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STD 12 - 13. probability — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 13. probability1 Mark
MCQ
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 $ o < m < n $ ) is
  • A
    $\frac{{(n - m)\,(n - m + 1)}}{{(n - 1)}}$
  • B
    $\frac{{(n - m)\,(n - m + 1)}}{{2n}}$
  • C
    $\frac{{(n - m)\,(n - m - 1)}}{{2n\,(n - 1)}}$
  • ✓
    $\frac{{(n - m)\,(n - m + 1)}}{{2n\,(n - 1)}}$
✓

Answer

Correct option: D.
$\frac{{(n - m)\,(n - m + 1)}}{{2n\,(n - 1)}}$
d
(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 first number chosen is $x,$

we must have $x - y \ge m$ or $y \le x - m.$

Therefore $x > m$ and $y < n - m.$

Thus $P({E_x}) = 0$ for $0 < x \le m$ and $P({E_x}) = \frac{1}{n}$ for $m < x \le n.$

Also $P(A/{E_x}) = \frac{{(x - m)}}{{(n - 1)}}$

Therefore, $P(A) = \sum\limits_{x = 1}^n {P({E_x})\,\,P(A/{E_x})} $

$ = \sum\limits_{x = m + 1}^n {P({E_x})\,\,P(A/{E_x})} = \sum\limits_{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)}}{{2n(n - 1)}}.$

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