Related Formula
Mean x = (Σ xᵢ)/(n)$$\text{Mean } \bar{x} = \frac{\sum x_i}{n}$$
Variance σ² = (Σ xᵢ²)/(n) - ( x)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$2, 4, 10, x, 12, 14, y$.
(2 + 4 + 10 + x + 12 + 14 + y)/(7) = 8$$\frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$$
⇒ x + y + 42 = 56 ⇒ x + y = 14 (1)$$\Rightarrow x + y + 42 = 56 \Rightarrow x + y = 14 \dots(1)$$
Given Variance = 16$= 16$:
16 = (2² + 4² + 10² + x² + 12² + 14² + y²)/(7) - 8²$$16 = \frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} - 8^2$$
16 + 64 = (4 + 16 + 100 + x² + 144 + 196 + y²)/(7)$$16 + 64 = \frac{4 + 16 + 100 + x^2 + 144 + 196 + y^2}{7}$$
80 × 7 = 460 + x² + y² ⇒ 560 = 460 + x² + y² ⇒ x² + y² = 100 (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 = x² + y² + 2xy$(x+y)^2 = x^2 + y^2 + 2xy$:
14² = 100 + 2xy ⇒ 196 - 100 = 2xy ⇒ 2xy = 96 ⇒ 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 t² - 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 × 5 = 30$6 \times 5 = 30$ permutations (or 62 = 15$\binom{6}{2} = 15$ combinations).
Let's use combinations. Total ways to choose 2 numbers = 62 = 15$\binom{6}{2} = 15$.
We need the probability that the smaller number is less than 4$4$.
P(smaller < 4) = 1 - P(smaller ≥ 4)$P(\text{smaller} < 4) = 1 - P(\text{smaller} \geq 4)$.
Step 3: Final Calculation via Complement
For the smaller number to be ≥ 4$\geq 4$, BOTH chosen numbers must be ≥ 4$\geq 4$.
The available numbers ≥ 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 32 = 3$\binom{3}{2} = 3$.
P(smaller ≥ 4) = (3)/(15) = (1)/(5)$$P(\text{smaller} \geq 4) = \frac{3}{15} = \frac{1}{5}$$
P(smaller < 4) = 1 - (1)/(5) = (4)/(5)$$P(\text{smaller} < 4) = 1 - \frac{1}{5} = \frac{4}{5}$$
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$1$.
Chapter Mix
Class 11 Maths: Statistics
Class 12 Maths: Probability
