The probability of a given event is 1/x . If the experiment is repeated n times, what is the probability that the event does not occur in any of the trials?

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Answer 1 · 2017-01-25T15:43:31

${(\frac{x - 1}{x})}^{n}$ $((x-1)/x)^n$

Let say $p$ $p$ is the probability and event occurs and $q$ $q$ an event does not occur.

$p = \frac{1}{x}, q = 1 - (\frac{1}{x}) = \frac{x - 1}{x}$ $p=1/x, q=1-(1/x)=(x-1)/x$

$P (X = r) =^{n} C_{r} \cdot p^{r} \cdot q^{n - r}$ $P(X=r)="^nC_r*p^r*q^(n-r)$

r=0 when the event does not occur

$P (X = 0) =^{n} C_{0} \cdot {(\frac{1}{x})}^{0} \cdot {(\frac{x - 1}{x})}^{n}$ $P(X=0)="^nC_0*(1/x)^0*((x-1)/x)^n$

$P (X = 0) = 1 \cdot 1 \cdot {(\frac{x - 1}{x})}^{n}$ $P(X=0)=1*1*((x-1)/x)^n$

$P (X = 0) = {(\frac{x - 1}{x})}^{n}$ $P(X=0)=((x-1)/x)^n$