Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, x >  y, be 8 and 16 respectively. Two numbers are chosen from {1, 2, 3, x - 4,y,5} one after an other without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is :

Mean $$\bar{x} = 8$$. Sum $$= 8 \times 7 = 56$$.

$$2+4+10+12+14+x+y = 56 \implies 42 + x + y = 56 \implies x+y = 14$$.

Variance $$\sigma^2 = 16$$. $$\frac{\sum x_i^2}{7} - \bar{x}^2 = 16 \implies \sum x_i^2 = 7(16 + 64) = 560$$.

$$4 + 16 + 100 + 144 + 196 + x^2 + y^2 = 560 \implies 460 + x^2 + y^2 = 560 \implies x^2 + y^2 = 100$$.

Solving $$x+y=14$$ and $$x^2+y^2=100$$: $$(14-y)^2 + y^2 = 100 \implies 196 - 28y + 2y^2 = 100 \implies y^2 - 14y + 48 = 0$$.

Roots are $$6, 8$$. Since $$x > y$$, $$x=8, y=6$$.

The set for probability is $$\{1, 2, 3, x-4, y, 5\} = \{1, 2, 3, 4, 6, 5\}$$ 

Total ways to pick 2 numbers from 6 without replacement $$= 6 \times 5 = 30$$.   $$< 4$$.

If smaller is $$\ge 4$$, both numbers must be from $$\{4, 5, 6\}$$. Ways $$= 3 \times 2 = 6$$.

$$P(\text{smaller} < 4) = 1 - \frac{6}{30} = 1 - \frac{1}{5} = \frac{4}{5}$$.