# Homework Help: Conditional Probability

1. Jan 20, 2016

### Zondrina

1. The problem statement, all variables and given/known data

The problem statement is given below:

2. Relevant equations

3. The attempt at a solution

Here is my attempt so far:

I'm sure questions 1 - 4 have been answered. Question 5 is what concerns me.

I need to find $P(C' | D')$, which is the probability a good item is identified, given a good item is chosen.

The probability a good item is chosen is $P(D') = 0.991$.

Now I need the probability the good item chosen is identified as a good item. From question 4), the chance a good item is identified is $P(C') = 0.9861$.

Would that mean of the $99.1$% of good items, $98.61$% of them will be identified as good items? So we would get:

$$P(C' | D') = 0.991 * 0.9861 = 0.977$$

Thank you.

2. Jan 20, 2016

### PeroK

To be honest, I would do a problem like this using a probability tree. I didn't check your answers to 1-4, but that answer to 5 is not right.

The answer should be very close to 100%.

3. Jan 20, 2016

### PeroK

Your answers to 1-4 are correct. Except you missed a digit on 3. The answers to 2 + 3 must equal 1.

4. Jan 20, 2016

### Zondrina

Yes there was precision error when using my calculator. 0.36 + 0.64 = 1 if we round everything nicely.

Following the logic in my first post, $P(D' | C') = \frac{P(D' \cap C')}{P(C')} = \frac{P(C' | D') P(D')}{P(C')} = \frac{0.977 * 0.991}{0.9861} ≈ 1$ barring some precision error.

EDIT: After some thought, I think this is the answer.

5. Jan 20, 2016

### PeroK

I'm not sure what you've calculated there. The answer should be:

$GG/(GG +DG)$

Where GG is probability that a good item is classified as good and DG is the probability that a defective item is classified as good.

Together they represent the restricted sample space of items classified as good.

I just read this off a probability tree. It's a much neater technique and less prone to error.

6. Jan 20, 2016

### PeroK

Here's how I did it. Apologies for the format.

D (0.009) - DD (0.009)(0.99)
------------- DG (0.009)(0.01)

G (0.991) - GD (0.991)(0.005)
------------- GG (0.991)(0.995)

So, for example, for question 1. You simply add the first and third lines. Those are the two cases where an item is classified as defective.

You can just read off the answer to almost any question just from that simple tree.

7. Jan 20, 2016

### PeroK

PS the tree extends to more complicated scenarios. Suppose you had two examiners who have different probabilities of getting things wrong and perhaps one does 60% of the items and the other 40%.

The tree would have more branches on it, but it would still be fundamentally simple.

I wouldn't like to do a problem like that by your method!

8. Jan 20, 2016

### Zondrina

I agree with the probability tree method, which will yield the same result. The method I used in post #4 based on post #1 was my intuition (it gives the same result as the tree).

I think from now on I'll start using probability trees more often though because I see how things can get very complicated from your comment in post #7.

9. Jan 20, 2016

### Ray Vickson

I usually find it much easier to solve such problems by using a more informative and more "immediate" notation, such as G = part is good, CG = classified as good, D = part is defective, CD = classified as defective. So, you are given P(D) = 0.9% = 0.009, P(CD|D) = 99% = 0.99 and P(CD|G) = 0.5% = 0.005.
You want (1) P(CD), (2) P(G|CD), (3) P(D|CD), (4) P(CG) and (5) P(G|CG).

Aside from being a bit hard to follow (just because of notation) your formulas seem OK. However, some of your numerical answers are a bit off, due to premature roundoff:
(2) should be 0.35737 ≈ 0.357 (not your 0.356), (3) should be 0.64263 ≈ 0.643 (not your 0.641). I recommend you either do exact rational arithmetic and round off at the end, or keep a few more significant figures during the computations themselves, and again, again round off at the end.

10. Jan 20, 2016

### Zondrina

Sorry, electronics practices have made me lazy.

Here is the solution with the precision taken into account:

I'll refrain from rounding off too early.

11. Jan 21, 2016

### HallsofIvy

Here's how I would do that problem: Imagine 1,000,000 items. "The company has evidence that its line produces 0.9% of defective items." So 0.009(1000000)= 9000 defective items and 1000000- 9000= 991000 good items. "An inspector working for a manufacturing company has a 99% chance of correctly identifying defective items and a 0.5 % chance of incorrectly classifying a good item as defective." So he will identify 0.99(9000)= 8910 of the defective items as defective and will incorrectly identify 9000- 8910= 90 defective items as good. He will incorrectly identity .005(991000)= 4955 good items as defective and will identify 991000- 4955= 986045 good items as good.

Use those numbers to answer the questions. For example, number 3: If an item is classified as defective, what is the probability that it is indeed defective?" There are a total of 8910+ 4955= 13865 items classified as defective of which 8910 are defective. The probability is 8910/13865.