Question

Question:Show that if\(m\)is a positive integer, then the probability that the\(m\)the success occurs on the\((m + n)\)the trial when independent Bernoulli trials, each with probability\(p\)of success, are run, is\(\left( {\begin{aligned}{{}{}}{n + m - 1}\\n\end{aligned}} \right){q^n}{p^m}\).

Short answer

Answer

\(\left( {\begin{aligned}{{}{}}{m + n - 1}\\{m - 1}\end{aligned}} \right) \cdot {q^n} \cdot {p^{m - 1}} \cdot p = \left( {\begin{aligned}{{}{}}{n + m - 1}\\n\end{aligned}} \right) \cdot {q^n} \cdot {p^m}\)

Step-by-step solution

Formula of Bernoulli trials

Statement The probability of exactly \(k\) successes in \(n\) independent Bernoulli trials, with probability of success \(p\) and probability of failure \(q = 1 - p\), is: \(C(n,k)\;{p^k}\;{q^{n - k}}\)

Use Bernoulli trials

The \(m - \) success occurs on the \((m + n)\) -the trial means there were exactly \((m - 1)\) successes and \(n\) failures before. So, the required probability as the Bernoullian trials is independent is: \(\left( {\begin{aligned}{{}{}}{m + n - 1}\\{m - 1}\end{aligned}} \right) \cdot {q^n} \cdot {p^{m - 1}} \cdot p = \left( {\begin{aligned}{{}{}}{n + m - 1}\\n\end{aligned}} \right) \cdot {q^n} \cdot {p^m}\).

Hence, proved.