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

Question

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 another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4 , is :

Accepted Answer

D. 1. Since the **mean** of the 7 observations is 8:
$$
\frac{2+4+10+x+y+12+14}{7}=8 \;\Rightarrow\; x+y=14
$$

2. Since the **variance** is 16, use the formula $	ext{Var}=\frac{\sum x_i^2}{n}-(	ext{mean})^2$:
$$
16=\frac{2^2+4^2+10^2+x^2+y^2+12^2+14^2}{7}-8^2
$$

3. Simplifying gives:
$$
x^2+y^2=100
$$

4. Solving $x+y=14$ and $x^2+y^2=100$ together with $x>y$ gives:
$$
x=8,\qquad y=6
$$

5. The set $\{1,2,3,x-4,y,5\}$ becomes $\{1,2,3,4,6,5\}=\{1,2,3,4,5,6\}$.

6. We want the probability that the smaller of two chosen numbers is less than 4. It's easier to find the complementary event: both chosen numbers are at least 4 (i.e., from $\{4,5,6\}$).
$$
P(	ext{both}\ge4)=\frac{{}^3C_2}{{}^6C_2}=\frac{3}{15}
$$

7. So the required probability is:
$$
1-\frac{3}{15}=\frac{4}{5}
$$

Hence, the answer is **Option D**.