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

### Related Formula textMean barx = fracsum x_in$$\text{Mean } \bar{x} = \frac{\sum x_i}{n}$$ textVariance sigma^2 = fracsum x_i^2n - (barx)^2$$\text{Variance } \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$ ### Core Logic Given Mean = 8$= 8$ for 7 observations: 2, 4, 10, x, 12, 14, y$2, 4, 10, x, 12, 14, y$. frac2 + 4 + 10 + x + 12 + 14 + y7 = 8$$\frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$$ Rightarrow x + y + 42 = 56 Rightarrow x + y = 14 dots(1)$$\Rightarrow x + y + 42 = 56 \Rightarrow x + y = 14 \dots(1)$$ Given Variance = 16$= 16$: 16 = frac2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^27 - 8^2$$16 = \frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} - 8^2$$ 16 + 64 = frac4 + 16 + 100 + x^2 + 144 + 196 + y^27$$16 + 64 = \frac{4 + 16 + 100 + x^2 + 144 + 196 + y^2}{7}$$ 80 times 7 = 460 + x^2 + y^2 Rightarrow 560 = 460 + x^2 + y^2 Rightarrow x^2 + y^2 = 100 dots(2)$$80 \times 7 = 460 + x^2 + y^2 \Rightarrow 560 = 460 + x^2 + y^2 \Rightarrow x^2 + y^2 = 100 \dots(2)$$ ### Step 1: Solve for x and y Using algebraic identity (x+y)^2 = x^2 + y^2 + 2xy$(x+y)^2 = x^2 + y^2 + 2xy$: 14^2 = 100 + 2xy Rightarrow 196 - 100 = 2xy Rightarrow 2xy = 96 Rightarrow xy = 48$$14^2 = 100 + 2xy \Rightarrow 196 - 100 = 2xy \Rightarrow 2xy = 96 \Rightarrow xy = 48$$ Since x+y=14$x+y=14$ and xy=48$xy=48$, roots of quadratic t^2 - 14t + 48 = 0$t^2 - 14t + 48 = 0$ are 8, 6$8, 6$. Given x > y$x > y$, we must select x = 8$x = 8$ and y = 6$y = 6$. ### Step 2: Construct the set and evaluate Probability The new set X$X$ is formed by \1, 2, 3, x-4, y, 5\$\{1, 2, 3, x-4, y, 5\}$. Substituting x=8$x=8$ and y=6$y=6$, we get \1, 2, 3, 4, 6, 5\$\{1, 2, 3, 4, 6, 5\}$. There are 6 distinct elements: \1, 2, 3, 4, 5, 6\$\{1, 2, 3, 4, 5, 6\}$. We choose two numbers without replacement. Total outcomes = 6 times 5 = 30$6 \times 5 = 30$ permutations (or binom62 = 15$\binom{6}{2} = 15$ combinations). Let's use combinations. Total ways to choose 2 numbers = binom62 = 15$\binom{6}{2} = 15$. We need the probability that the *smaller* number is less than 4$4$. P(textsmaller < 4) = 1 - P(textsmaller geq 4)$P(\text{smaller} < 4) = 1 - P(\text{smaller} \geq 4)$. ### Step 3: Final Calculation via Complement For the smaller number to be geq 4$\geq 4$, BOTH chosen numbers must be geq 4$\geq 4$. The available numbers geq 4$\geq 4$ in the set are \4, 5, 6\$\{4, 5, 6\}$ (Total 3 numbers). Ways to choose two numbers from these 3 is binom32 = 3$\binom{3}{2} = 3$. P(textsmaller geq 4) = frac315 = frac15$$P(\text{smaller} \geq 4) = \frac{3}{15} = \frac{1}{5}$$ P(textsmaller < 4) = 1 - frac15 = frac45$$P(\text{smaller} < 4) = 1 - \frac{1}{5} = \frac{4}{5}$$ ### Pattern Recognition When statistical problems ask for "at least one" or "minimum bounding", calculating the complement probability (e.g., both elements strictly exceeding the threshold) cuts combinatorial checks from 3+$3+$ cases down to exactly 1$1$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Statistics Class 12 Maths: Probability