91影视

The failure rate for a generator in a power outage at a hospital is 22%.

For a set of two or more events, the probability that they will occur together can be calculated as follows:

$P\left(AandB\right)=P\left(A\right)脳P\left(B|A\right)$

Here, $P\left(B|A\right)$shows the probability of occurrence of B when A has already occurred.

For an event A, the event of occurrence of 鈥渘ot A鈥� or 鈥�$\left(\stackrel{炉}{A}\right)$鈥漣s computed as follows:

$P\left(\stackrel{炉}{A}\right)=1-P\left(A\right)$

Here, A and $\left(\stackrel{炉}{A}\right)$can be considered as complementary events.

a.

Let A be the event of failure of the first generator during a power outage.

The probability of failure of the first generator during a power outage is given by:

$$\begin{gathered} P\left(A\right)=\frac{22}{100} \\ =0.22 \end{gathered}$$

Let B be the event of failure of the second generator.

It is known that events A and B appear independently.

Therefore,$P\left(B\left|A\right.\right)=P\left(B\right)$ .

The probability of failure of the second generator, once the first generator fails during a power outage, is given by:

$$\begin{gathered} P\left(B\right)=\frac{22}{100} \\ =0.22 \end{gathered}$$

The probability of failure of both the generators is given by:

$$\begin{gathered} P\left(AandB\right)=P\left(A\right)脳P\left(B\right) \\ =0.22脳0.22 \\ =0.0484 \end{gathered}$$

Therefore, the probability of failure of both the generators in a power outage is equal to 0.0484.

b.

If A is the event of failure of a generator, then $\left(\stackrel{炉}{A}\right)$is the event that the generator works.

$$\begin{gathered} P\left(\stackrel{炉}{A}\right)=1-P\left(A\right) \\ =1-0.22 \\ =0.78 \end{gathered}$$

As there are a total of two generators, the probability of having a working generator in a power outage can be divided into three cases:

$$\begin{gathered} P\left(\stackrel{炉}{A}and\stackrel{炉}{B}\right)=P\left(\stackrel{炉}{A}\right)脳P\left(\stackrel{炉}{B}\right) \\ =0.78脳0.78 \\ =0.6084 \end{gathered}$$

$$\begin{gathered} P\left(\stackrel{炉}{A}\text{and}B\right)=P\left(\stackrel{炉}{A}\right)脳P\left(B\right) \\ =0.78脳0.22 \\ =0.1716 \end{gathered}$$

$$\begin{gathered} P\left(Aand\stackrel{炉}{B}\right)=P\left(A\right)脳P\left(\stackrel{炉}{B}\right) \\ =0.22脳0.78 \\ =0.1716 \end{gathered}$$

Adding the above three probabilities, the probability of a working generator during a power outage is obtained.

$$\begin{gathered} P\left(workinggenerator\right)=0.6084+0.1716+0.1716 \\ =0.9516 \end{gathered}$$

Therefore, the probability of having a working generator in a power outage is equal to 0.952.

There is a 5% probability of total failure, with two backup generators at the hospital. Thus, there is an approximately 5% chance that none of the two generators will work during a power outage. The event cannot be considered rare.

This can be considered a very high probability in a hospital as there is a greater risk of losing lives if there is a power outage.