Mathematics, 18.01.2020 06:31 martina28

An electrical system has five circuit breakers designed to protect the circuit in the case of an overload. the probability that any circuit breaker will open on demand is 0.95 and the circuit breakers are assumed to operate independently of each other. if a fault condition occurs that requires all circuit breakers to open, what is the probability that at least one of the circuit breakers will fail to open?

1. 0.64
2. 0.999
3. 0.23
4. 0.05
5. 0.95

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Answer from: coolkitty35

Option 3: approximately 0.23 .

Step-by-step explanation:

Let represent the number of circuit breakers that did not open.

is the probability that none of the five circuit breakers did not open (i.e., all of them opened successfully.) P(n = 1)

is the probability that one out of the five circuit breakers did not open. P(n = 2)

is the probability that two out of the five circuit breakers did not open. P(n = 3)

is the probability that three out of the five circuit breakers did not open. P(n = 4)

is the probability that four out of the five circuit breakers did not open. P(n = 5)

is the probability that all five circuit breakers did not open.

There are only five circuit breakers in this circuit. As a result, these six situations represent all the possibilities that could happen to these circuit breakers. Therefore, the sum of these six possibilities should be equal to one:

$\begin{aligned} P(n = 0) + &P(n = 1) + P(n = 2) \cr & + P(n = 3) + P(n = 4) + P(n = 5) = 1\end{aligned}$ .

The question is asking for $P\left(n \ge 1\right)$ , the probability that at least one of these five circuit breakers does not open. Since must be an integer between and , $P\left(n \ge 1\right)$ would be the same as the sum of P(n = 1) , P(n = 2) , P(n = 3) , P(n = 4) , and P(n = 5) . In other words,

$\begin{aligned}P\left(n \ge 1\right) =& P(n = 1) + P(n = 2) + P(n = 3)\cr & + P(n = 4) + P(n = 5)\end{aligned}$ .

Calculating all five probabilities would likely be a very cumbersome task. That's how the following equation (which was just deduced) becomes very handy.

$\begin{aligned} P(n = 0) + &P(n = 1) + P(n = 2) \cr & + P(n = 3) + P(n = 4) + P(n = 5) = 1\end{aligned}$ .

Subtract P(n = 0) from both sides of this equality to obtain:

$\begin{aligned}& P(n = 1) + P(n = 2) \cr & + P(n = 3)+ P(n = 4) + P(n = 5) = 1 - P(n = 0)\end{aligned}$ .

In other words, to calculate $P\left(n \ge 1\right)$ , simply finding the value of P(n = 0) and doing a subtraction would be enough. Besides, the value of can be particularly easy to determine.

Recall that P(n = 0) gives the probability that all five circuit breakers open successfully. The question stated that the probability for each circuit breaker to open successfully is 0.95 . Besides, the question assumed that the five circuit breakers work independently from each other. Hence, the probability that all five of them worked should be found with the product rule of probability:

$\begin{aligned} P (n = 0) &= P(\text{all works}) \cr &= P(\text{1st one works}) \times P(\text{2nd one works})\times \cdots \times P(\text{5th one works}) \cr &=0.95 \times 0.95 \times 0.95 \times 0.95 \times 0.95 \cr &= 0.95^5 \cr &\approx 0.773781 \end{aligned}$ .

Hence,

$\begin{aligned} P(n \ge 1) &= P(n = 1) + P(n = 2) + P(n = 3) \cr &\phantom{=}~ + P(n = 4) + P(n = 5) \cr &= 1 - P(n = 0) \cr &\approx 1 - 0.773781 \cr &\approx 0.23 \end{aligned}$ .

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