Very trivial problem, but i dissagree with my professor

[shadowclown]

Ok homework problem for intro to prob + stat, that i find very easy and trivial, but my professor says is wrong.

Items are inspected for flaws by two quality inspectors. If a flaw is present, it will be detected by the first inspector with probability 0.9, and by the second inspector with probability 0.7. Assume the inspectors function independently.

a.) Assume that the second inspector examines only those items that have been passed by the first inspector. If an item has a flaw, what is the probability that the second inspector will find it?

If you find an answer please explain it, as I would like to see why my professor says I am wrong. I will post my answer in a little while.

[sebbb]

0.7; the fact that the 1st inspector inspected the item doesnt change the probabilty of inspector 2 finding the flaw

[sebbb]

or wait maybe I didnt get the question right

[sebbb]

ok got it, seems like the question is, "given any flawed item that goes thru the chain, what is the prob. that inspector 2 finds it"

we need to have the events: "inspector 1 doesnt see the flaw" and "guy 2 sees it".

the answer is p=0.1*0.7=0.07

[shadowclown]

Well I came up with the answer of 0.7, and my professor says it was 0.07.

My reasoning is that the two events have nothing to do with each other. Say for instance the items are bricks and that there are 100 of them. 10 of them are flawed ok. So the first inspector throws out 9 of them and 1 goes through.

The second inspector see's 90 bricks that are fine and then he comes upon the 1 that is flawed. The question asks what is the probability that he will see that one as flawed. And it should be 70% not 7%. His amount of flawed bricks may have changed from the first inspector, but he still see's the flaw 70% of the time.

I just do not see how 0.07 is the answer.

[sebbb]

I think you are misunderstanding the quesiton like I was at first. Question is hard to understand, maybe you have to take it out of the context:

"If an item has a flaw, what is the probability that the second inspector will find it?"

You have to realize that that item in the question goes through inspector 1 first, and if inspector 1 finds it, it doesn't go through inspect 2, and inspect 2 cannot find that item.


Say you have 100 flawed items. Inspector 1 finds 90 of them. Inspector 2 then finds 7 of them (70% of the remaining 10). He will have found only 7/100 items

The question is actually: "given any flawed item, you have three possibilities: guy 1 finds it, guy 2 finds it, or noone does. It goes through guy 1, then through guy 2. What is the probability that guy 2 finds it?"

[shadowclown]

Thanks for your responses sebbb, at least now I can see what my professor is saying. You are right, in that the wording is very poor for this problem, as it could be interpreted both ways.

I guess the way it is supposed to be thought of is that of the total flaws how many did the 2nd guy find. This includes the ones thrown out.

I originally looked at it as, "the first part of the question has no bearing on the answer," and thus looked at the second part.

Again thanks for the help.

ShadowNeedHelpInProbabilityClown

[irchans]

The question was stated:

" Assume that the second inspector examines only those items that have been passed by the first inspector. If an item has a flaw, what is the probability that the second inspector will find it?"

I think that sebbb and your professor are correct to interpret the question as "What is the probability that the first inspector will miss the flaw and the second inspector then finds the flaw?" Answer: 0.07.

The alternative interpretation of the question is "What is the probability that the second inspector will find the flaw given that the first inspector missed the flaw?" Answer 0.70.

Cheers,
Irchans

[SpaceAce]

[ QUOTE ]

a.) Assume that the second inspector examines only those items that have been passed by the first inspector. If an item has a flaw, what is the probability that the second inspector will find it?


[/ QUOTE ]

I don't actually think this question is poorly worded at all. To me, at least, it seems pretty clear what the part in bold is really asking. Since the second inspector only sees stuff passed by the first inspector and the item is drawn from the pool of all flawed items, you have a simple multiplication problem on your hands.

SpaceAce

[aggie]

This question is really more about reading comprehesnion than it is about probability or statistics.