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Probability

Two numbers a...

Question

Two numbers are selected at random from the numbers 1, 2, . .. . . n. The probability that the difference between the first and second is not less than m (where 0 < m < n), is

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Solution

The correct option is D
Let the first number be x and second is y.
Let A denotes the event that the difference between the first and second number is at least m.
let $E_{x}$ denote the event that the number chosen is x, we must have $x - y \geq m o r y \leq x - m .$
Therefore x > m and y < n – m. Thus $P (E_{x}) = 0$ for $0 < x \leq m$ and $P (E_{x}) = \frac{1}{n}$ for $m < x \leq n .$ Also $P (\frac{A}{E_{x}}) = \frac{(x - m)}{(n - 1)}$
Therefore, P(A) = $\sum_{x = 1}^{n} P (E_{x}) (P \frac{A}{E_{x}})$
$= \sum_{x = m + 1}^{n} P (E_{x}) (P \frac{A}{E_{x}}) = \sum_{x = m + 1}^{n} \frac{1}{n} . \frac{x - m}{n - 1}$
$= \frac{1}{n (n - 1)} [1 + 2 + 3 + . . . + (n - m)]$
$= \frac{(n - m) (n - m + 1)}{2 n (n - 1)}$

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