Question

Management system failures. Refer to the Process Safety Progress (December 2004) study of 83 industrial accidents caused by management system failures, Exercise 2.150(p. 142). A summary of the root causes of these 83 incidents is reproduced in the following table. One of the 83 incidents is randomly selected and the root cause is determined.

Management system cause category	Number of incidents
Engineering and design	27
Procedures and practices	24
Management and oversight	22
Training and communication	10
Total	83

a. List the sample points for this problem and assign reasonable probabilities to them.

b. Find and interpret the probability that an industrial accident is caused by faulty engineering and design.

c. Find and interpret the probability that an industrial accident is caused by something other than faulty procedures and practices.

Short answer

1. The sample points are E and D, P and P, M and O, T and C.
2. The probability is 0.325.
3. The probability is 0.711.

Step-by-step solution

Important formula

The formula for probability is $\mathbf{P}\mathbf{=}\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{out}\mathrm{comes}}$.

 Step 2: The possible probabilities. (a)

Here,

E and D=Engineering and design

P and P= Procedures and practices

M and O=Management and oversight

T and C= Training and communication

The sample points are E and D, P and P, M and O, T and C.

Now the probabilities are

$\begin{array}{c}\mathrm{P}\left(\mathrm{P}\text{and}\mathrm{P}\right)=\frac{24}{83}\\ =0.289\end{array}$

$\begin{array}{c}P\left(M\text{and}O\right)=\frac{22}{83}\\ =0.265\end{array}$

$\begin{array}{c}P\left(T\text{and}C\right)=\frac{10}{83}\\ =0.120\end{array}$

Thus, the sample points are E and D, P and P, M and O, T and C.

The probability for engineering and design.(b)

$\begin{array}{c}\mathrm{P}\left(\mathrm{E}\text{and}\mathrm{D}\right)=\frac{27}{83}\\ =0.325\end{array}$

Hence, the probability is 0.325.

The probability that an industrial accident is caused by something other than faulty procedures and practices.(c)

Now,

$\begin{array}{c}\mathrm{P}\left(\mathrm{P}\text{and}\mathrm{P}\right)=\frac{27+22+10}{83}\\ =0.711\end{array}$

Therefore, the probability is 0.711.