36 + 11×2/36 + 12×1/36

Simplifying this expression, we get:

E(X) = (2+3+4+5+6+7+8+9+10+11+12)/2

E(X) = 7

Therefore, the expected value of the sum of the numbers on the upper side of the dice is 7.

Hence, option B is the correct answer.