Question

Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of then!/[k!(n−k)!]possible arrangements of the k successes and n-k failures is equally likely.

Probability
ANSWERED
asked 2021-08-15
Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of the
\(\displaystyle{n}\frac{!}{{{k}!{\left({n}−{k}\right)}!}}\)
possible arrangements of the k successes and n-k failures is equally likely.

Expert Answers (1)

2021-08-16
Probability that exactly one of the event E or F occurs is
\(\displaystyle{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}\)
Now,
\(\displaystyle{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}{F}^{{c}}\right)}+{P}{\left({E}^{{c}}{F}\right)}-{P}{\left({E}{F}^{{c}}\cap{E}^{{c}}{F}\right)}\)
Since \(\displaystyle{P}{\left({E}{F}^{{c}}\cap{E}^{{c}}{F}\right)}={0}\)
\(\displaystyle\Rightarrow{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}{F}^{{c}}\right)}+{P}{\left({E}^{{c}}{F}\right)}\)
Also, \(\displaystyle{P}{\left({E}{F}^{{c}}\right)}={P}{\left({E}\right)}-{P}{\left({E}{F}\right)}\) and \(\displaystyle{P}{\left({E}^{{c}}{F}\right)}={P}{\left({F}\right)}-{P}{\left({E}{F}\right)}\Rightarrow{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}\right)}+{P}{\left({F}\right)}-{2}{P}{\left({E}{F}\right)}\)
Hence proved!
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