First slide

Introduction to probability

Question

If a and b are chosen randomly from the set consisting of numbers 1, 2, 3, 4, 5, 6 with replacement. Then the probability that $lim_{x \to 0} {[(a^{x} + b^{x}) / 2]}^{2 / x} = 6$ is

Moderate

A

1/3
B

1/4
C

1/9
D

2/9

Solution

Given limit,
$lim_{x \to 0} {(\frac{a^{x} + b^{x}}{2})}^{\frac{2}{x}}$
$\begin{matrix} = {(lim_{x \to 0} {(1 + \frac{a^{x} + b^{x} - 2}{2})}^{\frac{2}{a^{x} + b^{x} - 2}})}^{lim_{x \to 0} (\frac{a^{x} - 1 + b^{x} - 1}{x})} \\ = e^{\log a b} = a b = 6 \end{matrix}$
Total number of possible ways in which a, b can take values is 6 x 6 = 36. Total possible ways are (1, 6), (6, 1), (2, 3), (3, 2). The total number of possible ways is 4. Hence, the required probability is 4/36 = 1/9.

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