The given bar graph shows the results from rolling one white and one red dice.

Read Data from the Bar Chart**:

• Red Die Frequencies ($f_R$):

  • $1: 7$
  • $2: 8$
  • $3: 1$
  • $4: 2$
  • $5: 1$
  • $6: 1$
  • Total red rolls = $7 + 8 + 1 + 2 + 1 + 1 = 20$
  • Sum of red rolls = $1(7) + 2(8) + 3(1) + 4(2) + 5(1) + 6(1) = 7 + 16 + 3 + 8 + 5 + 6 = 45$

• White Die Frequencies ($f_W$):

  • $1: 3$
  • $2: 3$
  • $3: 4$
  • $4: 3$
  • $5: 3$
  • $6: 3$
  • Total white rolls = $3 + 3 + 4 + 3 + 3 + 3 = 19$
  • Sum of white rolls = $1(3) + 2(3) + 3(4) + 4(3) + 5(3) + 6(3) = 3 + 6 + 12 + 12 + 15 + 18 = 66$

2. Calculate Current Overall Mean Score ($\mu$):

   • Total number of rolls so far: $N = 20 + 19 = 39$
   • Total sum of scores so far: $S = 45 + 66 = 111$
   • Current mean score:
     $$\mu = \frac{111}{39} \approx 2.846$$

3. Condition for Increasing the Mean:

   • Let the next roll of the fair white die be $X \in {1, 2, 3, 4, 5, 6}$.
   • The new mean after $N+1$ rolls is $\mu_{\text{new}} = \frac{S + X}{N + 1}$.
   • The mean score increases ($\mu_{\text{new}} > \mu$) if and only if:
     $$\frac{S + X}{N + 1} > \frac{S}{N} \iff X > \mu$$
   • Since $\mu \approx 2.846$, any roll $X$ satisfying $X > 2.846$ will increase the mean.

4. Favorable Outcomes and Probability:

   • The possible outcomes of a standard 6-sided die are ${1, 2, 3, 4, 5, 6}$.
   • Outcomes strictly greater than $2.846$ are ${3, 4, 5, 6}$, giving $4$ favorable outcomes.
   • Since the die is fair, each face has probability $1/6$:
     $$P(X > \mu) = \frac{4}{6} = \mathbf{\frac{2}{3}}$$

Thus, the correct option is C.