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 n

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Solution

Q.
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: [2026]

1 $\frac{4}{5}$

2 $\frac{1}{3}$

3 $\frac{3}{5}$

4 $\frac{2}{5}$

Ans.
(1)

$Mean (\bar{x}) = 8 (Given)$

$\Rightarrow \frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$

$\Rightarrow x + y = 14 . . . (1)$

$Variance (σ^{2}) = 16 (Given)$

$\Rightarrow 16 = \frac{2^{2} + 4^{2} + 10^{2} + x^{2} + 12^{2} + 14^{2} + y^{2}}{7} - 8^{2}$

$\Rightarrow x^{2} + y^{2} = 100 . . . (2)$

$∵$ ${(x + y)}^{2} = x^{2} + y^{2} + 2 x y$

$\Rightarrow x y = 48 (sum is 14, product is 48)$

$Since problem states x > y$

$∴$ $x = 8 and y = 6$

$Now set X = {1, 2, 3, 4, 6, 5}$

Now we choose two numbers one after another without replacement

Total outcomes $= 6 \times 5 = 30$

We want the probability that the smaller number among the two is less than 4

$P (smaller < 4) = 1 - P (smaller \geq 4)$

$= 1 - \frac{6}{30} = \frac{4}{5}$

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