# Probability..ez but problem with the wording.

• Feb 8th 2007, 09:55 AM
Ruichan
Probability..ez but problem with the wording.
I have a couple of hours more before this is dued. I actually did complete all 20 questions but my prof said I did this question wrong.

A chemical engineer is in charge of a particular process at an oil refinery. Past experience indicates that 10% of all shutdowns are dued to equipment failure alone, 5% are dued to a combination of equipment failure and operator error, and 40% involve operator error. A shutdown occurs. Find the probability that
a) equipment failure or operator error is involved;
b) operator error alone is involved;
c) neither operator error nor equipment failure is involved.

Here's what I did:
Let A=equipment failure alone; P(A)=0.1
Let B=operator error; P(B)=0.40
P(AnB) = 0.05

a)P(AUB) = P(A) + P(B) - P(AnB) = 0.45

b)operator error alone = 0.4

c) neither operator error nor equipment failure is involved
= 1-0.45
= 0.55

Prof said it's wrong. Said wordings are tricky. I've been looking at it for the past hour, i know something is wrong but I can't figure out what exactly is wrong.
• Feb 8th 2007, 12:23 PM
CaptainBlack
Quote:

Originally Posted by Ruichan
I have a couple of hours more before this is dued. I actually did complete all 20 questions completed but my prof said I did this question wrong.

A chemical engineer is in charge of a particular process at an oil refinery. Past experience indicates that 10% of all shutdowns are dued to equipment failure alone, 5% are dued to a combination of equipment failure and operator error, and 40% involve operator error. A shutdown occurs. Find the probability that
a) equipment failure or operator error is involved;
b) operator error alone is involved;
c) neither operator error nor equipment failure is involved.

0.1 are due to equipment failure alone
0.05 due to equipment failure and op error
0.4 due to op error.

Therefore 0.35 are due to op error alone (the 0.4 includes the 0.05 which
are due to both causes).

So for a) you need 0.1+0.35+0.1
for b) we have 0.35
for c) we have 1 minus the answer to a).

RonL
• Feb 8th 2007, 01:59 PM
Ruichan
"Therefore 0.35 are due to op error alone (the 0.4 includes the 0.05 which
are due to both causes)."

I can just subtract the 0.05 out of the 0.4? Wouldn't the 0.05 also belong to A as well as B?
So if it's equipment failure, I just simply add 0.05 to 0.1 = 0.15?
• Feb 8th 2007, 09:52 PM
CaptainBlack
Quote:

Originally Posted by Ruichan
I can just subtract the 0.05 out of the 0.4?

The 40% that involve operator error consist of 5% which incolve both op error and equipment failure, and the remainder that are op error only.

So the percentage which are op error only is 40-5=35%.

RonL
• Feb 8th 2007, 09:53 PM
CaptainBlack
Quote:

Originally Posted by Ruichan
"So if it's equipment failure, I just simply add 0.05 to 0.1 = 0.15?

Yes.

RonL
• Feb 9th 2007, 12:44 AM
Ruichan
I guess I was kind of thinking too much but brain's not working coz too drugged with all the flu and allergy medications.

I actually thought that I couldn't subtract 0.05 directly from the 0.40 since the 0.05 belongs to both A and B. I thought that I would have to divide 0.05 by 2 since it's the intersection part of A and B. Dohz...

Thank you very much.