Conditional Probability

1. Aug 9, 2009

Orphen89

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

A particular brand of cars, say ABC, comes in only two colours, white and grey. Exactly
90% of ABC cars in a particular town are white and 10% are grey. Mrs Z, a witness
to a bank robbery, claims to have seen the thieves escaping in an ABC grey car, which
taking into account other witnesses' accounts would convict Mr X of the crime. However,
in an attempt to discredit Mrs Z's account, the defence lawyer questioned her ability to
distinguish between the two colours of ABC cars (there is no argument as to the brand and
make of the car), producing results of tests that showed that people (in Mrs Z's condition
and position at the time she witnessed the robbery) were able to correctly identify the
colour of an ABC car only 75% of the time. Given that Mrs Z claims to have witnessed
a grey car, what is the probability that she was indeed correct in identifying the colour
of the ABC car.

3. The attempt at a solution

For this question, I made P(White) = P(W) = 0.9, P(Grey) = P(G) = 0.1, P(Correctly Identified Color) = P(C) = 0.75, and P(Correctly Identified Color l Grey) = ?

Now, I honestly have no idea what to do with this question, since I don't know what equation I can use for it. I initially thought that I could just draw up a tree diagram and use someting like P(C) * P(G) + P(C$$^{C}$$)*P(G), but that doesn't seem to work. Can someone please help show me what to do with this question?

2. Aug 9, 2009

benorin

We want P(G | she said it was G). Against my nature (people lie), I'll assume that she would only have said it was G if either it really was G(10%) and she ID'd its color as G (75%), or if it was actually W (90%), and she ID'd its color as G anyway (25%). You finish...

3. Aug 9, 2009

Orphen89

Okay, I *think* I understand what you're trying to say. Does this mean that P(G | she said it was G) = 0.1 * 0.75 + 0.25 * 0.9 (instead of 0.1)?

4. Aug 10, 2009

Orphen89

Bump, could someone please check if I have the correct method down?

5. Aug 10, 2009

Billy Bob

No, but you can use that computation somewhere.

P(G | "she said it was G") = P(G and "she said it was G") / P("she said it was G").

Your computation is simply P("she said it was G").

You are almost done, so you should finish it.

There is another way to do this problem using a table, which I prefer, and I would use the table method to check your problem. This is really a Type I - Type II error problem in disguise. Have you studied that concept yet? If not, don't worry. Here's how you make the table. It takes longer to describe than actually to do. It's worth doing because you'll certainly have a question like this on your exams.

The table will have 3 rows and 3 columns of numbers. Row 1 is labeled "Really White," row 2 is labeled "Really Gray," and row 3 is "Total." Column 1 is labeled "She Said White," col 2 is labeled "She Said Gray," and col 3 is "Total." The entry in row 3 col 3 is the (hypothetical) grand total of people who could have been witnesses, and you just make up a number. You could try 100, but let's do 1000.

Now you can fill in row 1 col 3, and row 2 col 3, because you know the split of true colors: 90% vs 10%. (900 and 100)

Now look at row 1, with total 900. You can fill in row 1 col 1, and row 1 col 2, because you know the probability of saying the color correctly: 75% vs 25%. So row 1 col 1 is 75% of 900. Etc.

Now finish the table! Now you can instantly answer almost every reasonable question that could be asked, like P("really Gray" | "she said gray"), or P("really white" | "she said white"), or P("really white" | "she said gray"), etc.

Did you get the same answer with both methods?

6. Aug 11, 2009

Orphen89

I think I understand what you mean - I did what I think you meant (i.e. P(G | "she said it was G") = P(G and "she said it was G") / P("she said it was G").) and I got (.1*.75)/(.1*.75+.25*.9) = 0.25, which does seem a lot more accurate than my previous answer, and after drawing up the table you described I again got 0.25 (then again, knowing me I probably messed it up somewhere). Does this seem correct to you?

I actually remember learning how to use the table, but I actually learned about it back in high school and completely forgot about it (It makes things SO much easier).

Thanks for all the help Billy Bob.

7. Aug 11, 2009

Good job!