Bergen

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

$n\frac{!}{k!(n-k)!}$

possible arrangements of the k successes and n-k failures is equally likely.

possible arrangements of the k successes and n-k failures is equally likely.

Nichole Watt

Skilled2021-08-16Added 100 answers

Probability that exactly one of the event E or F occurs is

$P(E{F}^{c}\cup {E}^{c}F)$

Now,

$P(E{F}^{c}\cup {E}^{c}F)=P\left(E{F}^{c}\right)+P\left({E}^{c}F\right)-P(E{F}^{c}\cap {E}^{c}F)$

Since$P(E{F}^{c}\cap {E}^{c}F)=0$

$\Rightarrow P(E{F}^{c}\cup {E}^{c}F)=P\left(E{F}^{c}\right)+P\left({E}^{c}F\right)$

Also,$P\left(E{F}^{c}\right)=P\left(E\right)-P\left(EF\right)$ and $P\left({E}^{c}F\right)=P\left(F\right)-P\left(EF\right)\Rightarrow P(E{F}^{c}\cup {E}^{c}F)=P\left(E\right)+P\left(F\right)-2P\left(EF\right)$

Hence proved!

Now,

Since

Also,

Hence proved!

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