Probability Calculation
To calculate the probability of generating two distinct random numbers \(p\) and \(q\) between \(1\) and \(n\), such that \(p < q\) and \(p \leq m\) (\(m < n\)), we can follow the steps below:
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Total Possible Cases:
There are \(\binom{n}{2} = \frac{n(n-1)}{2}\) ways to choose two distinct numbers from \(1\) to \(n\). Since \(p\) and \(q\) are ordered (considering \((p, q)\) and \((q, p)\) as different cases), the total possible cases amount to \(2 \times \binom{n}{2} = n(n-1)\).
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Cases where \(p < q\):
As we require \(p < q\), there are \(\binom{n}{2} = \frac{n(n-1)}{2}\) possible cases where this condition is satisfied.
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Cases where \(p \leq m\) and \(p < q\):
- \(p\) can take values from \(1\) to \(m\), giving \(m\) choices for \(p\).
- For each fixed \(p\), \(q\) can be any number from \(p+1\) to \(n\), providing \(n - p\) choices for \(q\).
Therefore, the total cases where \(p \leq m\) and \(p < q\) are given by: \[ \sum_{p=1}^{m} (n - p) \]
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Calculating the Probability:
Combining the above analysis, the probability can be expressed as: \[ P(p < q \text{ and } p \leq m) = \frac{\sum_{p=1}^{m} (n - p)}{\binom{n}{2}} \]
Computing this specifically: \[ \sum_{p=1}^{m} (n - p) = n \cdot m - \sum_{p=1}^{m} p = n \cdot m - \frac{m(m+1)}{2} \]
Therefore, the probability \(P\) is: \[ P = \frac{n \cdot m - \frac{m(m+1)}{2}}{\frac{n(n-1)}{2}} = \frac{2(n \cdot m - \frac{m(m+1)}{2})}{n(n-1)} \]
Simplifying further: \[ P = \frac{2(n \cdot m - \frac{m(m+1)}{2})}{n(n-1)} = \frac{2nm - m(m+1)}{n(n-1)} \]
Further simplification yields: \[ P = \frac{2nm - m^2 - m}{n(n-1)} \]
Therefore, the probability of generating two distinct random numbers \(p\) and \(q\) between \(1\) and \(n\), satisfying \(p < q\) and \(p \leq m\), is:
\[ P = \frac{2nm - m^2 - m}{n(n-1)} \]