Given that the two numbers appearing on throwing the two dice are different.

When dice is thrown, number of observations in the sample space = 6 × 6 = 36

Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.

$\therefore \mathrm{A}=\{(1,3),(2,2),(3,1)\}$

$\mathrm{B}=\left\{\begin{array}{lllll}(1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5)\end{array}\right\}$

$\mathrm{A} \cap \mathrm{B}=\{(1,3),(3,1)\}$

$\therefore \mathrm{P}(\mathrm{B})=\frac{30}{36}=\frac{5}{6}$ and $P(A \cap B)=\frac{2}{36}=\frac{1}{18}$

Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.

$\therefore P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{18}}{\frac{5}{6}}=\frac{1}{15}$

Therefore, the required probability is $\frac{1}{15}$.