The trainee clowns at Bozo College are in a state of shock. A thief has deprived the college of 873 yellow balloons and a broken-down balloon pump. Happily, there was a witness to the crime, who was stated that the thief was wearing the College's clown uniform and had a red nose. Previous research has shown that on 80 percent of occasions witnesses will correctly identify the color of the nose of a clown involved in committing a crime. It is also known that 85 percent of the clowns at Bozo College have blue noses and that 15 percent have red noses. What is the probability that the thief had a red nose (assuming that the witness it telling the truth about what he thinks he saw)?
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7 comments:
I'm a little rusty on my conditional probabilities, so I'm going to intuit this one...
There are initially four possibilities: the clown had a blue nose or a red one, and the witness got it right or didn't.
Given the condition that the witness picked a red nose, there are two possibilities: the clown had a blue nose and the witness got it wrong, or the clown had a red nose and the witness got it right.
The chance of a red nose and a correct witness is .15*.8
and a red nose and an incorrect witness is .2*.85
So the chance the witness was correct is:
.15*.8
----------------- = .413
(.15*.8)+(.2*.85)
(Note: I did not mean to imply that the witness was picking his nose.)
Given the witness is telling the truth, or trying to tell the truth, I'll say the probability (of accuracy) is 80%. Tom
Tom, I'm not sure I have this right, but I think you have to account for the extra information you have.
Say, for example, you had a medical test that was 99% accurate for both positives and negatives. But the condition it tests for is very rare (1:1 million). If you test positive, does it mean you have a 99% chance of having the disease?
In this case, say 1 million people take the test. We know on average that only 1 person should have the disease, but the test will be wrong 1% of the time and it will come back saying that 10,000 people have the disease. Of those 10,000 only 1 is a real positive.
Kim is sparkling - a well-deserved 40th point!
This is indeed a very tricky puzzle and a surprising answer, yet we encounter similar scenarios everywhere and almost everyday in our lives. Seems like it is worth reading every single line of Kim's solution very slowly to make sure we won't jail anyone innocent or be misguided by some new drug statistics.
Kim is saying that we have one out of two scenarios:
1)Thief had a red nose and was correctly identified by the witness. Chances of this happening together are 15% x 80% = 0.15 x 0.8 = 0.12
2)Thief had a blue nose and was incorrectly identified by the witness. Chances of this happening together are 85% x 20% = 0.85 x 0.2 = 0.17
(I think Kim accidentally wrote "red" instead of "blue" for this case)
Note that chances of case#2 are higher than case#1!
So, the probability that thief actually had a red nose is:
probability of Case 1 / (probability of case 1 or case 2)
0.12 / (0.12 + 0.17) = 0.12/0.29 = 0.413 or 41%
Maria, thanks for the correction. Yes, I meant to say "blue" for the .85*.2 case (that is 85% chance of wearing a blue nose * 20% chance of identifying the nose as red even though it was blue)
Works for me, says Tom. I leaped to the suspicion that the extra information was extraneous, as in "What color was the bus?" Kim's tough!
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