Answer:
Assume that for the communication to be available means that at least one of the [tex]n[/tex] circuits is operational. It would take at least 3 circuits to achieve a [tex]0.99999[/tex] overall availability.
Step-by-step explanation:
The probability that one circuit is not working is [tex]1 - 0.99 = 0.01[/tex].
Since the circuits here are all independent of each other, the probability that none of them is working would be [tex]\displaystyle \underbrace{0.01 \times 0.01 \times \cdots \times 0.01}_{\text{$n$ times}}[/tex]. That's the same as [tex]0.01^n[/tex].
The event that at least one of the [tex]n[/tex] circuits is working is the complement of the event that none of them is working. To find the probability that at least one of the [tex]n[/tex] circuits is working, simply subtract the probability that none of the circuit is working from one. That is:
[tex]\begin{aligned}&P(\text{At least one working}) \cr &= 1 - P(\text{None is working}) \cr &= 1- 0.01^n\end{aligned}[/tex].
The question requests that
[tex]P(\text{At least one working}) \ge 0.99999[/tex].
In other words,
[tex]1- 0.01^n \ge 0.99999[/tex].
[tex]0.01^n \le 1 - 0.99999 = 0.000001 = 10^{-6}[/tex].
Note that [tex]0.01 = 10^{-2}[/tex]. Hence, the inequality becomes
[tex]\left(10^{-2}\right)^n \le 10^{-6}[/tex].
[tex]10^{-2\,n} \le 10^{-6}[/tex]
Take the natural log of both sides of the equation:
[tex]\ln\left(10^{-2\, n}\right) \le \ln \left(10^{-6}\right)[/tex].
[tex](-2\, n)\ln\left(10\right) \le (-6) \ln\left(10\right)[/tex].
[tex]10 > 1[/tex], hence [tex]\ln(10) > 0[/tex]. Divide both sides by [tex]\ln(10)[/tex]:
[tex]-2\,n \le -6[/tex].
[tex]n \ge 3[/tex].
In other words, at least three parallel circuits must be set up to achieve that availability.
