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 :

The 7 observations are $$2,\,4,\,10,\,x,\,12,\,14,\,y$$ with $$x \gt y$$.
Their mean is given as $$8$$ and their variance as $$16$$.

Step 1 : Mean equation
Mean $$\mu$$ of $$n$$ observations $$x_1,x_2,\ldots ,x_n$$ is $$\mu=\dfrac{\sum x_i}{n}$$.
For the present data, $$n=7$$, so $$\frac{2+4+10+x+12+14+y}{7}=8$$

Simplifying, $$\frac{42+x+y}{7}=8 \;\Longrightarrow\;42+x+y=56 \;\Longrightarrow\;x+y=14$$ $$-(1)$$

Step 2 : Variance equation
Population variance $$\sigma^{2}$$ is $$\sigma^{2}=\dfrac{\sum x_i^{2}}{n}-\mu^{2}$$.
Here $$\sigma^{2}=16,\;n=7,\;\mu=8$$, hence $$\dfrac{2^{2}+4^{2}+10^{2}+x^{2}+12^{2}+14^{2}+y^{2}}{7}-8^{2}=16$$

Compute the known squares: $$2^{2}=4,\,4^{2}=16,\,10^{2}=100,\,12^{2}=144,\,14^{2}=196$$.
Sum of known squares $$=4+16+100+144+196=460$$.
Therefore $$\dfrac{460+x^{2}+y^{2}}{7}-64=16$$

Multiplying by $$7$$: $$460+x^{2}+y^{2}=80\times7=560$$ so $$x^{2}+y^{2}=100$$ $$-(2)$$

Step 3 : Solving for $$x$$ and $$y$$
From $$(1)$$, $$(x+y)^{2}=14^{2}=196$$.
Using $$(2)$$, $$(x+y)^{2}=x^{2}+y^{2}+2xy \Longrightarrow 196=100+2xy$$ hence $$2xy=96\;\Longrightarrow\;xy=48$$.

Thus $$x$$ and $$y$$ are the roots of $$t^{2}-14t+48=0$$.
Discriminant $$D=14^{2}-4\cdot48=196-192=4\;,$$ $$\sqrt{D}=2$$.
Roots: $$t=\dfrac{14\pm2}{2}\; \Rightarrow\; t=8,\,6$$.
Given $$x\gt y$$, we get $$x=8,\;y=6$$.

Step 4 : Forming the new set
The set given is $$\{1,\,2,\,3,\,x-4,\,y,\,5\}$$.
Substituting $$x=8,\;y=6$$: $$x-4=4$$, so the set becomes $$\{1,\,2,\,3,\,4,\,6,\,5\}=\{1,2,3,4,5,6\}$$.

Step 5 : Required probability
Two numbers are drawn without replacement. We need the probability that the smaller of the two numbers is less than $$4$$, i.e. it is $$1,\,2$$ or $$3$$.

Total unordered pairs from $$6$$ distinct numbers $$= {}^{6}C_{2}=15$$.

Favourable pairs:
• Smaller $$=1$$: pairs $$\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{1,6\}$$  →  $$5$$ pairs.
• Smaller $$=2$$: pairs $$\{2,3\},\{2,4\},\{2,5\},\{2,6\}$$  →  $$4$$ pairs.
• Smaller $$=3$$: pairs $$\{3,4\},\{3,5\},\{3,6\}$$  →  $$3$$ pairs.

Total favourable pairs $$=5+4+3=12$$.

Therefore $$\text{Probability}=\dfrac{12}{15}=\dfrac{4}{5}$$.

Answer: The required probability is $$\frac{4}{5}$$ (Option C).